Introduction to State-Space Control Theory

I. E. KoseDept. of Mechanical Engineering

Bogazici University

December 11, 2003

Part I

Mathematical Background

1

Chapter 1

Vectors and Linear Vector Spaces

1.1 Linear Vector Spaces

Definition 1.1.1 (Linear Vector Space). A set X is said to be a linear vector space (LVS)if operations addition and scalar multiplication (over the scalar field1 IF) are defined such that

(a) x + y ∈ X for all x, y ∈ X(b) αx ∈ X for all x ∈ X and α ∈ IF

and for all x, y, z,∈ X and α, β ∈ IF, the following conditions hold:

(i) x + y = y + x

(ii) (x + y) + z = x + (y + z)

(iii) There is a null vector θ ∈ X such that x + θ = x

(iv) α(x + y) = αx + αy

(v) (α + β)x = αx + βx

(vi) (αβ)x = α(βx)

(vii) 0x = θ

(viii) 1x = x

In order to make explicit which scalar field we have in mind, we will use the notation (X , IF) todenote the LVS X associated with IF. Some examples of linear vector spaces are

(i) The real line, (IR, IR).

(ii) The complex plane, (C, IR), or (C,C).

(iii) The n-dimensional real Euclidean space, (IRn, IR).1In this definition, the field IF denotes either IR or C.

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(iv) The n-dimensional complex Euclidean space, (Cn, IR) or (Cn,C).

(v) The set of real-valued continuous functions defined on the [0, 1], (C[0, 1], IR).

In this class, we will mostly deal with IRn and Cn. However, it is best to keep things as generalas possible for flexibility in later definitions and manipulations. In the following definitions, let Xbe an LVS associated with the scalar field IF.

Definition 1.1.2 (Linear independence). A set of vectors {x1, · · · , xm} ⊆ X is said to belinearly independent if

α1x1 + α2x2 + · · ·+ αmxm = 0 implies αi = 0 ∀i = 1 : m. (1.1.1)

Definition 1.1.3 (Span). Given a set of vectors V 4= {x1, · · · , xm} ⊆ X , the span of V is the set

of all linear combinations of x1, · · · , xm, i.e.,

span(V)4=

{m∑

i=1

αixi : αi ∈ IF

}. (1.1.2)

Definition 1.1.4 (Basis). A linearly independent set of vectors is said to form a basis of X iftheir span is the whole of X .

Definition 1.1.5 (Dimension). The dimension of X , denoted dim(X ), is the number of elementsin a basis of X .

In the definition above, we have taken for granted one subtle point of great importance. Thatis the fact that any basis of X has the same number of elements in it. Can you prove it?

With these definitions, it is obvious that the dimension of (IRn, IR) is simply n. However, thedimension of C[0, 1] is infinite.

1.1.1 Linear Subspaces

Definition 1.1.6 (Linear Subspace). Given (X , IF), a set S ⊆ X is said to be a linear subspaceof X if

(i) x + y ∈ S for all x, y ∈ S(ii) α x ∈ S for all x ∈ S, α ∈ IF.

Linear subspaces of X can be generated as follows: Given a basis B for X, consider a set B′ ⊆ B.Then, span(B′) is a subspace of X. For instance, consider the basis

B =

100

,

010

,

001

of IR3. Then, the span of the set

B′ =

100

,

010

3

is a subspace of IR3. In fact, it is nothing but the x− y plane in the familiar x− y − z coordinateframe. Furthermore, a linear subspace of C[0, 1] is the set consisting of the elements of C[0, 1] suchthat f(a) = 0 for some given a ∈ [0, 1].

Definition 1.1.7 (Direct Sum). Let X1 and X2 be two LVSs defined over the same scalar fieldIF. The direct sum of X1 and X2, denoted X1 ⊕X2, is defined as ...

1.2 Inner Products and Inner Product Spaces

Let us define inner products:

Definition 1.2.1 (Inner product). An inner product over a given LVS X is any function< ·, · >: X × X → IF such that

(i) < x, x >≥ 0 for all x ∈ X(i’) < x, x >= 0 if and only if x = 0.

(ii) < x, y > =< y, x >.

(iii) < αx, y >= α < x, y >.

(iv) < x + y, z >=< x, z > + < y, z >.

Definition 1.2.2. A linear vector space, X , equipped with an inner product, < ·, · >, is called aninner product space, denoted by (X , < ·, · >).

Note that the usual inner product in Cn is defined as

< x, y >4=

n∑

i=1

yixi. (1.2.1)

Definition 1.2.3. Let (X , < ·, · >) be an inner product space. Two vectors x, y ∈ X are said to beorthogonal, denoted by x ⊥ y if

< x, y >= 0.

We say a basis of X is orthogonal if xi ⊥ xj for all i 6= j, where xi’s are the basis vectors ofX . An even more useful basis is an orthonormal basis, which is orthogonal and < xi, xi >= 1 also.For instance, the basis {e1, · · · , en}, where each vector ei consists of zeros except for a 1 in the ithentry, is an orthonormal basis for IRn. In fact, given any basis of a subspace, one can always obtainan orthonormal basis through the Gram-Schmidt orthonormalization process as follows:

Lemma 1.2.4 (Gram-Schmidt Orthonormalization). Let (X , < ·, · >) be an inner productspace and let S ⊆ X be a subspace of X . Let {s1, s2, · · · , sr} be a basis of S. Let {s1, · · · , sr} bedefined as

s14=

s1

‖s1‖ and sk =zk

‖zk‖ ∀k = 2 : r,

where

zk4= sk −

k−1∑

i=1

< sk, si > si.

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Lastly, we say two subspaces of X , say, Y and Z are orthogonal if y ⊥ z for all y ∈ Y and z ∈ Z.We denote orthogonality of subspaces Y and Z as Y ⊥ Z. Conversely, given a linear subspace S of(X , < ·, · >), we define its orthogonal complement as follows:

Definition 1.2.5 (Orthogonal Complement). Let (X , < ·, · >) be an inner product space andlet S ⊆ X be a subspace of X . Then,

S⊥ 4= {x ∈ X : < x, s >= 0 ∀s ∈ S} (1.2.2)

is called the orthogonal complement of S.

Theorem 1.2.6 (Orthogonal decomposition). Let (X , < ·, · >) be an inner product space andlet S ⊆ X be a subspace of X . Then, given any x ∈ X , there exists a unique decomposition

x = x1 + x2, where x1 ∈ S and x2 ∈ S⊥.

Proof. ¥

The vectors x1 and x2 are the so-called orthogonal projections of x onto S and S⊥, respectively.They can be calculated using the following theorem:

Theorem 1.2.7 (Orthogonal Projection). Let (X , < ·, · >) be an inner product space, S ⊆ Xa subspace of X and {s1, s2, · · · , sr} an orthonormal basis of S. Then, given x ∈ X ,

xS4=

r∑

i=1

< x, si > si

is the orthogonal projection of x onto S. That is, x− xS ⊥ S.

Inner products are most useful for defining orthogonality between vectors. In order to definethe “size” of vectors in any LVS, we define vector norms.

1.3 Vector Norms and Normed Linear Vector Spaces

Definition 1.3.1 (Vector Norm). Let X be an LVS associated with IF. A vector norm on Xis a function ‖ · ‖ : X → IR that satisfies the following conditions:

(i) ‖x‖ ≥ 0 for all x ∈ X(i’) ‖x‖ = 0 if and only if x = 0

(ii) ‖ax‖ = |a| ‖x‖ for all a ∈ IF, x ∈ X(iii) ‖x + y‖ ≤ ‖x‖+ ‖y‖ for all x, y ∈ X

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The most commonly used vector norms on Cn are the so-called p-norms, defined by

‖x‖p4=

(n∑

i=1

|xi|p)1/p

, for p ∈ [1,∞]. (1.3.1)

Among the p-norms, the most commonly used ones are the 1-, 2- and the ∞-norms:

(i) The 1-norm: ‖x‖14=

n∑

i=1

|xi|.

(ii) The 2-norm: ‖x‖24=

(n∑

i=1

|xi|2)1/2

.

(iii) The ∞-norm: ‖x‖∞ 4= max

i=1:n|xi|.

It is obvious that the 2-norm is the well-known Euclidean distance on IRn. It is also easy to seethat ‖x‖2 =< x, x >1/2. In order to see how the ∞-norm can be obtained from the definition for ageneral p-norm, consider the fact following for an arbitrary x ∈ IRn:

maxi=1:n

|xi| ≤ ‖x‖p ≤ n1/p maxi=1:n

|xi|.

Then, letting p → ∞, we obtain ‖x‖∞ = maxi=1:n

|xi|. Note, finally, that the p-norms are defined for

p ∈ [1,∞]. Once can verify easily that if p < 1, the triangle inequality, i.e., condition (iii) in thedefinition of a vector norm is violated. Note, also, that with this definition of a vector norm, wecan define the “distance” between two vectors as

d(x, y)4= ‖x− y‖.

When the norm is derived from an inner product (i.e., ‖x‖ =< x, x >1/2), we have a simplerelationship between the inner product of two vectors and their norms.

Theorem 1.3.2 (The Parallelogram Law). Let (X , < ·, · >) be an inner product space anddefine ‖x‖2 =< x, x >. Then, for any x, y ∈ X

‖x + y‖2 + ‖x− y‖2 = 2‖x‖2 + 2‖y‖2. (1.3.2)

Theorem 1.3.3 (Pythagorean Theorem). Let X be an inner product space. For any x, y ∈ X ,

if x ⊥ y, then ‖x + y‖2 = ‖x‖2 + ‖y‖2. (1.3.3)

Theorem 1.3.4. Let (X , ‖ · ‖) be a complex normed linear space such that

‖x + y‖2 + ‖x− y‖2 = 2(‖x‖2 + ‖y‖2

) ∀x, y ∈ X . (1.3.4)

Then,

< x, y >=14

{‖x + y‖2 − ‖x− y‖2 + j‖x + jy‖2 − j‖x− jy‖2}

(1.3.5)

defines an inner product on X such that ‖ · ‖2 =< ·, · >. Moreover, the inner product in (1.3.5) isthe only one that generates the norm ‖ · ‖.

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In words, what the theorem above says is the following: (i) Given (X , ‖ · ‖), the norm ‖ · ‖ isderived from an inner product (i.e., ‖x‖2 =< x, x >) if and only if condition (1.3.4) is satisfied, and(ii) when (1.3.4) is satisfied, the inner product given in (1.3.5) is the inner product which produces‖ · ‖.

For instance, consider Cn, ‖ · ‖2. it is easily shown that (1.3.4) is satisfied and when the right-hand-side of (1.3.5) is computed, it does give

∑ni=1 yixi. However, the ∞-norm on Cn is not derived

from an inner product, as the counter-example below given for n = 2 indicates:

x =[

11

]and y =

[1

−1

].

With these vectors, ‖x + y‖2∞ + ‖x− y‖2∞ = 8, but 2(‖x‖2∞ + ‖y‖2∞

)= 4.

1.4 Sequences and Convergence

Definition 1.4.1 (Cauchy Sequence). A sequence {xn} is said to be Cauchy if

for any ε > 0, there exists an N such that m,n > N implies ‖xm − xn‖ < ε.

Definition 1.4.2 (Convergence). A sequence {xk} is said to converge to x, denoted by {xk} → x,if

for any ε > 0, there exists an N such that n > N implies ‖xn − x‖ < ε.

It is obvious that every convergent sequence is Cauchy. However, not every Cauchy sequenceconverges to a vector in the vector space considered.Example 1.4.3. Let Q denote the space of rational numbers, equipped with the absolute valuefunction as the norm. Consider the sequence of rational numbers generated by the iterative rule

xk+1 =xk + 2

xk

2, x0 = 1.

Then, it can be shown that {xk} is a Cauchy sequence and converges to√

2. However, since√

2 isnot rational, we conclude that Q is not a complete vector space. ¤

Those spaces in which every Cauchy sequence converges to an element in the space are calledcomplete.

Definition 1.4.4 (Complete LVS). A normed LVS (X, ‖ · ‖) is said to be complete if everyCauchy sequence in X converges to an element in X .

A basic fact that follows from elementary algebra is that IR is complete. When we take thatfor granted, the following becomes easy to prove:

Theorem 1.4.5. (IRn, ‖ · ‖) is complete.

Proof. Let {xk} be a Cauchy sequence in IRn. ¥

Definition 1.4.6 (Banach Space). A complete normed linear space is called a Banach space.

Definition 1.4.7 (Hilbert Space). A complete inner-product space with the norm defined as

‖x‖ 4=< x, x >1/2 is called a Hilbert space.

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1.5 The Cauchy-Schwarz, Holder and Minkowski Inequalities

Theorem 1.5.1 (Cauchy-Schwarz Inequality). Let (X , < ·, · >) be an inner-product space with‖x‖ =< x, x >1/2. Then,

| < x, y > | ≤ ‖x‖ ‖y‖ ∀x, y ∈ X . (1.5.1)

Furthermore, equality holds if and only if x and y are linearly dependent.

Proof. Let t ∈ IR and x, z ∈ X , z 6= 0 (if z = 0, the statement becomes 0 ≤ 0, which is alwaystrue). Then,

p(t)4= ‖x + tz‖2 = ‖x‖2 + t < x, z > +t < z, x > +t2‖y‖2 = ‖x‖2 + 2tRe < x, z > +t2‖y‖2 ≥ 0.

Hence, p(t) is quadratic in t and is always nonnegative. This means its discriminant (the familiarb2 − 4ac) is always non-positive. Equivalently,

(Re < x, z >)2 ≤ ‖x‖2 ‖y‖2.

Since X is a linear vector space and z is arbitrary (6= 0), set z =< x, y > x for some y ∈ X , y 6= 0(if y = 0, 0 ≤ 0 again). Note that

Re 〈x,< x, y > y〉 = Re(< x, y > < x, y >) = < x, y > < x, y >= | < x, y > |2.

It follows from the ”discriminant inequality” above that

| < x, y > |4 ≤ ‖x‖2 〈< x, y > y, < x, y > y〉= ‖x‖2 < x, y > < x, y > ‖y‖2

= ‖x‖2 | < x, y > |2 ‖y‖2

⇒ | < x, y > |2 ≤ ‖x‖2‖y‖2.

Hence, we have shown the Cauchy-Schwarz inequality. As for the equality version, it occurs if andonly if the discriminant is equal to zero. This condition, combined with p(t) ≥ 0 implies that p(t)must have a repeated root for some t∗ ∈ IR. Hence, this t∗ must satisfy < x + t∗z, x + t∗z >= 0,which is true if and only if x + t∗z = x + t < x, y > y = 0, which means x and y are linearlydependent. We’re done. ¥

Theorem 1.5.2 (Holder Inequality). Suppose two real numbers p, q > 1 are given such that

1p

+1q

= 1.

Then, ∫ ∞

0|x(t)y(t)| dt ≤ ‖x‖p‖y‖q. (1.5.2)

Theorem 1.5.3 (Minkowski Inequality). Given p ≥ 1,

‖x + y‖p ≤ ‖x‖p + ‖y‖p. (1.5.3)

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1.6 Examples of Linear Vector Spaces

1.6.1 The Spaces IRn and Cn

The spaces IRn and Cn consist of n-tuples with real and complex entries, respectively. Simpleexamples are

123

∈ IR3 and

[1 + j3− j

]∈ C2.

1.6.2 The space C[0, 1]

Consider C[0, 1], for instance. The definitions for linear independence, span and basis carry onwithout any modification. We can define the inner product between f, g ∈ C[0, 1] as

< f, g >=∫ 1

0f(t)g(t) dt,

from which we can derive the norm

‖f‖ 4=< f, f >1/2 .

One can verify that this norm satisfies all the conditions in Definition 1.3.1 but we will not go intoany details for linear vector spaces other than Cn.

1.6.3 Lp Spaces

Definition 1.6.1 (Lp-spaces, Lp-norm). The LVS Lp(0,∞), for 1 ≤ p ≤ ∞, is defined as thecollection of all measurable functions x(t) such that

∫ ∞

0|x(t)|p dt < ∞.

The space Lp(0,∞) is equipped with the norm

‖x‖p4=

(∫ ∞

0|x(t)|p dt

)1/p

1.6.4 lp Spaces

Definition 1.6.2 (lp-spaces, lp-norm). The LVS lp, for 1 ≤ p ≤ ∞, is defined as the collectionof all sequences {xk} such that

∞∑

k=1

|xk|p < ∞.

The space lp is equipped with the norm

‖x‖p4=

( ∞∑

k=1

|xk|p)1/p

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1.7 Exercises

Problem 1.1. Let S ⊂ IRn be the set whose only element is the zero vector in IRn, i.e., S =

0...0

. Is S linearly independent? Why/why not?

Problem 1.2. What is the dimension of (C, IR)? How about (C,C)? Verify your answer.

Problem 1.3. Prove that ‖ · ‖1, ‖ · ‖2 and ‖ · ‖∞ all satisfy the conditions for being a norm on Cn.Hint: Verify and use the fact that Re(z) ≤ |z| ∀z ∈ C.

Problem 1.4. In IR2, sketch the unit balls for the 1-, 2- and ∞-norms. That is, the sets

IBp4= {x ∈ IR2 : ‖x‖p ≤ 1} for p = 1, 2,∞.

Problem 1.5. In any real inner product space (X , < ·, · >), the “angle” between two vectors isdefined as

θ = cos−1

(< x, y >

‖x‖‖y‖)

, θ ∈ [0, π]

where ‖x‖ =< x, x >1/2. Show that θ is always well-defined for non-zero x, y and that the “cosinelaw” of plane geometry can be deduced from this definition of the angle between two vectors.

Problem 1.6. Let X be a linear vector space equipped with a vector norm ‖ · ‖. Show that

|‖x‖ − ‖y‖| ≤ ‖x− y‖ ∀x, y ∈ X .

Problem 1.7. Let X be an LVS. Show that any linear subspace of X contains the origin (i.e., thenull element θ) of X .

Problem 1.8. Verify the parallelogram law (Theorem 1.3.2).

Problem 1.9. Verify the Pythagorean theorem (Theorem 1.3.3).

Problem 1.10. Is the 1-norm on Cn derived from an inner product? Justify your answer.

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Chapter 2

Matrices

2.1 Basics

For the purposes of this class, we define a matrix as a rectangular array of scalars. We say A ∈ Cm×n

if A is an m − by − n array of complex scalars and use the notation A = [aij ]m×n, to denote thematrix in terms of its entries. We define basic arithmetic operations on matrices as follows:

(i) Matrix summation: Given A,B ∈ Cm×n,

A + B = [aij + bij ]. (2.1.1)

(ii) Scalar multiplication: Given A ∈ Cm×n and α ∈ C,

αA = [α aij ]. (2.1.2)

(iii) Matrix multiplication: Given A ∈ Cm×k and B ∈ Ck×n,

AB = C ∈ Cm×n, where cij =k∑

q=1

aiqbqj . (2.1.3)

Note that for two matrices A,B ∈ Cn×n, AB 6= BA in general. If AB = BA, then A and Bare said to commute. Moreover, since a vector x ∈ Cn can be viewed as a matrix in Cn×1, themultiplication Ax must be interpreted as a matrix multiplication as in (iii) above. In addition tothe operations above, adjoint of a matrix A = [aij ] ∈ Cm×n is A∗ = [aji] ∈ Cn×m. Similarly, the

transpose of a matrix A = [aij ] ∈ Cm×n is the matrix AT 4= [aji] ∈ Cn×m. Obviously, for real

matrices, the adjoint and the transpose are equal. In order to keep things general, we prefer to usethe adjoint, which naturally covers the transpose for real matrices.w

Two basic facts about adjoints are that

(A + B)∗ = A∗ + B∗ (2.1.4)

and(AB)∗ = B∗A∗ (2.1.5)

for matrices A and B of appropriate dimensions.Two basic linear subspaces associated with a matrix are the range and null space.

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Definition 2.1.1 (Range, null space). Given A ∈ Cm×n,

(i) The range of A is the set

R(A)4= {y ∈ Cm : Ax = y for some x ∈ Cn}. (2.1.6)

(ii) The null space of A is the set

N (A)4= {x ∈ Cn : Ax = 0}. (2.1.7)

The range of A is also called the image of A, denoted by Im(A) and the null space of A isalso called the kernel of A, denoted by Ker(A). It is clear that Im(A) and Ker(A) are linearsubspaces themselves. Moreover, both the range and the null space are never empty, since the zerovector (of appropriate dimension) is always a member of both. sets.

The dimensions of the range and null space of a matrix are related as follows:

Theorem 2.1.2. Given A ∈ Cm×n,

n = dim(R(A)) + dim(N (A)). (2.1.8)

Proof. ¥

Theorem 2.1.3. Given A ∈ Cm×n,R(A) ⊥ N (A∗). (2.1.9)

That is, y∗x = 0 for any x ∈ R(A) and y ∈ N (A∗).

Definition 2.1.4 (Rank). The rank of a matrix A ∈ Cm×n, denoted by rank(A), is the numberof linearly independent columns (or equivalently, rows) of A.

It follows from the definition that

rank(A) = dim(R(A)). (2.1.10)

It is also immediate that rank(A) ≤ min{m,n}. An even more useful inequality regarding the rankof a matrix product is provided by the following lemma.

Lemma 2.1.5 (Sylvester’s Rank Inequality). For any A ∈ Cm×k and B ∈ Ck×n,

rank(A) + rank(B)− k ≤ rank(AB) ≤ min{rank(A), rank(B)}. (2.1.11)

Example 2.1.6. As an application of Sylvester’s rank inequality, consider the matrix product Z =XY , where X ∈ Ck×k, Y ∈ Ck×m, rank(X) = k and rank(Y ) = m ≤ k. Then,

k + m− k ≤ rank(Z) ≤ min{k, m} ⇒ rank(Z) = m.

For square matrices only, we define the determinant:

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Definition 2.1.7 (Determinant). Given A ∈ Cn×n, the determinant of A is defined as

det (A)4=

n∑

i=1

(−1)i+jaij det(A[i,j]) for any j = 1 : n (2.1.12)

=n∑

j=1

(−1)i+jaij det(A[i,j]) for any i = 1 : n, (2.1.13)

where A[i,j] denotes the submatrix of A obtained by deleting the ith row and jth column and thedeterminant of a scalar is defined as its scalar value.

The expressions above are called the Laplace expansion formulas for the determinant. Beloware some essential facts about determinants:

Proposition 2.1.8. The determinant function satisfies the following:

(i) det(AT ) = det(A) for any A ∈ Cn×n.

(i’) det(A∗) = det(A) for any A ∈ Cn×n.

(ii) det(AB) = det(A) det(B) for any A,B ∈ Cn×n.

(iii) det(In + AB) = det(Im + BA) for any A ∈ Cn×m and B ∈ Cm×n.

(iv) det(α A) = αn det A for any α ∈ C, A ∈ Cn×n.

Among the many theoretical uses of these properties of the determinant, item (ii) above has animmediate numerical use:

Example 2.1.9. Suppose we are given two matrices A ∈ Cn×1 and B ∈ IR1×n and we need tocompute det(In + AB). Instead, we can equivalently compute

det(I1 + BA) = 1 + BA,

which involves much less arithmetic operations.

Definition 2.1.10 (Invertibility, inverse). A matrix A ∈ Cn×n is said to be invertible (ornonsingular) if there exists a unique matrix in Cn×n, denoted by A−1, such that

AA−1 = A−1A = I. (2.1.14)

The matrix A−1 is then called the inverse of A.

Invertibility of a given matrix can be checked in various ways:

Proposition 2.1.11. Given A ∈ Cn×n, the following statements are equivalent:

(i) A is invertible

(ii) rank(A) = n

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(iii) The rows of A are linearly independent

(iv) The columns of A are linearly independent

(v) det(A) 6= 0

(vi) R(A) = Cn

(vii) N (A) = {0}(viii) Ax = 0 implies x = 0.

(ix) For any b ∈ Cn, there exists a unique x such that Ax = b

(x) 0 is not an eigenvalue of A (see the next section).

Remark 2.1.12. Given an invertible A ∈ Cn×n, the inverse of A can be computed as

A−1 =1

detACT , (2.1.15)

where C is the cofactor matrix of A, where

cij = (−1)i+j det A[i,j].

However, this is not how the inverse is computed numerically. What is done, instead, is to cast theinversion problem as a solution of system of equations. That is, solve for xi ∈ Cn×1 such that

Axi = ei ∀i = 1 : n,

where ei denotes the ith column of the n-dimensional identity matrix In. After solving the aboveproblem (via the Gaussian elimination method, or any other), the inverse of A is constructed as

A−1 =[

x1 x2 · · · xn

].

This procedure requires much less operations than the formula for A−1.

Remark 2.1.13. There are useful formulas for the inversion of a sum of matrices and a matrixdefined by matrix blocks:

(i) (A + BCD)−1 = A−1 −A−1B(C−1 + DA−1B)−1DA−1

(ii) Define ∆4= A−BD−1C and ∇ 4

= D − CA−1B, whenever the inverses exist. Then,

[A BC D

]−1

=[

∆−1 −∆−1BD−1

−D−1C∆−1 D−1 + D−1C∆−1BD−1

]

=[

A−1 + A−1B∇−1CA−1 −A−1B∇−1

−∇−1CA−1 ∇−1

].

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Using matrix inverses, we can express determinants of block-partitioned matrices in terms oftheir submatrices.

Theorem 2.1.14 (Schur’s determinant identity). Given matrices A ∈ Cn×n, B ∈ Cn×m,C ∈ Cm×n and D ∈ Cm×m, where A invertible, the following holds:

det[

A BC D

]= det A · det(D − CA−1B). (2.1.16)

Proof. Express the “big” matrix as follows:[

A BC D

]=

[I 0

CA−1 I

] [A 00 D − CA−1B

] [I A−1B0 I

].

Therefore,

det[

A BC D

]= det

[I 0

CA−1 I

]det

[A 00 D − CA−1B

]det

[I A−1B0 I

].

By direct expansion for the determinants, it is easy to see (but make sure you’re convinced) that

det[

I 0CA−1 I

]= det

[I A−1B0 I

]= 1.

Furthermore, [A 00 D − CA−1B

]=

[A 00 I

] [I 00 D − CA−1B

].

Going back to determinants,

det[

A BC D

]= 1 · det

[A 00 I

]det

[I 00 D − CA−1B

]· 1 = detA · det(D − CA−1B).

¥

Definition 2.1.15 (Trace). The trace of A ∈ Cn×n, denoted trace(A), is the sum of its diagonalelements. That is,

trace(A) =n∑

i=1

aii. (2.1.17)

It goes without saying that trace(·) is a linear function. That is, for any A,B ∈ Cn×n

(i) trace(α A) = α trace(A) for any α ∈ C(ii) trace(A + B) = trace(A) + trace(B).

We also have the following very useful result regarding the trace of the product of two matrices:

Proposition 2.1.16. Given A ∈ Cm×n and B ∈ Cn×m,

trace(AB) = trace(BA). (2.1.18)

Proof. trace(AB) =m∑

i=1

n∑

j=1

aijbji

=

n∑

j=1

(m∑

i=1

bjiaij

)= trace(BA). ¥

15

2.2 Eigenvalues and Eigenvectors

Consider a matrix A ∈ Cn×n. The equation

Ax = λx (2.2.1)

where x ∈ Cn, x 6= 0, λ ∈ C is called the eigenvalue-eigenvector equation of A. The scalars λ andthe corresponding non-zero vectors x that satisfy (2.2.1) are called the eigenvalues and eigenvectorsof A, respectively. The set of eigenvalues of A (including repetitions) is called the spectrum of A,denoted by σ(A).

Equation (2.2.1) can be rearranged as

(λI −A)x = 0, x 6= 0.

There exists a nontrivial solution to the equation above if and only if the matrix λI − A is rank-deficient, that is

pA(λ)4= det(λI −A) = 0. (2.2.2)

The polynomial pA(λ), which is an nth-order polynomial in λ, is called the characteristic polynomialof A. Since A is real, so are all the coefficients of pA(λ). It follows that pA(λ) has n roots, whichare the eigenvalues of A. Moreover, if A is real, the spectrum of A is symmetric about the real axis.In other words, if λ∗ is an eigenvalue of A, then so is λ∗. To see this, suppose λ∗ is an eigenvalueof A. Then,

pA(λ∗) = 0 ⇐⇒ pA(λ∗) = 0 ⇐⇒ pA(λ∗) = 0.

Note that we can express the characteristic polynomial as

pA(λ) = (λ− λ1)(λ− λ2) · · · (λ− λn).

Definition 2.2.1 (Minimal Polynomial). Given a square matrix A, the minimal polynomialof A is the monic polynomial ψ(λ) of least degree such that ψ(A) = 0.

We have two simple facts regarding the eigenvalues of a matrix:

Proposition 2.2.2. Given A ∈ Cn×n,

(i) det(A) =n∏

i=1

λi

(ii) trace(A) =n∑

i=1

λi

Proof. (i) Recall that pA(λ) = det(λI −A) = (λ− λ1)(λ− λ2) · · · (λ− λn). Now let λ = 0. Then,

pA(0) = det(0I −A) = det(−A) = (−1)n det(A)

and

pA(0) = (0− λ1)(0− λ2) · · · (0− λn) = (−1)nn∏

i=1

λi.

Hence, det(A) =n∏

i=1

λi.

(ii) We need to introduce the spectral decomposition of a matrix to prove this. ¥

16

2.2.1 Spectral Decomposition

Definition 2.2.3 (Similarity). Two matrices A, B ∈ IRn×n are said to be similar if there existsa nonsingular S ∈ IRn×n such that

B = S−1AS. (2.2.3)

The transformation A → S−1AS is called a similarity transformation of A under S.

Definition 2.2.4 (Diagonalizability). A square matrix is said to be diagonalizable if it is similarto a diagonal matrix.

Theorem 2.2.5. A matrix A ∈ IRn×n is diagonalizable if and only if it has n linearly independenteigenvectors.

We will next show that any square matrix A can be decomposed as

A = MJM−1 (2.2.4)

for some J ∈ Cn×n, in the so-called Jordan canonical form and some invertible M ∈ Cn×n. TheJordan canonical form is

J1 0 · · · 00 J2 · · · 0...

.... . .

...0 0 · · · Jk

,

where k ≤ n and each Ji ∈ Cni×ni is in the form

Ji =

λi 1 0 · · · 00 λi 1 · · · 0...

.... . . . . .

...0 0 · · · λi 10 0 · · · 0 λi

∈ Cni×ni .

When each Jordan block is a scalar (i.e., ni = 1 for all i = 1 : k), A is said to be diagonalizable.It turns out that A is diagonalizable if and only if it has n linearly independent eigenvectors. Wenext discuss the two cases where A does and does not have linearly independent eigenvectors.

(i) A has linearly independent eigenvectors: Now suppose A has n linearly independenteigenvectors. Then, we can express (2.2.1) as a matrix equation as follows:

AU = UΛ,

where

U4=

[x1 x2 · · · xn

]and Λ

4=

λ1 0 · · · 00 λ2 · · · 0...

.... . .

...0 0 · · · λn

17

and xi is the eigenvector corresponding to the eigenvalue λi. Since U is now invertible byTheorem 2.2.6, we obtain

A = UΛU−1.

Obviously, we now have k = n, i.e., each Jordan block is a scalar. When compared with thegeneral expression for the spectral decomposition, we have U = M and Λ = J .

A sufficient condition for A to have linearly independent eigenvectors is that it have distincteigenvalues, as shown in the lemma below.

Lemma 2.2.6. Given A ∈ IRn×n, let λ1, . . . , λk (k ≤ n) be distinct eigenvalues of A and letx1, . . . , xk be the associated eigenvalues. Then, the set {x1, . . . , xk} is linearly independent.

Proof. Suppose not, i.e., suppose {x1, . . . , xk} is linearly dependent. Then, there exists anontrivial combination of xi’s that produces 0. Let such a combination with the fewestnonzero coefficients be given by

α1x1 + α2x2 + · · ·+ αrxr = 0, 1 < r ≤ k.

(We can assume without loss of generality that x1, . . . , xr are the first r of the k eigenvectors.)We also have

A(α1x1 + α2x2 + · · ·+ αrxr) = α1λ1x1 + α2λ2x2 + · · ·+ αrλrxr = 0.

Now multiply the first equation by λr and subtract it from the second. The result is

α1(λ1 − λr)x1 + · · ·+ αr−1(λr−1 − λr)xr−1 = 0.

Hence, we have a nontrivial linear combination of xi’s that gives 0 with fewer nonzero coef-ficients than r. This is a contradiction, since we assumed that r was the fewest. Therefore,contrary to our initial supposition, the eigenvectors x1, . . . , xk are linearly independent. ¥

Remark 2.2.7. Note that the reverse implication is not necessarily true. That is, there may bemore than one independent eigenvector associated with a repeated eigenvalue. An immediateexample is In, the identity matrix of dimension n. All n of its eigenvalues are equal to1. However, it does have n linearly independent eigenvectors, which can be taken as its ncolumns. In fact, when we talk about the multiplicity of an eigenvalue, we usually meanits algebraic multiplicity, meaning its multiplicity as a root of the characteristic polynomial.The number of linearly independent eigenvectors associated with an eigenvalue is definedas its geometric multiplicity. Hence, in the case of In, algebraic multiplicity and geometricmultiplicity of the only eigenvalue, 1, are both n.

Remark 2.2.8. When A is real, but it has complex eigenvalues (and hence complex eigenvec-tors), we may prefer to work with real matrices M and J . We can accomplish this as follows:First note that since A is real, its eigenvalues are symmetric about the real axis (see Prob-lem 2.11) and the corresponding eigenvectors are complex conjugates of each other. Now, forsimplicity, suppose n = 2 and the two eigenvalues of A are complex conjugates of each other.

18

Denote the eigenvalues as α ± βj ∈ C and the corresponding eigenvectors as x ± yj ∈ Cn.Then,

A = TΛT−1 =[

x + yj x− yj] [

α + βj 00 α− βj

] [x + yj x− yj

]−1

=[

x y] [

1 1j −j

] [α + jβ 0

0 α− jβ

] [1 1j −j

]−1 [x y

]−1

=[

x y] [

1 1j −j

] [α + jβ 0

0 α− jβ

] [1/2 −j/21/2 j/2

] [x y

]−1

=[

x y] [

α β−β α

] [x y

]−1.

(ii) A does not have linearly independent eigenvectors: Now suppose A does have repeatedeigenvalues and the eigenvectors associated with them are not all linearly independent. Inthis case, we have to resort to generalized eigenvectors and the Jordan form of A.

Suppose λ is an eigenvalue of algebraic multiplicity q, but with geometric multiplicity 1. Thatis, λ is a q-time-repeated eigenvalue of A, but there is only one eigenvector associated withit. Then, we first solve for the eigenvector associated with it,

Ax1 = λx1,

and then solve for x2, x3, . . . , xq such that

Ax2 = x1 + λx2

Ax3 = x2 + λx3

...Axq = xq−1 + λxq.

We call x2, . . . , xq the generalized eigenvectors associated with λ. We can then show that{x1, x2, . . . xq} is a linearly independent set. And obviously,

A[

x1 · · · xq

]=

[x1 · · · xq

]

λ 1 · · · 0...

. . . . . ....

.... . . 1

0 · · · · · · λ

.

Repeating the same process for each of the eigenvalues, we obtain the form (2.2.4).

We can now prove that trace(A) =n∑

i=1

λi:

19

Proof of Proposition 2.2.2, (ii). Let A be spectrally decomposed as A = MJM−1. Then,

trace(A) = trace(MJM−1) = trace(JM−1M) = trace(J) =n∑

i=1

λi.

(Note that this is independent of whether A has real, complex, distinct or repeated eigenvalues.) ¥

Theorem 2.2.9 (Cayley-Hamilton). Every square matrix satisfies its own characteristic poly-nomial.

Proof. Let the square matrix of concern be A ∈ Cn×n and let its characteristic polynomial bedenoted by pA(·). We will divide the proof into two parts:

(a) Diagonalizable A: In this case A = MΛM−1, where Λ = diag{λ1, . . . , λn}.pA(A) = pA(MΛM−1)

= (MΛM−1 − λ1I) · · · (MΛM−1 − λnI)

= [M(Λ− λ1I)M−1] · · · [M(Λ− λnI)M−1]

= M(Λ− λ1I) · · · (Λ− λnI)M−1

= MpA(Λ)M−1.

Therefore, pA(A) = 0 ⇐⇒ pA(Λ) = 0. But pA(Λ) has the following form:

pA(Λ) = diag(0, λ2 − λ1, . . . , λn − λ1) · · ·diag(λn − λ1, . . . , λn−1 − λn, 0) = 0.

(b) Non-diagonalizable A: Let there be q(< n) distinct eigenvalues, each of which is repeated nq

times. Then, A = MJM−1, where J = diag{J1, J2, . . . , Jq} and each Ji ∈ Cnq×nq is in theJordan canonical form.

The characteristic polynomial is in the form

pA(λ) = (λ− λ1)n1 · · · (λ− λq)nq

Therefore,

pA(A) = (MJM−1 − λ1I)n1 · · · (MJM−1 − λqI)nq

= M(J − λ1)n1 · · · (J − λq)nqM−1

= M diag{(Ji − λiI)n1 · · · (Ji − λq)nq}qi=1M

−1.

The term (Jk − λkI)nk , 1 ≤ k ≤ q is of the form

(Jk − λkI)nk =

0 1 0 · · · 00 0 1 · · · 0...

......

. . ....

0 0 0 · · · 10 0 0 · · · 0

nk

,

which is equal to 0 (Problem 2.1).

20

¥

Example 2.2.10. Consider the matrix

A =[

1 23 4

].

The characteristic polynomial of A is pA(λ) = λ2 − 5λ− 2. Then,

pA(A) =[

1 23 4

]2

− 5[

1 23 4

]− 2 I =

[7 1015 22

]−

[5 1015 20

]−

[2 00 2

]=

[0 00 0

].

The most important implication of the Cayley-Hamilton theorem is that every term Ak, wherek ≥ 0 can be expressed as a linear combination of I, A, A2, ..., An−1. In order to see how thisworks, consider the polynomial division of λk by pA(λ), assuming k ≥ n:

λk = q(λ)pA(λ) + r(λ),

where q(λ) is called the quotient and r(λ), the remainder. The order of r(λ) is less than or equalto n− 1. Then, since pA(A) = 0, we obtain Ak = r(A), which is a combination of Ai, i = 0 : n− 1.

We now show that similar matrices have the same eigenvalues.

Lemma 2.2.11. If matrices A,B ∈ IRn×n are similar, then σ(A) = σ(B).

Proof. Let B = S−1AS for some invertible S. Since the eigenvalues of A and B are the roots ofthe polynomials pA(λ) and pB(λ),

pB(λ) = det(λI − S−1AS) = det[S−1(λI −A)S] = det(S−1) det(λI −A) det(S) = det(λI −A).

But det(λI−A) is nothing but pA(λ). Therefore, A and B have the same characteristic polynomials,hence the same spectra. ¥

Remark 2.2.12. Having the same spectrum is necessary, but not sufficient for similarity. Consider,for instance, the matrices

A =[

0 10 0

]and B =

[0 00 0

].

The eigenvalues of both are 0, with multiplicity 2 but they are not similar.

2.2.2 Left-eigenvectors*

Let A ∈ Cn×n. A scalar λ ∈ C and a vector x ∈ Cn are said to satisfy the eigenvalue, left-eigenvectorequation if

x∗A = λx∗, x 6= 0. (2.2.5)

One can easily show that there is no distinction between left- and right-eigenvalues of A.Moreover, if (λ, x) is an eigenvalue, right-eigenvector pair of A, then (λ, x) is an eigenvalue, left-eigenvector pair of A∗.

21

In order to see how the right- and left-eigenvectors are related, assume A has linearly indepen-dent right-eigenvectors. Then, A can be decomposed as

A = SΛS−1,

where S is the matrix of right-eigenvectors, namely

S4=

[x1 x2 · · · xn

].

One can also show that A can also be decomposed as

A = T−1ΛT,

where T is the matrix of left-eigenvectors, namely

T4=

z∗1z∗2...

z∗n

wherez∗i A = λiz

∗i .

In short, the left-eigenvectors can be obtained from the rows of the inverse of the right-eigenvectormatrix.

Therefore, A can also be decomposed as

A = SΛT =n∑

i=1

xiz∗i . (2.2.6)

2.3 The Singular Value Decomposition

Theorem 2.3.1. Given A ∈ IFm×n, there exist unitary matrices

U =[

u1 u2 · · · um

] ∈ IFm×m and V =[

v1 v2 · · · vn

] ∈ IFn×n

and a matrix Σ, where

Σ4=

[Σ1 00 0

], Σ1 = diag{σ1, σ2, · · · , σk},

k4= rank(A), and σ1 ≥ σ2 ≥ · · · ≥ σk > 0 such that

A = UΣV ∗. (2.3.1)

In IR2, the singular value decomposition has a very simple geometric interpretation. Considerthe action of a matrix A on vectors forming the unit circle. If A is decomposed as A = UΣV T , thetransformation of the unit circle through A can be visualized as shown in Figure 2.1

22

−2 0 2−2

−1

0

1

2

x

−2 0 2−2

−1

0

1

2

VTx

−2 0 2−2

−1

0

1

2

Σ VTx

−2 0 2−2

−1

0

1

2

Ax=UΣ VTx

Figure 2.1: Matrix multiplication.

Remark 2.3.2. When we define

U14=

[u1 · · · uk

]and V1

4=

[v1 · · · vk

],

we can decompose A asA = U1Σ1V

∗1 .

Note that U1 and V1 are not unitary now, but only satisfy U∗1 U1 = I and V ∗

1 V1 = I.

2.4 Pseudo-inverses

Definition 2.4.1. Given A ∈ Cm×n, the (Moore-Penrose) pseudo-inverse of A is the uniquematrix A+ ∈ Cn×m such that

(i) (A+A)∗ = A+A

(ii) (AA+)∗ = AA+

(iii) A+AA+ = A+

(iv) AA+A = A

The definition is not a stand-alone definition, since it involves the assertion that the pseudo-inverse A+ is unique. The following reasoning (courtesy of Orkan Akcan) proves uniqueness: Sup-pose the pseudo-inverse is not unique. That is, there exist A+

1 and A+2 such that

(A+1 A)∗ = A+

1 A (A+2 A)∗ = A+

2 A (2.4.1a)(AA+

1 )∗ = AA+1 (AA+

2 )∗ = AA+2 (2.4.1b)

A+1 AA+

1 = A+1 A+

2 AA+2 = A+

2 (2.4.1c)AA+

1 A = A AA+2 A = A (2.4.1d)

23

Now pre-multiply the A+1 -equation of (2.4.1d) by A+

2 . Then, we have

A+2 AA+

1 A = A+2 A

⇒ [(A+2 A)(A+

1 A)]T = (A+2 A)T

⇒ (A+1 A)T (A+

2 A)T = (A+2 A)T

⇒ A+1 AA+

2 A = A+2 A by (2.4.1a)

⇒ A+1 A = A+

2 A by (2.4.1d)⇒ A+

1 AA+1 = A+

2 AA+1

⇒ A+1 = A+

2 AA+1 by (2.4.1c)

Now post-multiply the A+1 -equation of (2.4.1d) by A+

2 . Then,

AA+1 AA+

2 = AA+2

⇒ [(AA+1 )(AA+

2 )]T = (AA+2 )T

⇒ (AA+2 )T (AA+

1 )T = (AA+2 )T

⇒ AA+2 AA+

1 = AA+2 by (2.4.1b)

⇒ AA+1 = AA+

2 by (2.4.1d)⇒ A+

2 AA+1 = A+

2 AA+2

⇒ A+2 AA+

1 = A+2 by (2.4.1c)

The conclusion is that A+1 = A+

2 and uniqueness is proven.

Proposition 2.4.2. The pseudo-inverse of A ∈ Cm×n is given by

A+ = V1Σ−11 U∗

1 .

Theorem 2.4.3. Given A ∈ Cm×n and B ∈ Cm×p, there exists an X ∈ Cn×p such that

AX = B (2.4.2)

if and only if(I −AA+)B = 0. (2.4.3)

When condition (2.4.3) is satisfied, all solutions to equation (2.4.2) are given by

X = A+B + (I −A+A)Z, (2.4.4)

where Z ∈ Cn×p is arbitrary.

2.5 Hermitian Matrices

A matrix A ∈ Cn×n is said to be Hermitian if A = A∗. Hermitian matrices have several very usefulproperties. Some are listed below:

Theorem 2.5.1. All eigenvalues of a Hermitian matrix are real.

24

Proof. Take an arbitrary A = A∗ ∈ Cn×n and consider its eigenvalue-eigenvector equation Ax = λx.Now pre-multiply by x∗ to obtain x∗Ax = λx∗x. Now take the complex conjugate transposes ofboth sides to obtain (x∗Ax)∗ = (λx∗x)∗. But (x∗Ax)∗ = x∗AT x = x∗Ax = λx∗x since A = A∗.Further, since (λx∗x)∗ = λx∗x, we have λx∗x = λx∗x. Since x is nonzero, we obtain λ = λ, i.e., λis real. ¥

We say a matrix U ∈ Cn×n is unitary if U∗U = UU∗ = I. An extremely useful fact about thespectral decomposition of Hermitian matrices is the following:

Theorem 2.5.2. Any Hermitian matrix A = A∗ ∈ Cn×n can be spectrally decomposed as

A = UΛU∗, (2.5.1)

where U is unitary and Λ = diag{λ1, . . . , λn}.Proof. The proof is by induction over the dimension n. For a Hermitian matrix in C1×1, the resultis obvious. Because if A ∈ C1×1 is Hermitian, it is a real scalar. Then, Λ is equal to itself andU = 1, which is unitary.

Now assume the statement is true for all Hermitian matrices in C(n−1)×(n−1) and considerA = A∗ ∈ Cn×n. Let λ1 be an eigenvalue of A and x1 an eigenvalue associated with λ1. ByTheorem 2.5.1, λ1 is real and x1 can be chosen such that x∗1x1 = 1 without loss of generality. Nowlet X be a unitary matrix with x1 as its first column,

X4=

[x1 x2 · · · xn

] ∈ Cn×n.

Then, the first column of the product X∗AX gives

X∗Ax1 = X∗(λ1x1) = λ1X∗x1 = λ1e1,

where e1 denotes the first column of In. Moreover, the first row of X∗AX is

x∗1AX = (Ax1)∗X = λ1x∗1X = λ1e

∗1

since A = A∗. We then have

X∗AX =[λ1 00 A

],

where A = A∗ ∈ C(n−1)×(n−1). By the induction hypothesis, we can express A as

A = U ΛU∗,

where U is unitary and Λ is real diagonal. It follows that

A = X

[I 00 U

]

︸ ︷︷ ︸U

[λ1 00 Λ

]

︸ ︷︷ ︸Λ

[I 00 U

]∗X∗

︸ ︷︷ ︸U∗

.

Defined as such, U is unitary and Λ is real diagonal. Its diagonal entries are, of course, theeigenvalues of A. ¥

25

Remark 2.5.3. This decomposition is in fact possible for a larger class of matrices, namely thosethat are normal, i.e., matrices A ∈ Cn×n that satisfy A∗A = AA∗. Hermitian matrices are clearlynormal.

Theorem 2.5.4 (Rayleigh-Ritz Inequality). Given any A = A∗ ∈ Cn×n,

λmin‖x‖22 ≤ x∗Ax ≤ λmax‖x‖2

2 ∀x ∈ Cn (2.5.2)

where λmin and λmax denote the minimum and maximum eigenvalues of A.

Proof. Since A is hermitian, there exist a unitary matrix U and a diagonal matrix Λ = diag{λ1, . . . , λn}such that A = UΛU∗. For any x ∈ Cn,

x∗Ax = x∗UΛU∗x = (U∗x)∗Λ(U∗x) =n∑

i=1

λi|(U∗x)i|2.

Hence, we have the following lower and upper bounds for x∗Ax:

λmin

n∑

i=1

|(U∗x)i|2 ≤ x∗Ax ≤ λmax

n∑

i=1

|(U∗x)i|2.

Butn∑

i=1

|(U∗x)i|2 = (U∗x)∗(U∗x) = x∗UU∗x = x∗x = ‖x‖2

since U is unitary. We have established the desired inequality. ¥

Definition 2.5.5 (Congruence). Two matrices A,B ∈ Cn×n are said to be congruent to eachother if there exists a nonsingular T ∈ Cn×n such that

B = T ∗AT. (2.5.3)

The transformation A → T ∗AT is called a congruence transformation of A under T .

Since all eigenvalues of a Hermitian matrix are real, we can group the ones that are positive,negative and zero. This brings us to the definition of the inertia of a Hermitian matrix.

Definition 2.5.6 (Inertia). Given a matrix A = A∗ ∈ Cn×n, the inertia of A is the triplet

in(A) = {n+, n−, n0}, (2.5.4)

where n+, n− and n0 denote the number of positive, negative and zero eigenvalues (counting mul-tiplicities), respectively.

Analogous to similar matrices having the same spectrum, congruent matrices have the sameinertia as stated below:

Lemma 2.5.7. Given two Hermitian matrices A,B ∈ Cn×n,

A and B are congruent ⇐⇒ in(A) = in(B).

26

Proof. (⇒) Let A and B be congruent and let in(A) = {n+(A), n−(A), n0(A)}. First, observethat rank(A) = rank(B) by Sylvester’s rank inequality, since there exists a nonsingular T ∈ Cn×n

such that B = T ∗AT . It then follows that n0(A) = n0(B). So, we only have to show thatn+(A) = n+(B).

Let v1, · · · , vn+(A) be the orthogonal eigenvectors of A associated with the positive eigenvalues

λ1, · · · , λn+ . Define S+(A)4= span{v1, · · · vn+(A)}. Then, dim(S+(A)) = n+(A). Now, for any

nonzero x =∑n+(A)

i=1 αivi, we have x∗Ax =∑n+(A)

i=1 λi|αi|2 > 0. In this case, the fact that

x∗T−∗BT−1x = (T−1x)∗B(T−1x) > 0

implies y∗By > 0 for all nonzero y ∈ span{T−1v1, · · · , T−1vn+(A)}, which also has dimensionn+(A). Then, it can be shown (which we will not) that B must have at least n+(A)-many positiveeigenvalues, i.e., n+(B) ≥ n+(A). By reversing the roles of A and B in the argument above, weobtain n+(A) ≥ n+(B), implying n+(A) = n+(B). Together with n0(A) = n0(B), this leads to theconclusion that in(A) = in(B).

(⇐) Let in(A) = in(B) = {n+, n−, n0}. Then,

A = UALA diag(In+ ,−In− , 0n0)LAU∗A and B = UBLB diag(In+ ,−In− , 0n0)LBU∗

B,

where UA and UB are the (unitary) eigenvector matrices of A and B and

LA4= diag

{λ

1/21 , · · · , λ1/2

n+, (−λn++1)1/2, · · · , (−λn++n−)1/2, 1, · · · , 1

}

andLB

4= diag

{µ

1/21 , · · · , µ1/2

n+, (−µn++1)1/2, · · · , (−µn++n−)1/2, 1, · · · , 1

}

and λ’s and µ’s denote the eigenvalues of A and B, respectively. Then,

B = T ∗AT, where T4= UAL−1

A LBU∗B.

Clearly, T is invertible. Hence, A and B are congruent. ¥

In words, the inertia of a symmetric matrix remains unchanged under congruence transforma-tion.

2.6 Sign Definiteness

Definition 2.6.1 (Positive/Negative definiteness/semidefiniteness). A matrix A = A∗ ∈Cn×n is said to be

(i) Positive definite ifxT Ax > 0 ∀x ∈ Cn, x 6= 0 (2.6.1)

(ii) Negative definite ifxT Ax < 0 ∀x ∈ Cn, x 6= 0 (2.6.2)

27

(iii) Positive semidefinite ifxT Ax ≥ 0 ∀x ∈ Cn, x 6= 0 (2.6.3)

(iv) Negative semidefinite ifxT Ax ≤ 0 ∀x ∈ Cn, x 6= 0 (2.6.4)

It follows from this definition and the Rayleigh-Ritz inequality that the sign-definiteness of aHermitian matrix is determined completely by the signs of its eigenvalues.

Lemma 2.6.2. A matrix A = A∗ ∈ Cn×n is

(i) Positive definite if and only if λi > 0 for all i = 1 : n

(ii) Negative definite if and only if λi < 0 for all i = 1 : n

(iii) Positive semi-definite if and only if λi ≥ 0 for all i = 1 : n

(iv) Negative semi-definite if and only if λi ≤ 0 for all i = 1 : n

where λi denote the eigenvalues of A.

Lemma 2.6.3. P = P ∗ ≥ 0 if and only if all principal submatrices of P are also positive semi-definite.

Proof. Problem 2.17. ¥

Corollary 2.6.4. Given A = A∗ ∈ Cn×n and B = B∗ ∈ Cm×m,[

A 00 B

]≥ 0 ⇐⇒

{A ≥ 0B ≥ 0

Lemma 2.6.5. Let P = P ∗ > 0. Then, the matrix T ∗PT is also symmetric positive definite forall nonsingular T ∈ Cn×n.

Proof. If P > 0, then in(P ) = {n, 0, 0}. Since inertia of a Hermitian matrix is invariant undercongruence transformations, in(T ∗PT ) = in(P ), hence T ∗PT ≥ 0. ¥

Similar to positive (or non-negative) scalars, we can define the square root of a positive semidef-inite matrix.

Definition 2.6.6 (Matrix square root). The square root of P = P ∗ ≥ 0 is defined as thematrix

P 1/2 4= UT Λ1/2U, (2.6.5)

where P = U∗ΛU is the spectral decomposition of P and

Λ1/2 4= diag{

√λ1,

√λ2, · · · ,

√λn}.

Note that P 1/2 is positive semidefinite as well.

28

Lemma 2.6.7 (Schur Complement Formula). Given A = A∗ ∈ Cn×n, B ∈ Cn×m and C =C∗ ∈ Cm×m, the following statements are equivalent:

(i)[

A BB∗ C

]> 0

(ii) A > 0 and C −B∗A−1B > 0

(iii) C > 0 and A−BC−1B∗ > 0

Proof. (i) ⇐⇒ (ii): Use the congruence transformation

T ∗[

A BB∗ C

]T =

[A 00 C −B∗A−1B

], where T

4=

[I −A−1B0 I

]

(i) ⇐⇒ (iii): Note that

J∗[

A BB∗ C

]J =

[C B∗

B A

], where J

4=

[0 II 0

]

and repeat the proof of (i) ⇐⇒ (ii). ¥

Theorem 2.6.8.P ≥ 0 ⇐⇒ trace(PS) ≥ 0 ∀S ≥ 0.

Proof. (⇒) Assume P ≥ 0. Let S ≥ 0 be given and decompose it as

S = UΛUT =n∑

i=1

λiuiuTi .

Then,0 ≤ uT

i Pui = trace(PuiuTi )

for each i. Since each λi ≥ 0, we obtain

0 ≤∑

λi trace(PuiuTi ) = trace

(P

∑

i

λiuiuTi

)= trace(PS).

(⇐) Assume trace(PS) ≥ 0 ∀S ≥ 0. Since xxT ≥ 0 for each x ∈ IRn, we have

0 ≤ trace(PxxT ) = xT Px ∀x ∈ IRn.

Hence, P ≥ 0 by definition. ¥

29

2.7 Decomposition of Matrices

2.7.1 Spectral Decomposition

2.7.2 Singular Value Decomposition

2.7.3 Cholesky Decomposition

2.7.4 Polar Decomposition

2.8 Matrix Norms

Similar to vectors, we frequently need a way of measuring the “size” of a matrix. An appropriatemeasure is called a “matrix norm”, defined as follows:

Definition 2.8.1. A matrix norm is any function ‖ · ‖ : Cn×n → IR that satisfies

(i) ‖A‖ ≥ 0 for all A ∈ Cn×n

(i’) ‖A‖ = 0 if and only if A = 0

(ii) ‖α A‖ = |α| ‖A‖ for all A ∈ Cn×nand α ∈ C(iii) ‖A + B‖ ≤ ‖A‖+ ‖B‖ for all A,B ∈ Cn×n

(iv) ‖AB‖ ≤ ‖A‖ ‖B‖ for all A,B ∈ Cn×n.

Here are some examples of matrix norms:

(i) Similar to the 1−norm for vectors in Cn, we define

‖A‖14=

n∑

i=1

n∑

j=1

|aij |.

That this expression satisfies conditions (i), (ii) and (iii) requires no detailed proof. As forcondition (iv),

‖AB‖1 =∑

i,j

∣∣∣∣∣∑

k

aikbkj

∣∣∣∣∣ ≤∑

i,j,k

|aikbkj | ≤∑

i,j,k,l

|aikblj |

=

∑

i,k

|aik|

∑

l,j

|blj | = ‖A‖1 ‖B‖1,

where the first inequality is obtained by the scalar triangle inequality, and the second one byadding positive terms to the sum.

(ii) The 2−norm is defined as

‖A‖24=

n∑

i=1

n∑

j=1

|aij |2

1/2

.

30

This norm is also called the Frobenius norm, denoted ‖A‖F . Note that it can be obtained as‖A‖2

F = trace(A∗A). That the Frobenius norm is a valid matrix norm can be verified by thefollowing:

‖AB‖2F =

∑

i,j

∣∣∣∣∣n∑

k=1

aikbkj

∣∣∣∣∣2

≤∑

i,j

(∑

k

|aik|2)(∑

m

|bmj |2)

=

∑

i,k

|aik|2

∑

m,j

|bmj |2 = ‖A‖2

F ‖B‖2F .

The inequality used above is nothing but the Cauchy-Schwarz inequality.

(iii) In the case of the ∞−norm, it can be shown that ‖A‖∞ 4= maxi,j |aij | is a vector norm on

Cn×n, but not a matrix norm. This is because the submultiplicativity condition ‖AB‖∞ ≤‖A‖∞‖B‖∞ is violated, for instance, in the case A = B = ( 1 1

1 1 ).

However, if we define a modified ∞−norm as

‖A‖∼∞ 4= n‖A‖∞ = n max

1≤i,j≤n|aij |,

then, ‖A‖∼∞ is an appropriate matrix norm. Because

‖AB‖∼∞ = n maxi,j

∣∣∣∣∣∑

k

aikbkj

∣∣∣∣∣ ≤ nmaxi,j

∑

k

|aikbkj |

≤ nmaxi,j

∑

k

‖A‖∞‖B‖∞

= n‖A‖∞ n‖B‖∞ = ‖A‖∼∞‖B‖∼∞

2.8.1 Induced Matrix Norms

In addition to matrix norms, we can define ”induced matrix norms”:

Definition 2.8.2. Let ‖ ·‖∗ be a vector norm on Cn. Then, the induced *-norm on Cn×n is definedas

‖A‖i∗4= max

x 6=0

‖Ax‖∗‖x‖∗ . (2.8.1)

In fact, it can be shown that

‖A‖i∗ = max‖x‖∗≤1

‖Ax‖∗ = max‖x‖∗=1

‖Ax‖∗.

Moreover, it is easy to show that induced matrix norms are legitimate matrix norms as defined inDefinition 2.8.1. Condition (i) is clearly satisfied since the induced norm is always nonnegative bydefinition and Ax = 0 for all nonzero x ∈ Cn if and only if A = 0. Condition (ii) is also satisfiedsince (let’s use the alternative form above for simplicity)

‖α A‖i∗ = max‖x‖∗=1

‖α Ax‖∗ = max‖x‖∗=1

|α| ‖Ax‖∗ = |α| max‖x‖∗=1

‖Ax‖∗ = |α| ‖A‖i∗.

31

Condition (iii) is also easy to show:

‖A + B‖i∗ = max‖x‖∗=1

‖(A + B)x‖∗ ≤ max‖x‖∗=1

(‖Ax‖∗ + ‖Bx‖∗)

≤ max‖x‖∗=1

‖Ax‖∗ + max‖x‖∗=1

‖Bx‖∗ = ‖A‖i∗ + ‖B‖i∗.

Lastly, condition (iv), namely the property of submultiplicativity, can be demonstrated as

‖AB‖i∗ = maxx6=0

‖ABx‖∗‖x‖∗ = max

x 6=0

‖ABx‖∗‖Bx‖∗

‖Bx‖∗‖x‖∗ ≤ max

y 6=0

‖Ay‖∗‖y‖∗ max

x 6=0

‖Bx‖∗‖x‖∗ = ‖A‖i∗‖B‖i∗.

It also follows from the definition of an induced norm that

‖Ax‖∗ ≤ ‖A‖i∗‖x‖∗ ∀x ∈ Cn.

Expectedly, the most commonly-used induced norms are the induced p-norms for p = 1, 2,∞:

‖A‖i14= max

x 6=0

‖Ax‖1

‖x‖1, ‖A‖i2

4= max

x6=0

‖Ax‖2

‖x‖2, and ‖A‖i∞

4= max

x6=0

‖Ax‖∞‖x‖∞ .

Simple characterizations for the induced 1−, 2− and ∞−norms are available:

Proposition 2.8.3. The induced 1−, 2− and ∞−norms can be calculated as follows:

(i) The induced 1−norm is the maximum column-sum. That is,

‖A‖i1 = maxj=1:n

n∑

i=1

|aij |.

(ii) The induced 2−norm, also called the spectral norm, is the square-root of the maximum eigen-value of A∗A. That is,

‖A‖i2 =√

λmax(A∗A).

(iii) The induced ∞−norm is the maximum row-sum. That is,

‖A‖i∞ = maxi=1:n

n∑

j=1

|aij |

Proof. We appeal directly to the definition of induced norms:

(i) Express A in terms of its columns as A =[

a1 · · · an

]. Then, ‖A‖i1 = maxi ‖ai‖1. It

follows that

‖Ax‖1 = ‖n∑

i=1

xiai‖1 ≤n∑

i=1

‖xiai‖1 =∑

i

|xi|‖ai‖1 ≤∑

i

|xi|maxi‖ai‖1 =

∑

i

|xi|‖A‖i1

= ‖x‖1‖A‖i1.

32

Hence,

maxx 6=0

‖Ax‖1

‖x‖1≤ ‖A‖i1.

On the other hand, if we denote the kth column of In by ek, we have

maxx6=0

‖Ax‖1

‖x‖1≥ ‖Aek‖1

‖ek‖1= ‖ak‖1 for all k = 1 : n.

Therefore,

‖A‖i1 = maxx6=0

‖Ax‖1

‖x‖1≥ max

k‖ak‖1 = max

k

∑

i

|aik|.

Since nonstrict inequality in both directions hold, we have

‖A‖i1 = maxj

n∑

i=1

|aij |.

(ii) This follows from the Rayleigh-Ritz inequality:

‖A‖i2 = maxx6=0

‖Ax‖2

‖x‖2= max

x 6=0

√x∗A∗Ax√

x∗x=

√maxx 6=0

x∗A∗Ax

x∗x=

√λmax(A∗A).

(iii) We prove nonstrict inequality in both directions again:

‖Ax‖∞ = maxi

∣∣∣∣∣∣∑

j

aijxj

∣∣∣∣∣∣≤ max

i

∑

j

|aijxj | ≤ maxi

∑

j

|aij |‖x‖∞.

Therefore,

‖A‖i∞ = maxx6=0

‖Ax‖∞‖x‖∞ ≤ max

i

∑

j

|aij |.

Now assume A 6= 0 (if A = 0, the statement is trivially true). Suppose the kth column of Ais nonzero and define z ∈ Cn as

zi =

aki

|aki| if aki 6= 0

1 if aki = 0.

Then, ‖z‖∞ = 1 and akjzj = |akj | for all k = 1 : n. Moreover,

maxx 6=0

‖Ax‖∞‖x‖∞ ≥ ‖Az‖∞ = max

i

∣∣∣∣∣∣∑

j

aijzj

∣∣∣∣∣∣≥

∣∣∣∣∣∣∑

j

akjzj

∣∣∣∣∣∣=

∑

j

|akj |

for any k = 1 : n. Hence,

‖A‖i∞ = maxx6=0

‖Ax‖∞‖x‖∞ ≥ max

k

∑

j

|akj |.

33

We conclude‖A‖i∞ = max

k

∑

j

|akj |.

¥

2.9 Ellipsoids

Definition 2.9.1 (Ellipsoid). Given a matrix P ∈ IRn×n, where P = P T ≥ 0, the set

EP4= {x ∈ IRn : xT Px ≤ 1} (2.9.1)

is called the ellipsoid associated with P .

The ellipsoid EP as defined above is centered at the origin of IRn. We can also define ellipsoidscentered at different points as

EP,xc

4= {x ∈ IRn : (x− xc)T P (x− xc) ≤ 1}

If P > 0, the volume of an ellipsoid is

vol EP = αn (detP−1)1/2, (2.9.2)

where αn is the volume of the n-dimensional unit 2-ball, i.e.,

αn =

πn/2

(n/2)!if n is even

2nπ(n−1)/2((n− 1)/2)!n!

if n is odd.

Proposition 2.9.2. Given two matrices A ≥ 0 and B ≥ 0,

EA ⊆ EB ⇐⇒ B ≤ A.

Proof. (⇐) Suppose B ≤ A and let x ∈ EA. Then, xT Bx ≤ xT Ax ≤ 1. Therefore, x ∈ EB, so thatEA ⊆ EB.(⇒) Suppose EA ⊆ EB but that B � A, i.e., there exists a nonzero x∗ such that xT∗Bx∗ > xT∗Ax∗.Now scale x∗ to x∗ so that1 xT∗Ax∗ = 1. Then, x∗ ∈ EA but x∗ /∈ EB since xT∗Bx∗ > xT∗Ax∗ = 1.This is a contradiction due to the assumption that EA ⊆ EB. ¥

Example 2.9.3. Consider two matrices

A =

1.3244 1.1302 0.93641.1302 1.6114 0.86430.9364 0.8643 0.8834

and B =

1.0400 1.0502 0.76661.0502 1.3019 0.93040.7666 0.9304 0.6726

.

It is easy to check that B ≤ A, in fact B < A. Thus, EA ⊆ EB. The inclusion is illustrated inFigure 2.2.

1We assume Ax∗ 6= 0. For if Ax∗ = 0, there are two possibilities: a) If A = 0, then the statement becomes trivial.b) If A 6= 0, we can perturb x∗ gently enough to make Ax∗ 6= 0 and still satisfy xT

∗Bx∗ > xT∗Ax∗.

34

−3−2

−10

12

3

−10−5

05

10

−15

−10

−5

0

5

10

15

x2x

1

x 3

Figure 2.2: The bigger ellipsoid is EB, the smaller one EA. Note: B ≤ A.

Remark 2.9.4. When P > 0 the definition Definition 2.9.1 can be thought of as the unit ball of thevector norm on Cn defined as

‖x‖ 4= (x∗Px)1/2 .

But when P ≥ 0, then one can find nonzero vectors x ∈ Cn such that ‖x‖ = 0, violating condition(i’) for being a vector norm in Definition 1.3.1. It is then called a semi-norm on Cn.

Lemma 2.9.5. Given P = P T =[

P1 P2

P T2 P3

]> 0, let E 4

={x : xT Px ≤ 1

}. Then, the projection

of E onto the x1-subspace is given by

Ex1 ={z ∈ IRn : zT (P1 − P2P

−13 P T

2 )z ≤ 1}

.

Proof. The projection of E onto the x1 subspace is given by

minQ

vol{z : zT Qz ≤ 1

}

s.t. x ∈ E ⇒ xT1 Qx1 ≤ 1.

The last condition above is equivalent to

{x : xT Px ≤ 1

} ⊆{

x : xT

[Q 00 0

]x ≤ 1

},

or by Proposition 2.9.2,[

Q 00 0

]≤

[P1 P2

P T2 P3

]⇐⇒

[Q− P1 −P2

−P T2 −P3

]< 0.

By the Schur complement formula, this is equivalent to

Q− (P1 − P2P

−13 P T

2

) ≤ 0.

The minimum volume is then given by Q = P1 − P2P−13 P T

2 . ¥

35

2.10 Cm×n as a Linear Vector Space

It is obvious that Cm×n, the set of m × n-dimensional complex matrices, associated with C, is alinear vector space. What’s more, we can define an inner product on Cm×n, and generalize all theresults about inner product spaces to Cm×n.

Theorem 2.10.1 (Inner product on Cm×n). The function < ·, · >: Cm×n ×Cm×n → C definedby

< X,Y >4= traceY ∗X

is an appropriate inner product on the Cm×n.

2.11 MATLAB Commands

(i) For computing the norm of a vector or a matrix, use the command norm(.). Which normyou want can be specified by additional arguments.

(ii) For matrices, scalar multiplication, matrix addition and matrix multiplication are executed asa*A, A+B and A*B, respectively. The matrix summation aI +A must be typed as a*eye(n)+A,where A is n×n. In matrix summations, if the dimensions don’t match, you will get an errormessage. The command a+A will not give you an error, but add a to each and every entry inthe matrix A. The transpose of A is computed by A’.

(iii) A basis for the range of A is given by range(A). A basis for the null space is given by null(A).Both bases are orthonormal.

(iv) Rank of A is rank(A), determinant of A is det(A).

(v) The inverse of a square A is inv(A).

(vi) The trace of A is trace(A).

(vii) The coefficients (in a decreasing order) of the characteristic polynomial of a square matrix isgiven by poly(A).

(viii) The eigenvalues and eigenvectors of A are computed from eig(A). A more reliable result isgiven by the Jordan decomposition, namely jordan(A).

2.12 Exercises

Problem 2.1. Let Φ ∈ IRn×n be upper-triangular with all diagonal elements φii = 0. Show thatΦn = 0. (A square matrix M is called nilpotent if Mk = 0 for some positive integer k.)

Problem 2.2. Prove (2.1.4) and (2.1.5).

Problem 2.3. True or false?

• Scalar matrices commute with all matrices.

• Diagonal matrices commute with all matrices.

36

• Diagonal matrices commute with diagonal matrices.

If true, prove. If false, give a counterexample.

Problem 2.4.

Problem 2.5. Prove Theorem 2.1.3.

Problem 2.6. Given a matrix A ∈ IRm×n with rank k, show that A can be expressed as

A = ALAR

for some matrices AL ∈ IRm×k and AR ∈ IRk×n, both of rank k.

Problem 2.7. For any two matrices A,B ∈ IRn×n, prove the following:

B(I + AB)−1 = (I + BA)−1B

assuming that the inverses exist.

Problem 2.8. Given A, B ∈ IRn×n, both invertible, prove the following:

rank([

A II B

])= n ⇐⇒ A = B−1.

Hint: Note that [A II B

]=

[I 0

A−1 I

] [A 00 B −A−1

] [I A−1

0 I

]

and use Sylvester’s rank inequality.

Problem 2.9. Show that for any A ∈ IRm×n,

trace(AT A) =m∑

i=1

n∑

j=1

a2ij .

The value√

trace(AT A) is called the Frobenius norm of matrix A, denoted ‖A‖F . It is a commonmeasure of the size of a matrix.

Problem 2.10. Show that similar matrices have the same trace. Do not refer to Lemma 2.2.11.

Problem 2.11. In showing that the eigenvalues of a real matrix are symmetric about the real axis,we used the fact

pA(λ∗) = pA(λ∗).

Prove it.

Problem 2.12. Given A ∈ IRn×n, show that

(i) If A is diagonal, its eigenvalues are its diagonal entries

(ii) If A is upper- or lower-triangular, its eigenvalues are its diagonal entries.

(iii) If λ ∈ σ(A), then, λ + α ∈ σ(αI + A) for any α ∈ IR

37

(iv) λ ∈ σ(A) ⇒ λ−1 ∈ σ(A−1) if A is invertible.

(v) A is singular if and only if 0 is an eigenvalue of A.

(vi) If λ ∈ σ(A), then, λk ∈ σ(Ak) for any integer k ≥ 1.

(vii) If A is nilpotent, i.e., there exists an integer k ≥ 1 such that Ak = 0, then, all eigenvalues ofA are equal to zero.

(viii) If A is unitary, then, |λ| = 1 for all λ ∈ σ(A).

(ix) If A is skew-symmetric, i.e., A = −AT , then Re(λ) = 0 ∀λ ∈ σ(A).

(x) Given a polynomial p(·), a matrix A ∈ IRn×n, an eigenvalue of A, λ and the correspondingeigenvector x, show that p(λ) is an eigenvalue of p(A) and the corresponding eigenvector isstill x.

Problem 2.13. Compute A17 for the following matrix by using the Cayley-Hamilton theorem andonly I3, A and A2.

A =

1 2 32 5 8

−3 6 −9

.

Hint: Use the command deconv in MATLAB.

Problem 2.14. Let A = A∗ be spectrally decomposed as A = UΛU∗. Obtain a singular valuedecomposition of A based on its spectral decomposition.

Problem 2.15. Show that the eigenvectors corresponding to distinct eigenvalues of a Hermitianmatrix are orthogonal.

Problem 2.16. Determine whether the statements below are true or false. If true, prove; if false,give a counter-example:

(i) If all eigenvalues of A are 0, then A = 0.

(ii) If A ≥ 0, then Aij ≥ 0 for all i, j = 1 : n.

(ii’) If A ≥ 0, then Aii ≥ 0 for all i = 1 : n.

(iii) If Aij ≥ 0 for all i, j = 1 : n, then A ≥ 0.

(iii’) If Aii ≥ 0 for all i = 1 : n, then A ≥ 0.

Problem 2.17. Show, using the definition, that given A = AT ∈ IRn×n and A > 0, all principalsubmatrices are also positive definite. (A principal submatrix is formed by deleting a number ofrows and columns from A symmetrically. That is, if the ith row is deleted, so is the ith column.)

Problem 2.18. Given any M ∈ IRm×n, show that

MT M ≥ 0 and MMT ≥ 0.

What are the conditions for MT M > 0 and MMT > 0 to hold?

38

Problem 2.19. Without using the Schur complement formula, show the following:[

a bb c

]> 0 if and only if

{c > 0

ac− b2 > 0

Problem 2.20. Given two symmetric matrices A, B, both in IRn×n, show that

λmax(A + B) ≤ λmax(A) + λmax(B)

andλmin(A + B) ≥ λmin(A) + λmin(B)

Hint: Rayleigh-Ritz inequality.

Problem 2.21. For a given positive definite P = P T ∈ IRn×n, show that

(detP )1/n ≤ trace(P )n

Hint: Geometric mean, arithmetic mean... Remember?

Problem 2.22. Let ‖ · ‖ denote any matrix norm on Cn×n. Show that ‖I‖ ≥ 1.

Problem 2.23. Let L =[

a bb c

], where a, b, c ∈ IR and a, c > 0. Show that

‖L‖i2 =12

[a + c +

√(a− c)2 + 4b2

].

Problem 2.24. (Computer) Write computer programs in MATLAB that plot

(i) The circle of radius r, centered at (xc, yc) for given r and (xc, yc).

(ii) The ellipse EP ={x ∈ IR2 : (x− xc)T P (x− xc) ≤ 1

}for a given P = P T ∈ IR2×2, P > 0 and

xc ∈ IR2.

(iii) The ellipsoid EP ={x ∈ IR3 : (x− xc)T P (x− xc) ≤ 1

}for a given P = P T ∈ IR3×3, P > 0

and xc ∈ IR3.

39

Part II

Analysis

40

Chapter 3

Linear State Space Descriptions

3.1 Basic Definitions

3.1.1 Linearity/Non-Linearity

3.1.2 Time-Variance-Time-Invariance

3.2 State-Space Descriptions in Continuous Time

In this class, the topic of interest are dynamical systems that can be represented by the systemequations

x(t) = Ax(t) + Bu(t) (3.2.1a)y(t) = Cx(t) + Du(t), (3.2.1b)

where x(t) ∈ IRn is the state vector, u(t) ∈ IRm is the input, and y(t) ∈ IRp is the output of thesystem. This representation of a dynamical system is the so-called state-space representation. Wedefine the “state” of a system as any property of the system which relates the input to the outputsuch that knowledge of the input time function for t ≥ t0 and state at time t = t0 completelydetermines a unique output for t ≥ t0.

The system given in (3.2.1) is a so-called linear time-invariant (LTI) system. Roughly speaking,a linear time-varying (LTV) system is one in the form

x(t) = A(t)x(t) + B(t)u(t) (3.2.2a)y(t) = C(t)x(t) + D(t)u(t), (3.2.2b)

that is, the system matrices themselves are functions of time1. A nonlinear system, on the otherhand, has the following general state-space representation:

x(t) = f((t, x(t), u(t)

)(3.2.3a)

y(t) = h(t, x(t), u(t)

). (3.2.3b)

1One can construct a system with time-varying system matrices whose input/output behavior is time-invariant.

41

Depending on explicit dependence on time in f and h, one can define nonlinear time-invariant ornonlinear time-varying systems.

The 1st-order vector differential equation x = Ax + Bu would not have attracted so muchattention had it not arisen naturally from scalar ordinary differential equations of an arbitraryorder. To see this, consider an nth-order ordinary differential equation:

dny(t)dtn

+ an−1dn−1y(t)dtn−1

+ · · ·+ a1dy(t)dt

+ a0y(t) = bu(t). (3.2.4)

(We assume an = 1 without loss of generality). Let us define

x14= y, x2

4=

dy

dt, · · · xn

4=

dn−1y

dtn−1.

Then, we have

x1 = x2, x2 = x3, · · · xn =dny

dtn.

Sincedny

dtn= −an−1

dn−1y

dtn−1− · · · − a1

dy

dt− a0y, we can express (3.2.4) equivalently as

x =

0 1 0 · · · 00 0 1 · · · 0...

......

. . ....

0 0 0 · · · 1−a0 −a1 −a2 · · · −an−1

x +

00...0b

u, (3.2.5)

where x4=

[x1 x2 · · · xn

]T .

Example 3.2.1. Consider a spring-mass-damper system with forcing u(t) on the mass.

k

c

m

-

-F (t)

x

Figure 3.1: SDOF spring-mass-damper system.

The differential equation of motion is:

mz + cz + kz = u(t),

where z denotes the position of the mass. By defining x1 = z and x2 = z, we can recast the systemequation as

x =[

0 1−k/m −c/m

]x +

[0

1/m

]u.

42

If we want to study the behavior of the position in particular, we can choose the output to be

y = x1 =[

1 0]x = Cx.

If, moreover, we are interested in the acceleration of the mass, we can take

y = x2 =[ −k/m −c/m

]x + [1/m]u = Cx + Du.

We can also express several coupled linear ordinary differential equations in the form (3.2.1).

Example 3.2.2. Suppose we have a 2-DOF spring-mass-damper system with forces f1(t) and f2(t)acting on the two masses shown below.

k1

c1

m1

-x1

c2

k2

m2

- F1

- F2

-x2

Figure 3.2: 2DOF spring-mass-damper system

The differential equations of motion are

m1x1 = −k1x1 − c1x1 − k2(x1 − x2)− c2(x1 − x2) + F1

m2x2 = k2(x1 − x2) + c2(x1 − x2) + F2,

where x1 and x2 denote the absolute positions of m1 and m2. By defining

x4=

x1

x2

x1

x2

and u

4=

[F1

F2

],

we obtain the vector differential equations of the system

x =

0 0 1 00 0 0 1

−(k1 + k2)/m1 k2/m1 −(c1 + c2)/m1 c2/m1

k2/m2 −k2/m2 c2/m1 −c2/m2

x +

0 00 0

1/m1 00 1/m2

u.

If we assume that the accelerations of the two masses are measured, we obtain the output

y =[ −(k1 + k2)/m1 k2/m1 −(c1 + c2)/m1 c2/m1

k2/m2 −k2/m2 c2/m1 −c2/m2

]x +

[1/m1 0

0 1/m2

]u.

43

3.2.1 Linearity/Nonlinearity and Time-variance/Time-invariance

We have been using the terms linearity/nonlinearity and time-variance/time-invariance withouthaving defined them. Let us now clarify what we mean.

In order to define linearity and nonlinearity in our context, we will view dynamical systems as“mappings” from the “input space” to the “output space”. Then, if we denote the relation betweenthe input, the system and the output as

y = S(u),

then we say the system is linear if for zero initial conditions,

S(αu1 + βu2) = αS(u1) + βS(u2) ∀u1, u2 ∈ U ,

where U denotes the input space, i.e., the set of possible inputs. In that sense, the systems in(3.2.1) and (3.2.2) are linear, yet (3.2.3) need not be.

Example 3.2.3. Consider the scalar system

x = ax + bu

with the output y = x. The output for x(0) = 0 can be computed as

y(t) =∫ t

0ea(t−τ)bu(τ) dτ.

Obviously, the output is linear in the input. Consider, now, the system

x = ax + bu2.

The solution, now, is

y(t) =∫ t

0ea(t−τ)bu2(τ) dτ.

Then, y is clearly not linear in u.

Time-variance/time-invariance is another issue. We say the system is time-invariant if the timeaxis can be shifted without altering the system. That is, if we can set τ = t+T for some T ∈ IR andstill obtain the same system description, then the system is time-invariant. This can be checked byfirst shifting the system T -seconds and then checking whether the output y(t + T ) resulting fromthe initial condition x0 at time t0 + T is equal to the original output y(t), resulting from the sameinitial condition at time t0.

Here is an example:

Example 3.2.4. Consider the system

dx/dt = x + u, x(0) = x0

If we set τ4= t + T , we obtain

dx/dτ = x + u, x(T ) = x0

44

which is a carbon-copy of the original system description, hence this system is time-invariant. Nowconsider

dx/dt = tx + u, x(0) = x0

When the time-axis is shifted, we obtain

dx/dτ = τx + u + Tx, x(T ) = x0

which clearly describes a different system and therefore, the system is time-varying. One canactually verify that the solution of the first time-invariant system is not a solution of the secondone.

Remark 3.2.5. Time-variations in the system matrices do not imply nonlinearity. As an example,consider the following scalar time-varying system:

x = a(t)x + b(t)uy = c(t)x + d(t)u

It can be verified by the so-called Leibniz rule that y(t) can be expressed as

y(t) = c(t)∫ t

0eR t

τ a(σ) dσb(τ)u(τ) dτ + d(t)u(t) (3.2.6)

for zero initial conditions. Obviously, the output is linear in the input u.

Let us stress once again, that the rest of the class deals with linear, time-invariant (LTI) systemsonly.

3.2.2 State Transformations

We can use different states in order to describe the same dynamical system. In fact, the statevector that may be used to describe a linear system is never unique. For instance, suppose we wantto obtain equations for system (3.2.1) using a new state vector

z4= Tx

for some invertible T ∈ IRn×n. We then have x = T−1z, so that x = T−1z. We then obtain

T−1z = AT−1z + Bu

y = CT−1z + Du.

By pre-multiplying the first equation by T , we obtain the equivalent system description

z = Az + Bu

y = Cz + Du,

whereA4= TAT−1, B

4= TB, C

4= CT−1 and D

4= D. (3.2.7)

45

Note that the direct feedthrough matrix D is unchanged in the two representations. In fact, we haveonly changed the internal description of the system, but not the essential input/output description.State transformations are also commonly called coordinate transformations.

In certain cases, state transformations can be given physical interpretations as shown in thenext example.

Example 3.2.6. Consider the 2DOF spring-mass-damper system in Example 3.2.2. Let us use therelative positions z1 and z2 in deriving the equations of motion, that is, z1 = x1 and z2 = x2− x1.

k1

c1

m1

-z1

c2

k2

m2

- F1

- F2

-z2

Figure 3.3: The system in Example 3.2.2 described in terms of relative positions.

We obtain

m1z1 = −k1z1 + k2z2 − c1z1 + c2z2 + F1

m2(z1 + z2) = −k2z2 − c2z2 + F2

Now define z4=

[z1 z2 z1 z2

]T . In state-space, we have

z =

0 0 1 00 0 0 1

−k1/m1 k2/m1 −c1/m1 c2/m1

k1/m1 −k2(1/m1 + 1/m2) c1/m1 −c2(1/m1 + 1/m2)

z +

0 00 0

1/m1 0−1/m1 1/m2

F

We could have obtained the same result by applying the similarity transformation

z =

1 0 0 0−1 1 0 0

0 0 1 00 0 −1 1

x.

3.3 Linearization

Consider now a nonlinear system represented by the differential equation

x = f(x), (3.3.1)

where f is some known, continuous function. We define an equilibrium point of (3.3.1) as a pointx0 ∈ IRn such that f(x0) = 0. We want to obtain a linear system x = Ax that approximates thebehavior of the original system around some given equilibrium point.

46

3.3.1 Linearization About an Equilibrium Point

Recall that for an analytic scalar function f(x), Taylor’s Theorem gives

f(x) =∞∑

k=0

f (k)(x0)k!

(x− x0)k. (3.3.2)

If we choose x0 = xeq to be an equilibrium point of the system x = f(x), we have f(xeq) = 0 and

x = f(x) = f ′(xeq)(x− xeq) +∞∑

k=2

f (k)(xeq)k!

(x− xeq)k.

For x’s close to xeq, the terms f (k)(xeq)k! (x − xeq)k with k ≥ 2 go to zero with increasing k, so they

are negligible compared to the term f ′(xeq)(x− xeq). Then,

x ∼= f ′(xeq)(x− xeq).

Define x4= x− xeq and A

4= f ′(xeq). Then, ˙x = x, so that

˙x ∼= Ax.

Example 3.3.1. Consider the one-dimensional nonlinear system

x = V −Ae−α/x = f(x).

An equilibrium point, xeq, of the system is computed from f(xeq) = 0, or equivalently,

V = Ae−α/xeq .

Solving for xeq, we obtainxeq = − α

ln(V/A).

Then, a linear approximation is obtained as

˙x = f ′(xeq)x,

where x4= x− xeq and

f ′(xeq) = −Aα

x2eq

e−α/xeq = −V α

x2eq

.

Example 3.3.2 (Boyd). Nonlinear systems that have very different actual behavior may have thesame linear approximations. Consider, for instance the systems

x = −x3 and z = z3.

The solutions of the nonlinear equations are

x(t) = (x−20 + 2t)−1/2 and z(t) = (z−2

0 − 2t)−1/2.

47

When we linearize the two systems about the common equilibrium point xeq = zeq = 0, we obtainthe same linear models, namely

˙x = 0 and ˙z = 0.

The solution, obviously, isx(t) = x0 and z(t) = z0 ∀t ≥ 0.

Hence, the linearized systems behave in the very same way. However, as the following figure demon-strates, the characteristics of the original nonlinear systems and their common linear approximationare wildly disparate.

0 10 20 30 40 50 60 70 80 90 1000

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

dx/dt=−x3

dz/dt=z3 dy/dt=0

Figure 3.4: Simulation results for x = −x3, z = z3 and the linear approximation y = 0. In allcases, the initial condition is 0.1.

We can now extend the idea to systems with control inputs. That is,

x = f(x, u).

Let xeq and ueq be such thatf(xeq, ueq) = 0.

Then, expanding f(x, u) about (xeq, ueq), we obtain

x = f(x, u) =[

∂f∂x

∂f∂u

](xeq,ueq)

[x− xeq

u− ueq

]+ h.o.t.

Or, by defining x4= x− xeq, u

4= u− ueq, A

4= ∂f

∂x

∣∣∣(xeq ,ueq)

and B4= ∂f

∂u

∣∣∣(xeq,ueq)

,

x ∼= Ax + Bu.

Example 3.3.3. Consider the following mechanical system with an adjustable damping coefficient:

48

m1

k2 cvar

m2

k1 6

6

6

xg

x2

x1

Figure 3.5:

3.3.2 Linearization about a Solution Trajectory

Suppose xsol, usol satisfyxsol = f(xsol, usol).

Then, the Taylor series expansion of f around (xsol, usol) gives

f(x, u) ∼= f(xsol, usol) + f ′(xsol, usol)[

x− xsol

u− usol

],

which leads to

x− xsol = f(x, u)− f(xsol, usol) ∼= fx(xsol, usol)(x− xsol)− fu(xsol, usol)(u− usol).

Defining x4= x − xsol, u

4= u − usol, A

4= fx(xsol, usol) and B

4= fu(xsol, usol), we obtain the linear

system˙x = Ax + Bu.

An interesting fact is that linearization of a time-invariant nonlinear system along a solutiontrajectory can lead to time-varying linear model.Example 3.3.4. Consider the system described by the equation

y + 3y + 2y = uy3.

When put into state-space form, we have

x =[

x2

−2x1 − 3x2 + x31u

]= f(x, u),

where x1 = y and x2 = y. It can be easily shown that ysol = e−t and usol = 0 is a solution of thesystem. When we linearize the system about this solution, we obtain

˙x =[

0 1−2 + 3ux2

1 −3

]

(xsol,usol)

x +[

0x3

1

]

(xsol,usol)

u

=[

0 1−2 −3

]x +

[0

e−3t

]u.

Note that even though there was no explicit time-dependence in the original nonlinear system, theresulting linear system is time-varying.

49

3.3.3 (Exact) Feedback Linearization*

Although we will not go into the details of exact linearization, the basic idea can be demonstratedby a simple example. Consider the differential equation of motion of an inverted pendulum:

Jθ −mgL/2 sin θ = T (t),

where T (t) is the torque applied to the pendulum. By defining T (t) = T (t) + mgL/2 sin θ, weobtain the linear differential equation

Jθ = T (t).

The manipulation above is a simple example of what is referred to as exact feedback linearization.The exact conditions for exact feedback linearizability and its generalizations will not be discussedhere, but they are one of the focal points of any class on nonlinear control theory. Reference [7] isan excellent source for the details of the exact linearization procedure.

3.3.4 Quasi-Linearization*

If f(0) = 0 and f is continuously differentiable, we can obtain a quasi-linear description of system(3.3.1) in the form

x = A(x)x.

The matrix-valued function A(x) can be obtained from the fact that

f(x) =(∫ 1

0

∂f

∂s

∣∣∣∣s=σx

dσ

)x

Note that this description is valid everywhere, independent of any equilibrium point.

Example 3.3.5. Consider the equation of a pendulum in state-space:

x =

[x2

− g

Lsinx1

], (3.3.3)

where x4=

[θθ

]and L is the length of the pendulum. The procedure mentioned above yields the

quasi-linear system

x =

0 1

− g

L

sinx1

x10

x

3.4 State-Space Descriptions in Discrete-Time*

Another useful way of describing dynamical systems is by investigating their behavior at discreteinstants in time. Such an approach results in “difference” equations, rather than “differential”equations. The general form of for a discrete-time dynamical system representation is:

x((k + 1)T

)= f

(kT, x(kT ), u(kT )

)

y(kT ) = h(kT, x(kT ), u(kT )

),

50

where T is the sampling period of the system and k stands for the sampling point. For ease ofnotation, it is much more preferable to use

xk+1 = f(k, xk, uk)y = h(k, xk, uk).

An LTI system in discrete time is described by

xk+1 = Axk + Buk

y = Cxk + Duk.

3.5 Exercises

Problem 3.1. Approximate equations of motion for a hot air balloon are given by

h = v

v = − 1τ2

v + σδθ

δθ = − 1τ1

δθ + u,

where

h : height of the balloon measured from equilibrium altitudev : vertical velocity of the balloon

δθ : difference from equilibrium temperature of air in the balloonu : (a variable proportional to) the heat added into the balloon

Put the system equations in state-space form.

Problem 3.2.

x =[

et 10 −2

]x +

[01

]u

y =[

0 1]x

Is the system time-varying?

Problem 3.3. Verify that the state transformation in Example 3.2.6 yields the same system equationsas those derived from Newton’s law for relative positions z1 and z2.

Problem 3.4. Consider system (3.2.1). What is the system you obtain after applying a time-varyingcoordinate transformation z = T (t)x, where T (t) is invertible for all t ≥ 0?

Problem 3.5. The linearized equations of a predator-prey model are fox-rabbit-carrot ecosystemare given by

r = arf + bc

f = afr

where r denotes the rabbit population, f denotes the fox population and c stands for the amountof carrot supply.

51

(i) What should the signs of ar, af and b be for a realistic model?

(ii) In what way does this model resemble that of a spring-mass model with force F applied tothe mass?

(iii) Now suppose we are environmentalists trying to keep the rabbit population at a constantlevel R by adjusting the amount of carrots in the area. Our adjustment will be based on ourobservation of the rabbit population. Cast the problem stated above as a control problem.

Problem 3.6. Verify (3.2.6) using Leibniz’ rule:

d

dt

∫ b(t)

a(t)f(t, s) ds =

∫ b(t)

a(t)

∂f(t, s)∂t

ds + f(t, b(t)

)b′(t)− f

(t, a(t)

)a′(t)

Problem 3.7. We define a system as a mapping between an input and an output. Yet, we alwaysrefer to an equation x = f(x) as a system, even though there is no explicit input or output. Discusshow one can modify the definition of a system to include x = f(x) and define “linearity” forx = f(x). Hint: Think of what “drives” the system, i.e., what initiates dynamics.

Problem 3.8. Consider an unforced pendulum with the dynamic equation of motion

θ +g

Lsin θ = 0.

Obtain an approximate linearization about θeq = 0 and simulate the response of the original systemand the linear approximation to different initial conditions using SIMULINK.

Problem 3.9. Consider the inverted pendulum equation

θ − g/L sin θ = T.

Linearize the system about θ∗ = π/4 and the corresponding T ∗ that keeps the system at that angle.

Problem 3.10 (Kailath). The dynamics of a communications satellite of mass m are given in termsof the state vector

x =[

r r θ θ φ φ]T

and the input vectoru =

[ur uθ uφ

]T.

The differential equations of motion are given in vector form as

x = f(x, u) =

r

rθ2 cos2 φ + rφ2 − k/r2 + ur/m

θ

−2rθ/r + 2θφ sinφ/ cosφ + uθ/(mr) cos φ

φ

−θ2 cosφ sinφ− 2rφ/r + uφ/(mr)

.

Verify that the solution

xsol(t) =[

r0 0 ω0t ω0 0 0]T and usol(·) = 0

52

where r30ω

20 = k (constant) is an equilibrium condition and that linearization of the system equations

about this solution yields

A =

0 1 0 0 0 03ω2

0 0 0 2ω0r0 0 00 0 0 1 0 00 −2ω0/r0 0 0 0 00 0 0 0 0 10 0 0 0 −ω2

0 0

and B =

0 0 01/m 0 0

0 0 00 1/(mr0) 00 0 00 0 1/(mr0)

.

53

Chapter 4

The Autonomous System

Before the solution of the vector differential equation x = Ax, we begin by considering the scalarequation

x = ax, x(0) = x0.

From elementary differential equations, the solution can be obtained as

x(t) = eatx0.

The equation of interest in this section is the vector differential equation

x = Ax, x(0) = x0 (4.0.1)

where A ∈ IRn×n and x0 ∈ IRn. Considering the semblance between the scalar and the vectorequations, one is tempted to conclude that the solution of (4.0.1) is the same as its scalar version,with matrix A replacing the scalar a. It turns out that this is true, yet the term eAt needs to bedefined. Before the definition of this so-called matrix exponential, consider the definition of thescalar exponential function: Given a real scalar x,

ex 4= 1 + x +

x2

2!+

x3

3!+ · · · =

∞∑

i=0

xi

i!. (4.0.2)

Here are some facts about ex:

(i) ea is well-defined for all a ∈ IR; i.e., the infinite summation in (4.0.2) converges.

(ii) ea > 0 for all a ∈ IR.

(iii) ddte

at = aeat for all a ∈ IR.

(iv) eae−a = 1.

(v) ea+b = eaeb

It turns out that some, but not all, of the properties of ex carry on to the matrix case.

54

4.1 The Matrix Exponential

Definition 4.1.1 (eA). Given a matrix A ∈ IRn×n, the matrix exponential of A is defined as

eA 4= I + A +

A2

2!+

A3

3!+ · · · =

∞∑

i=0

Ai

i!, (4.1.1)

where A0 4= I.

Now, similar to the facts about the scalar exponential function, we have the following regardingthe matrix exponential function:

Proposition 4.1.2. The following hold for all A ∈ IRn×n:

(i) eA is well-defined.

(ii) eA is nonsingular.

(iii) ddte

At = AeAt = eAtA.

(iv) eAe−A = I.

Proof. We verify statements (i) and (ii) only. The other two statements are a consequence ofProposition 4.1.3, which is proved independently next.

(i) Given A, such that ‖A‖ < ∞, (assume the matrix norm is such that ‖I‖ = 1 without loss ofgenerality)

∥∥eA∥∥ =

∥∥∥∥∥∞∑

i=0

Ai

i!

∥∥∥∥∥ ≤∞∑

i=0

∥∥∥∥Ai

i!

∥∥∥∥ =∞∑

i=0

∥∥Ai∥∥

i!≤

∞∑

i=0

‖A‖i

i!= e‖A‖ < ∞.

Therefore, eA is well-defined for any A ∈ Cn×n.

(ii) Differentiate the infinite summation term by term:

d

dteAt = A + A2t +

A3t2

2!+ · · · = A

(I + At +

A2t2

2!+ · · ·

)=

(I + At +

A2t2

2!+ · · ·

)A

= AeAt = eAtA.

¥

Although the matrix exponential is defined in a similar manner to the scalar exponential, someproperties we take for granted for scalars do not apply to matrices. For instance, unlike the scalarproperty ea+b = eaeb, the exponential eA+B is not equal to eAeB unless A and B commute.

Proposition 4.1.3. Given A,B ∈ IRn×n,

if AB = BA, then eA+B = eAeB.

55

Proof. The proof is a direct analog of the proof of the scalar version.

eA+B =∞∑

i=0

1i!

(A + B)i =∞∑

i=0

1i!

i∑

j=0

(i

j

)AjBi−j =

∞∑

i=0

i∑

j=0

1j!(i− j)!

AjBi−j

=∞∑

q=0

∞∑

p=0

1q!

1p!

AqBp =

∞∑

q=0

1q!

Aq

∞∑

p=0

1p!

Bp

= eAeB

¥

With Proposition 4.1.3, it is now easy to show that eAe−A = I. Since A and −A commute,eAe−A = eA−A = I. Therefore, for any A ∈ IRn×n, the matrix exponential eA is always nonsingular,with the inverse e−A. This proves items (ii) and (iv) of Proposition 4.1.2.Remark 4.1.4. The converse of the statement in Proposition 4.1.3 is not true in general. Forinstance, consider

A =

0 0 0 00 0 0 00 0 0 −2π0 0 2π 0

and B =

0 0 1 00 0 0 10 0 0 −2π0 0 2π 0

Then, A and B do not commute, yet eA+B = eAeB. You can see this for yourself in MATLABusing the command expm for computing matrix exponentials. (Can you actually prove this?)

4.1.1 Calculation of eAt:

Fortunately, we do not have to carry out the infinite summation in the definition of the matrixexponential to compute eAt. There are other ways of calculating eA which not only simplify thecomputational task involved, but also provide invaluable insight into the nature of the system.

(i) Inverse Laplace Transforms: We will now show that, in agreement with the scalar case,

eAt = L−1{(sI −A)−1

}. (4.1.2)

We only need to note that1

(sI −A)−1 = s−1

(I − A

s

)−1

= s−1

[I +

A

s+

(A

s

)2

+(

A

s

)3

+ · · ·]

=[I

s+

A

s2+

A2

s3+

A3

s4+ · · ·

]

Therefore,

L−1{(sI −A)−1

}= I + At +

A2

2!t2 +

A3

3!t3 + · · · = eAt.

Hence, the matrix exponential eAt can be computed through the inverse Laplace transformof (sI −A)−1. The matrix (sI −A)−1 is usually referred to as the resolvent matrix.

1Remember: For a scalar r < 1, 11−r

=P∞

i=0 ri. The same idea applies here for all s values for which (sI −A)−1

is well-defined.

56

(ii) Vandermonde Matrix: Recall, from the Cayley-Hamilton theorem, that for a given A ∈IRn×n and any k ≥ 0,

Ak =n−1∑

i=0

αiAi

for some αi ∈ IR, i = 1 : n. Then, the matrix exponential eAt can be expressed as

eAt =n−1∑

i=0

αi(t)Ai

for all t ≥ 0, for some scalar functions of time αi(t). In order to compute αi(t), we need todo some work. We first invite the reader to verify the following fact:

Proposition 4.1.5. If λ is an eigenvalue of A, then eλ is an eigenvalue of eA and thecorresponding eigenvector is unchanged.

Then, if λ ∈ σ(A) and x ∈ Cn is the corresponding eigenvector, we have

eAx = eλx

⇒n−1∑

i=0

αiAix = eλx

⇒n−1∑

i=0

αiλix = eλx

Then, (n−1∑

i=0

αiλi − eλ

)x = 0

But since x 6= 0, we obtain

eλ =n−1∑

i=0

αiλi.

When we repeat this at any time t, we get

eλt =n−1∑

i=0

αi(t)λi.

Now suppose A has n distinct eigenvalues. Then, the expression above gives n linear equationsin unknowns, which can be put in the form

1 λ1 · · · λn−11

1 λ2 · · · λn−12

......

. . ....

1 λn · · · λn−1n

α0(t)α1(t)

...αn−1(t)

=

eλ1t

eλ2t

...eλnt

.

57

Solving for αi(t)’s, we can construct eAt. The matrix on the left is a so-called Vandermondematrix. It is in a standard form encountered in polynomial curve fitting, etc.

On the other hand, if an eigenvalue is repeated q times, in order to obtain an independentequation, all we have to do is differentiate eλt =

∑αi(t)λi with respect to λ up to q−1 times.

We will thus obtain the additional q − 1 independent equations we need to solve for αi(t)’s.

(iii) Modal summation: Note that for the spectral decomposition A = MJM−1, the matrixexponential eAt becomes

eAt = MeJtM−1.

Now J is in the formJ = diag{J1, · · · , Jq},

where each Ji has dimensions ni × ni and is in the form

Ji =

λi 1 0 · · · 00 λi 1 · · · 0...

.... . . . . .

...

0 0 0. . . 1

0 0 0 · · · λi

= λiI + Ei, where Ei4=

0 1 0 · · · 00 0 1 · · · 0...

.... . . . . .

...

0 0 0. . . 1

0 0 0 · · · 0

.

Then, since λiI and Ei commute, we have

e(λiI+Ei)t = eλitIeEit = eλiteEit

since, clearly, eλitI = eλitI. Now, one can show that (see Problem 2.1)

eEit = I + Eit +E2

i

2!t2 + · · ·+ Eni−1

i

(ni − 1)!tn1−1 =

1 t t2

2 · · · tni−1

(ni−1)!

. . . . . . . . ....

.... . . . . . t2

2. . . t

0 · · · 1

.

Hence,eAt = M diag{eλ1teE1t, · · · , eλqteEqt}M−1.

If we partition the columns of M and the rows of N4= M−1 as

M =[

M1 M2 · · · Mq

], and N =

N1

N2...

Nq

,

we can express eAt as

eAt =q∑

i=1

eλitMieEitNi. (4.1.3)

58

But since eEit is a polynomial matrix in t of order ni − 1, this reduces to

eAt =q∑

i=1

eλitPi(t), (4.1.4)

where Pi(t) is polynomial in t, given by

Pi(t)4= Mie

EitNi.

Example 4.1.6. Consider the system

x =[

0 1−1 −2

]x

let us find the state transition matrix eAt using the methods described above.

(i) Inverse Laplace Transform: First compute the resolvent matrix.

(sI −A)−1 =[

s −11 s + 2

]−1

=1

s2 + 2s + 1

[s + 2 1−1 s

]

By taking inverse Laplace transforms, we obtain

eAt =[

(t + 1)e−t te−t

−te−t (1− t)e−t

].

(ii) Vandermonde Matrix: The eigenvalues of A are −1, repeated twice. Then, we are searchingfor α0(t) and α1(t) such that

e−t = α0(t) + α1(t)(−1) and te−t = α1(t).

The solution gives α0(t) = e−t + te−t and α1(t) = te−t. Hence,

eAt = α0(t)I + α1(t)A =[

(1 + t)e−t te−t

−te−t (1− t)e−t

].

(iii) Modal Summation: When put in the Jordan form, A becomes

A =[

1 1−1 0

] [ −1 10 −1

] [1 1

−1 0

]−1 4= MJM−1.

Therefore,

eAt = e−tMeEtM−1, where E4=

[0 10 0

].

Since

eEt = I + Et =[

1 t0 1

],

the matrix product gives

eAt = e−t

[1 + t t−t 1− t

].

59

Example 4.1.4 continued. We can now prove that for the A and B matrices given in Example 4.1.4,we have eA+B = eAeB even though AB 6= BA.

The eigenvalues of A are {0, 0,±2πj}. The eigenvalue-eigenvector decomposition of A is

A =

1 0 0 00 1 0 00 0 1 10 0 −j j

0 0 0 00 0 0 00 0 2πj 00 0 0 −2πj

1 0 0 00 1 0 00 0 1 10 0 −j j

−1

.

Therefore,

eA =

1 0 0 00 1 0 00 0 1 10 0 −j j

1 0 0 00 1 0 00 0 e2πj 00 0 0 e−2πj

1 0 0 00 1 0 00 0 1/2 1/2j0 0 1/2 −1/2j

=

1 0 0 00 1 0 00 0 e2πj+e−2πj

2 j(

e2πj−e−2πj

2

)

0 0 −j(

e2πj−e−2πj

2

)e2πj+e−2πj

2

=

1 0 0 00 1 0 00 0 cos 2π − sin 2π0 0 sin 2π cos 2π

=

1 0 0 00 1 0 00 0 1 00 0 0 1

The eigenvalues of B are {0, 0,±2πj}. The eigenvalue-eigenvector decomposition of B is

B =

1 0 −j j0 1 −1 −10 0 2π 2π0 0 −2πj 2πj

0 0 0 00 0 0 00 0 2πj 00 0 0 −2πj

1 0 −j j0 1 −1 −10 0 2π 2π0 0 −2πj 2πj

−1

.

Therefore,

eB =

1 0 −j j0 1 −1 −10 0 2π 2π0 0 −2πj 2πj

1 0 0 00 1 0 00 0 e2πj 00 0 0 e−2πj

1 0 0 −1/2π0 1 1/2π 00 0 1/4π j/4π0 0 1/4π −j/4π

=

1 0 12π

(e2πj−e−2πj

2j

)− 1

2π + 12π

(e2πj+e−2πj

2

)

0 1 12π − 1

2π

(e2πj+e−2πj

2

)12π

(e2πj−e−2πj

2j

)

0 0 e2πj+e−2πj

2 j(

e2πj−e−2πj

2

)

0 0 −j(

e2πj−e−2πj

2

)e2πj+e−2πj

2

60

=

1 0 12π sin 2π − 1

2π + 12π cos 2π

0 1 12π − 1

2π cos 2π 12π sin 2π

0 0 cos 2π − sin 2π0 0 sin 2π cos 2π

=

1 0 0 00 1 0 00 0 1 00 0 0 1

One can now easily show that

eA+B =

1 0 14π sin 4π − 1

4π + 14π cos 4π

0 1 14π − 1

4π cos 2π 14π sin 4π

0 0 cos 4π − sin 4π0 0 sin 4π cos 4π

=

1 0 0 00 1 0 00 0 1 00 0 0 1

Hence, even though AB 6= BA, we still have eA+B = eAeB. ¥

Remark 4.1.7. When combined with the solution x(t) = eAtx0, the solution becomes

x(t) =q∑

i=1

eλitMieEitNix0. (4.1.5)

Note that the solution of the system becomes a linear combination of the generalized eigenvectorsof A, which form the columns of Mi’s. When A is diagonalizable, we have an even simpler form,namely

x(t) =n∑

i=1

α(i)eλitxi, (4.1.6)

where xi’s are the eigenvectors of A and α(i)’s are the entries of M−1x0. In both cases, the normof x(t) goes to zero as t →∞ if and only if all eigenvalues of A have negative real parts. Moreover,if all eigenvalues have negative real parts, then the rate at which ‖x(t)‖ goes to zero is dictated bythe real part that is closest to the imaginary axis. this value, namely maxi Re(λi(A)) is called thedecay rate of the system.

Example 4.1.8. Consider an autonomous system described by

x =[

0 1−1 −c

]x,

where c > 0.

Remark 4.1.9. By choosing the initial conditions appropriately, one can excite one particular modein the system.

Example 4.1.10. Consider the following two-degree-of-freedom mechanical system:

61

6 m, Jx

k k

θµ

For small values of θ, the equations of motion for the system are

mx = −2kx and Jθ = −2kL2θ.

In the state-space form, we have

x =

0 1 0 0−2k/m 0 0 0

0 0 0 10 0 −2kL2/J 0

x,

where x4=

[x x θ θ

]T. Let us use the numerical values k = 1, m = 1, L = 1 and J = 1. One

can easily show that the eigenvalues of the system are:

σ(A) = {√

2k/mj,−√

2k/mj,√

2kL2/Jj,−√

2kL2/Jj}4= {aj,−aj, bj,−bj}

and the corresponding eigenvectors are

1aj00

,

1−aj00

,

001bj

,

001−bj

When we define M4=

[x1 x2 x3 x4

], then M−1 is easily calculated as

M−1 =

1/2 −j/(2a) 0 01/2 j/(2a) 0 00 0 1/2 −j/(2b)0 0 1/2 j/(2b)

.

Remark 4.1.11. (On obtaining A from eA)At this point, the reader may be tempted to ask the question “Is it possible to obtain A from the

expression for eA?”. In the scalar case, the answer is provided simply by the expression a = ln b,where b = ea. Note, however, that b has to be a positive scalar for the answer to be real-valued.The matrix case is much more involved.

Now consider the problem of computing the “natural logarithm” of B ∈ IRn×n.

62

4.2 Stability

Definition 4.2.1 (Stability). The system is said to be asymptotically stable if for all initialconditions x0,

‖x(t)‖ → 0 as t →∞.

Example 4.2.2. Several real-life examples of stable systems can be given. Here are some:

(i) Spring-mass-damper system

(ii) Automobile taking a turn ... Unstable when backing up.

Since we cannot verify the statement of the theorem by checking all possible initial conditionsone by one, we need equivalent, verifiable characterizations of stability. We give two below:

Theorem 4.2.3. A is stable if and only if all of its eigenvalues have negative real parts.

Definition 4.2.4 (Marginal stability). The system x = Ax is said to be marginally stable ifA has nonrepeated eigenvalues on the imaginary axis and the rest of the eigenvalues have negativereal parts.

Hence, we reserve the word unstable for system which are neither stable nor marginally stable.Put differently, the system x = Ax is unstable if it has at least one eigenvalue in the open right-halfplane, or repeated eigenvalues on the imaginary axis.

Theorem 4.2.5 (Lyapunov Equation). The system x = Ax is asymptotically stable if and onlyif there exists a solution P = P T > 0 to the equation

AT P + PA = −Q (4.2.1)

for any Q = QT > 0.

Proof. (i) ⇒ (ii) Assume A is stable. For an arbitrary Q = QT > 0, define

P4=

∫ ∞

0eAT tQeAt dt > 0.

Then,

AT P + PA =∫ ∞

0

(AT eAT tQeAt + eAT tQeAtA

)dt =

∫ ∞

0

d

dteAT tQeAt dt = eAT tQeAt

∣∣∣t=∞

t=0.

Now, since A is stable, limt→∞ eAt = 0. Therefore, AT P + PA = −Q < 0.(ii) ⇒ (i) Assume equation (4.2.1) holds. Let x ∈ Cn be an eigenvector of A associated witheigenvalue λ. Pre- and post-multiplication of (4.2.1) by x∗ and x gives

x∗AT Px + x∗PAx = −x∗Qx

⇒ λx∗Px + λx∗Px = −x∗Qx

⇒ (λ + λ)x∗Px = −x∗Qx

⇒ 2Re(λ)x∗Px = −x∗Qx

⇒ Re(λ) = − x∗Qx

2x∗Px< 0

Repeating the same procedure for all eigenvalues of A, we conclude that A is stable. ¥

63

4.3 The Autonomous System in Discrete-time

Now consider an autonomous, linear discrete-time system described by the difference equation

x(k + 1) = Adx(k), (4.3.1)

where x is an infinite sequence of vectors in IRn and x(k) denotes the kth entry in the sequence.Such a description can be obtained from the continuous-time system description as follows: Supposethe system x = Ax is sampled at intervals of T seconds. If we denote x(kT ) as x(k), we obtain

x(k + 1) = eAT x(k)4= Adx(k).

Asymptotic stability for discrete time-time systems is defined in much the same way as incontinuous-time systems:

Definition 4.3.1. (Stability in Discrete-time) The system described by (4.3.1) is said to be asymp-totically stable if for any initial condition x0,

‖x(k)‖ → 0 as k →∞.

Note that the state vector after N steps can be obtained as

x(k + N) = ANd x(k).

Hence, the stability condition for system (4.3.1) is the same as requiring

‖Akdx0‖ → 0 for any x0.

The relationship between the eigenvalues of Ad and stability of the discrete-time system x(k +1) = Adx(k) is provided by the following theorem:

Theorem 4.3.2. The discrete-time system (4.3.1) is stable if and only if all eigenvalues of Ad liein the open unit disc, i.e.,

|λ| < 1 ∀λ ∈ σ(Ad). (4.3.2)

Proof. Let Ad be spectrally decomposed as Ad = SJS−1. Then, Akd = SJkS−1. But since

Jk = diag{Jk1 , · · · , Jk

q },

it suffices to show that for each i = 1 : q, ‖Jki zi‖ → 0 as k → ∞ for arbitrary zi ∈ IRni . If Ji is

scalar, i.e., Ji = λi, then the result is obvious. If, however, Ji = λiIni + Ei, with ni > 1, we have

Jki = (λiIni + Ei)k =

k∑

j=0

(k

j

)λk

i Ek−ji =

k∑

j=k−ni+1

(k

j

)λk

i Ek−ji

Now since |λi| < 1, we have ‖Jki zi‖ → 0 as k →∞ for any zi. ¥

64

Theorem 4.3.3. All eigenvalues of Ad lie in the open unit disc if and only if for any arbitraryQ = QT > 0, there exists a matrix P = P T > 0 such that

ATd PAd − P = −Q. (4.3.3)

Proof. (i) ⇒ (ii) Suppose all eigenvalues of Ad lie in the open unit disc. Let Q = QT > 0 be given.Define

P4=

∞∑

k=0

(ATd )kQAk

d > 0.

Then,

ATd PAd − P =

∞∑

k=0

((AT

d )k+1QAk+1d − (AT

d )kQAkd

)= lim

q→∞(ATd )qQAq

d −Q.

But since all eigenvalues of Ad lie in the open unit disc, we have limq→∞(ATd )qQAq

d = 0, so thatAT

d PAd − P = −Q.(ii) ⇒ (i) Suppose

ATd PAd − P = −Q < 0

for some P > 0. Let λ be any eigenvalue of Ad, with the corresponding eigenvector x. Then, pre-and post-multiplication by x∗ and x gives

|λ|2x∗Px− x∗Px < 0.

Since P > 0 and x 6= 0, we can rewrite the inequality above as

|λ|2 <x∗Px

x∗Px= 1.

¥

4.4 Exercises

Problem 4.1. Let A = MJM−1 be the spectral decomposition of A. Show that

eAt = MeJtM−1.

Problem 4.2. Given an invertible A ∈ IRn×n, show that∫ t

0eAτdτ = A−1(eAt − I) = (eAt − I)A−1.

Problem 4.3. Claim: e2A = (eA)2. True or false? Justify your answer.

Problem 4.4. Consider the system x = Ax, where

A4=

−0.1005 1.0939 2.0428 4.4599−1.0880 −0.1444 5.9859 −3.0481−2.0510 −5.9709 −0.1387 1.9229−4.4575 3.0753 −1.8847 −0.1164

.

65

(a) Compute the transition matrix Z such that Zx(t) = x(t + 15).

(b) Compute the transition matrix Y such that Y x(t) = x(t− 20)

(c) Find x(0) such that x(10) =[

1 1 1 1]T .

Problem 4.5. Consider the system x = Ax, where A4=

[ −1 1−1 1

].

(a) Compute eAt (Hint: A is nilpotent).

(b) Suppose x1(0) = 1 and x2(1) = 2. Find x(2).

Problem 4.6. Consider the system

x =[

0 ω−ω 0

]x.

Such a system is called a harmonic oscillator.

(a) Find the eigenvalues and the solution x(t) in terms of x(0) = x0.

(b) Prove that the state trajectories describe circles, i.e., that ‖x(t)‖24= (x(t)T x(t))1/2 is constant.

(c) Show, also, that x(t) is perpendicular to x(t) at all times, so that the state trajectories arecircular (analogous to velocity being perpendicular to position in circular motion).

Problem 4.7. Consider the following system:

x =[ −1 + 3

2 cos2 t 1− 32 sin t cos t

−1− 32 sin t cos t −1 + 3

2 sin2 t

]x

(a) Show that the eigenvalues of A(t) have negative real parts for all t ≥ 0.

(b) Verify that x∗(t) = et/2

[ − cos tsin t

]satisfies the system equations.

(c) Is the system stable?

(d) What is the moral of the story?

Problem 4.8. Consider the systems

x =

2.2762 −2.7304 −0.96512.7976 −3.9151 −0.32861.5634 −1.4201 −1.3610

x and x =

0.5858 0.3848 −1.91610.4963 0.3260 −1.62340.5844 0.3839 −1.9118

x.

Find the eigenvalues for both dynamic matrices. Are the systems stable or not? Explain in termsof the solutions x(t) and the role of the positions of the eigenvalues.

Problem 4.9. Let A ∈ IRn×n be given. Show that Re(λ) < −α for all λ ∈ σ(A) if and only if thereexists a P > 0 such that

AT P + PA + 2αP < 0.

66

Problem 4.10. Consider a spring mass damper system with no input. Let x4=

[x1

x2

], where x1 is

the position of the mass and x2 is the velocity. The state-space equation is then given by

x =

[0 1

− k

m− c

m

]x.

(a) Plot the phase portrait of the system (x1 vs. x2) for x(0) =[

11

], c = 0 and

k

m= 0.5, 1, 2.

Plot them on the same graph to see the differences.

(b) Plot the phase portrait for the same initial conditions,k

m= 1 and

c

m= 0.5, 1, 2.

Use the initial command in MATLAB.

Problem 4.11. Show that the condition

AT PA− P < 0, P > 0

is equivalent toASAT − S < 0, S = P−1 > 0.

67

Chapter 5

The System with Inputs and Outputs

5.1 Solving the System Equations

We now consider the systemx = Ax + Buy = Cx + Du,

(5.1.1)

where x(t) ∈ IRn is the state vector, u(t) ∈ IRm is the input and y(t) ∈ IRp is the output.We can obtain an expression for the output in the Laplace domain or in the time domain:

(i) Laplace Domain Solution: Consider (5.1.1) and suppose x(0) = x0. Taking the Laplacetransform of the dynamics equation gives

sX(s)− x0 = AX(s) + BU(s). (5.1.2)

Solving for X(s), we have

X(s) = (sI −A)−1x0 + (sI −A)−1BU(s). (5.1.3)

When we plug this in the output equation, we obtain

Y (s) = C(sI −A)−1x0 +[C(sI −A)−1B + D

]U(s) (5.1.4)

For zero initial conditions, the direct relation between the input and the output is given by

Y (s) =[C(sI −A)−1B + D

]U(s)

4= G(s)U(s),

where G(s) is called the transfer function (or more explicitly, the transfer function matrix) ofthe system. We use the notation

G =[

A B

C D

](5.1.5)

to express a transfer function in terms of the system matrices. Since Y (s) = G(s)U(s) is alinear mapping, we conclude that system (5.1.1) is a linear system.

68

(ii) Time Domain Solution: Using (5.1.3), we can obtain the time-domain solution for x as

x(t) = eAtx0 +∫ t

0eA(t−τ)Bu(τ)dτ, (5.1.6)

since L−1{(sI −A)−1

}= eAt and multiplication in the s-domain corresponds to convolution

in the time-domain. Therefore,

y(t) = C

[eAtx0 +

∫ t

0eA(t−τ)Bu(τ)dτ

]+ Du(t). (5.1.7)

Just like in the Laplace-domain solution, when x0 = 0, the output becomes a linear functionin the input, implying that the system is linear.

Remark 5.1.1. Suppose u(t) = δ(t), where δ(t) denotes the unit impulse function. Then, L{δ(t)} =1 and for zero initial conditions, we obtain

Y (s) = C(sI −A)−1B + D

in the Laplace domain andy(t) = CeAtB 1(t) + Dδ(t)

in the time-domain, where 1(t) denotes the unit step function,

1(t) =

{1 for t ≥ 00 for t < 0.

The expressions for the output above are called the impulse response of the system.

Two fundamental concepts regarding dynamical systems are controllability and observability.Roughly speaking, controllability implies that any final state can be reached from any initial statearbitrarily fast, using some appropriate control input. Observability, on the other hand, implies thatthe complete time history of the state can be obtained from any output. A moment’s thought willreveal that controllability and observability are in fact dual concepts in the sense that controllabilitydeals with driving the state forward in time through the input and observability deals with followingthe state backward in time through the output. Luckily, the mathematics of the two are fairlysimilar also.

We now proceed to introduce the concepts of controllability and observability and the weakerconditions, stabilizability and detectability.

5.2 Controllability and Stabilizability

We now concentrate on the equationx = Ax + Bu (5.2.1)

only.

69

Definition 5.2.1 (Controllability). The system in (5.2.1) –or the pair (A,B)– is said to becontrollable if for any initial condition x0, final condition xf and time T , there exists a (piecewisecontinuous) input u : [0, T ] → IRm such that the solution of (5.1.1) satisfies x(T ) = xf . Otherwise,the system –or the pair (A,B)– is uncontrollable.

In words, the system is controllable if any point in IRn can be driven to any other point xf

arbitrarily fast. Similar to stability, controllability of a system cannot be verified by appealing tothe definition directly. Instead, we have the following characterizations of controllability.

Theorem 5.2.2. The following statements are equivalent:

(i) (A,B) is controllable.

(ii) WC(t)4=

∫ t

0eAτBBT eAT τ dτ > 0 for all t > 0.

(iii) rank(C) = n, where

C 4=

[B AB A2B · · · An−1B

].

(iv) rank([

A− λI B])

= n for all λ ∈ C.

(v) If x∗A = λx∗ for some nonzero x ∈ Cn, then x∗B 6= 0.

(vi) There exists a K ∈ IRm×n such that the eigenvalues of A + BK can be assigned arbitrarily(as long as they are symmetric about the real axis).

Proof. (ii) ⇒ (i) Given x0, xf and T > 0, suppose Wc(t) > 0 for all t ≥ 0. Then, Wc(t)−1 existsfor all t ≥ 0. Let

u(t) = BT eAT (T−t)Wc(T )−1(xf − eAT x0).

Then, plugging the control input above into (5.1.6), after routine manipulations, we obtain x(T ) =xf .

(i) ⇒ (ii) Suppose (A,B) is controllable, yet there exists a T > 0 such that WC(T ) is singular.Then, since the integrand eAtBBT eAT t ≥ 0 for all t, there exists a nonzero v ∈ IRn such that

vT eAtB = 0 ∀t ∈ [0, T ].

Now let x(T ) = xf = 0. Then, pre-multiplication of (5.1.6) by vT gives

0 = vT eAT x0.

By choosing x0 = e−AT v, we obtain vT v = 0, or equivalently, v = 0, which is a contradiction sincewe assumed v to be nonzero. Hence, (i) implies (ii).

(ii) ⇒ (iii) Suppose Wc(t) > 0 for all t ≥ 0, but rank(C) < n. Then, there exists a nonzerov ∈ IRn such that

vT AiB = 0 ∀i = 1 : n− 1.

70

But since eAt is a combination of (At)i by the Cayley-Hamilton Theorem, this implies vT AiB = 0for all i ≥ 0, and therefore

vT eAtB = 0 ∀t ≥ 0.

Hence, vT WC(t) = 0 for all t ≥ 0, which is a contradiction. Therefore, if WC(t) > 0 for all t ≥ 0,then rank(C) = n.

(iii) ⇒ (ii) Let rank(C) = n, but assume there exists a T > 0 such that WC(T ) is singular.Then, there exists a nonzero v ∈ IRn such that

vT eAtB = 0 ∀t ∈ [0, T ].

Taking time-derivatives of orders 0 up to n− 1, and evaluating them at t = 0, we have

vT AiB = 0 ∀i = 0 : n− 1,

or equivalently,vTC = 0.

Since this indicates that rank(C) < n, we arrive at a contradiction and conclude that (iii) implies(ii).

(iii) ⇒ (iv) Suppose rank(C) = n, but there exists a nonzero x ∈ Cn such that

x∗[

A− λI B]

= 0

for some λ ∈ C. Equivalently,x∗A = λx∗ and x∗B = 0.

But this implies

x∗C =[

x∗B x∗AB · · · x∗An−1B]

=[

x∗B λx∗B · · · λn−1x∗B]

= 0.

But since we assumed rank(C) = n, this is a contradiction. Hence, (iii) implies (iv).

(iv) ⇒ (v) Obvious from (iii) ⇒ (iv).

(v) ⇒ (iii) Assume (v) holds, but rank(C) 4= k < n. Then, it is shown in Section 5.5 that thereexists a coordinate transformation matrix T such that

TAT−1 =[

AC A12

0 AC

]and TB =

[BC

0

],

where AC ∈ IR(n−k)×(n−k). Now let λ1 and xC 6= 0 be such that

x∗CAC = λ1x

∗C.

Then, x∗TB = 0, where

x4=

[0

xC

]

71

is an eigenvector of TAT−1 corresponding to λ1. This implies that (TAT−1, TB) is uncontrollable,contradicting the initial assumption that (A,B) is controllable (Problem 5.12). Hence, (v) ⇒ (iii).

(vi) ⇒ (i) Suppose the eigenvalues of A + BK are freely assignable, but that the system isuncontrollable. Then, the decomposition in Section 5.5 leads to uncontrollable modes, which areunaffected by the control input. Then the eigenvalues cannot be assigned arbitrarily, leading to acontradiction. Therefore, (vi) ⇒ (i).

(i) ⇒ (vi) In the next chapter, we will explicitly construct the matrix K that places theeigenvalues of A + BK at the desired locations. ¥

Remark 5.2.3. Each one of the characterizations above have its strengths and weaknesses. Wediscuss some below:

(i) We are trying to verify condition (i). Obviously, the easiest way to check it is via (iii), sinceit involves a simple computation only. The others –except (v)– can be checked only througha sweep over infinitely many “things”, such as all λ ∈ C in (iv).

(ii) The condition Wc > 0 is essential in the construction of u(t) that steers x0 to xf . It is shownin the proof of theorem Theorem 5.2.2 in Appendix B that the control input needed to drivex0 to x(T ) = xf is given by

u(t) = BT eAT (T−t)Wc(T )−1(xf − eAT x0). (5.2.2)

From this construction of u(t), it is clear that the state transfer x0 → xf can be accomplishedin any specified time, no matter how small (with a price to pay, of course).

(iii) Consider condition (v). If x∗A = λx∗ for some x 6= 0, then λ is an eigenvalue of A (and so isλ) and x is a left-eigenvector of A. Now, if x∗B = 0, then the mode associated with λ is saidto be an uncontrollable mode of the system. The reason for this becomes clear by consideringthe modal summation expression for eAt and the solution for the state in the time domaingiven in (5.1.6). Apparently, the system is controllable if and only if it has no uncontrollablemodes. To see this more clearly, consider the following: Let A be decomposed as

A = T−1ΛT, where T4=

z∗1z∗2...

z∗n

and zi’s are the left-eigenvectors of A. Now consider

x = Ax + Bu = T−1ΛTx + Bu.

Now define x4= Tx. The system becomes

˙x = Λx + TBu.

72

For each xi, the solution is

xi(t) = eλitxi0 +∫ t

0eλi(t−τ)z∗i Bu(τ) dτ.

Now suppose z∗i B = 0 for some i. Then, the solution for xi(t) becomes

xi(t) = eλitxi0

and hence we do not have any control over how xi evolves in time. Therefore, the system isuncontrollable.

Conversely, if z∗i B 6= 0 for any left-eigenvector of A, one can always solve for a u such that

xif = eλiT xi0 +∫ T

0eλi(T−τ)z∗i Bu(τ) dτ. (5.2.3)

(iv) If A is stable, the limit limt→∞WC(t) is well-defined and called the controllability gramian ofthe system. Written explicitly,

LC4=

∫ ∞

0eAtBBT eAT t dt. (5.2.4)

If A is stable, one can show that (A,B) is controllable if and only if

ALC + LCAT + BBT = 0. (5.2.5)

Even if the system is uncontrollable, there are some states that we can reach through someappropriate control input and some that we cannot, no matter what we use. We define the set ofstates that can be reached from x0 as the reachable set associated with x0.

Definition 5.2.4 (Reachable Set). Given system (5.1.1), the reachable set from x0 is the setof states that can be reached at some time t starting from x(0) = x0, using some input u. That is,

Rx0

4=

{x ∈ IRn : x = eAT x0 +

∫ T

0eA(T−τ)Bu(τ) dτ for some T, u(·)

}. (5.2.6)

Clearly, when the system is controllable, Rx0 = IRn for any x0. But if the system is uncon-trollable, not every point is reachable, and we could be lucky enough to estimate the reachable setroughly, as demonstrated in the example below.

Example 5.2.5. Consider the following mechanical system: It is intuitively clear that if we startfrom zero initial conditions, we cannot find a control input u such that the positions and velocitieswill be anything other than x1 = −x2 and x1 = −x2 at any point in time. Hence, for instance, thestate

[1 1 0 0

]T is not in R0.

The reachable set is obviously determined by the nature of the system, which is reflected bythe system matrices A and B only. An exact characterization of the reachable set from the originis given in the following, simple result.

73

-u

m

-

m

¾-x1 x2

Figure 5.1: Uncontrollable mechanical system.

Proposition 5.2.6. R0 = Im(C).Proof. We prove the statement by showing the following:

(i) R0 ⊆ Im(C)(ii) Im(C) ⊆ Im(WC(t))

(iii) Im(WC(t)) ⊆ R0

(i) R0 ⊆ Im(C)Let x∗ ∈ R0. Then, there exists a T > 0 and u : [0, T ] → IRm such that

x∗ =∫ T

0eA(T−τ)Bu(τ) dτ.

By the Cayley-Hamilton theorem, eA(T−τ) =∑n−1

i=0 αi(T − τ)Ai for some scalar functions αi.Therefore,

x∗ =∫ T

0

(n−1∑

i=0

αi(T − τ)Ai

)Bu(τ) dτ =

n−1∑

i=0

AiB

∫ T

0αi(T − τ)u(τ) dτ

Now let

ci(T )4=

∫ T

0αi(T − τ)u(τ) dτ for i = 0 : n− 1.

Then,

x∗ =[

B AB · · · An−1B]

c0(T )c1(T )

...cn−1(T )

.

Hence, x∗ ∈ Im(C), so that R0 ⊆ Im(C).(ii) Im(C) ⊆ Im(WC(t))

Assume x1 ∈ Im(C), but x1 is not in Im(WC).

74

(iii) Im(WC(t)) ⊆ R0

Let x∗ ∈ Im(WC(t)). Then, there exists v ∈ IRn such that

x∗ =(∫ t

0eAτBBT eAT τ dτ

)v,

or equivalently,

x∗ =(∫ t

0eA(t−τ)BBT eAT (t−τ) dτ

)v.

Definingu(τ)

4= BT eAT (t−τ)v,

we obtain

x∗ =(∫ t

0eA(t−τ)Bu(τ) dτ

).

Hence, x0 ∈ R0 and therefore, Im(WC(t)) ⊆ R0.

The proof is complete. ¥

Hence, we can characterize the reachable set as the span of the linearly independent columnsof C. We now return to Example 5.2.5 and use Proposition 5.2.6.

Example 5.2.5 (Revisited). Consider the system in Figure 5.1 again. For simplicity, let’s assumem = 1, k = 1 and c = 1. Then, the state-space equation for the system is

x =

0 0 1 00 0 0 1

−1 1 −1 11 −1 1 −1

x +

00

−11

u.

The controllability matrix then becomes

C =

0 1 −2 20 −1 2 −21 −2 2 0

−1 2 −2 0

.

Now, rank(C) = 2 and we can take columns 1 and 4 as the linearly independent columns. Theimage of C is, then,

span

1−1

00

,

001

−1

=

x ∈ IR4 : x =

a−a

b−b

for some a, b ∈ IR

,

in agreement with intuition.

75

Remark 5.2.7. It is reasonable to ask how “big” a control force is needed to steer the origin (orany x0) to xf in time T . But in order to answer the question, we first have to quantify the “size”of the control input. We use the function norm

‖u‖ 4=(∫ T

0‖u(t)‖2

2 dt

)1/2

.

It can be shown that ‖u‖ is minimized by the control input

u(t) = BT eAT (T−t)WC(T )−1xf .

Furthermore, it is easy to show (Problem 5.14) that this u satisfies

‖u‖2 = xTf WC(T )−1xf .

Hence, when a bound ‖u‖ ≤ 1 is imposed, the set of reachable points from the origin in time Tbecomes the ellipsoid

EWC(T )−1 ={x ∈ IRn : xT WC(T )−1x ≤ 1

}.

One can now show (Problem 5.15) that if T2 ≥ T1, then∫ T2

0‖u(t)‖2

2 dt ≤∫ T1

0‖u(t)‖2

2 dt.

This means when the time limit for reaching xf is relaxed, the amount of control input necessaryto achieve the state transfer decreases. Obviously, the minimum control effort is obtained whenT →∞. Hence, when there is no limitation on how fast we want to move, the minimum norm forthe control input that drives the origin to xf is

∫ ∞

0‖u(t)‖2

2 dt = xTf L−1

C xf ,

where LC is the controllability gramian defined in (5.2.4).

A condition weaker than controllability is stabilizability.

Definition 5.2.8 (Stabilizability). The system in (5.2.1) –or the pair (A,B)– is said to bestabilizable if there exists a matrix K ∈ IRm×n such that A + BK is stable.

Equivalent characterizations of stabilizability are given in the following theorem:

Theorem 5.2.9. The following statements are equivalent:

(i) (A,B) is stabilizable.

(ii) If x∗A = λx∗ for x 6= 0 and Re(λ) ≥ 0, then x∗B 6= 0.

(iii) rank([

A− λI B])

= n for all Re(λ) ≥ 0.

(iv) There exists a K ∈ IRm×n such that A + BK is stable.

Hence, stabilizability implies that unstable modes are controllable. In fact, we could havedefined stabilizability as follows: The system is stabilizable if any x0 can be driven to the originas t → ∞. Clearly, controllability implies stabilizability, but not vice versa. This is why we saystabilizability is a “weaker” condition than controllability.

76

5.3 Observability and Detectability

Instead of the pair (A,B), our focus is now on (C, A) and the system

x = Ax (5.3.1a)y = Cx (5.3.1b)

The time-domain solution for the output is,

y(t) = CeAtx0. (5.3.2)

Observability is a problem of obtaining the initial condition x0 from the output, as made precisebelow:

Definition 5.3.1 (Observability). The system in (5.3.1) –or the pair (C, A)– is said to be ob-servable if for any time T > 0 and the output history y(t) over [0, T ], the initial condition x0 canbe obtained. Otherwise, the system –or the pair (C, A)– is unobservable.

Remark 5.3.2. A different, yet equivalent definition of observability concerns the system

x = Ax + Bu

y = Cx + Du.

The definition of observability for this system is as follows: The system above is said to beobservable if for any time T > 0 and the input and output histories u(t) and y(t) over [0, T ], theinitial condition x0 can be obtained.

Now it is easy to show that the two definitions are equivalent by noting the following: Theoutput for the controlled system is

y(t) = eAtx0 +∫ t

0CeA(t−τ)Bu(τ) dτ + Du(t).

Since we have the entire time-history for the input u, we can define

y(t)4= y(t)−

∫ t

0CeA(t−τ)Bu(τ) dτ −Du(t)

and work with the system (5.3.1) where the input is zero.

Similar to controllability, we need equivalent characterizations for observability.

Theorem 5.3.3. The following statements are equivalent:

(i) (C, A) is observable.

(ii) WO(t)4=

∫ t

0eAT τCT CeAτ dτ > 0 for all t > 0.

77

(iii) rank(O) = n, where

O 4=

CCACA2

...CAn−1

.

(iv) rank([

A− λIC

])= n for all λ ∈ C.

(v) If Ax = λx for some nonzero x ∈ Cn, then Cx 6= 0.

(vi) There exists an L ∈ Rn×p such that the eigenvalues of A + LC can be assigned arbitrarily (aslong as they are symmetric about the real axis).

(vii) (AT , CT ) is controllable.

Proof of Theorem 5.3.3. We will prove the equivalence of statements (i) and (iii) only. From this,the equivalence of observability of (C,A) and the controllability of (AT , CT ) follows. The restfollows from Theorem 5.2.2.

(iii) ⇒ (i) At t = 0, we have

y(0) = Cx0, y(0) = CAx0, · · · , y(n−1)(0) = CAn−1x0,

or equivalently,

y(0)y(0)

...y(n−1)(0)

=

CCA...

CAn−1

x0 = Ox0.

Since x0 ∈ IRn and rank(O) = n, there exists a unique solution x0 to the equation above.

(i) ⇒ (iii) Now assume (C, A) is observable, but rank(O) < n. Then, there exists a nonzerox0 ∈ Cn such that Ox0 = 0. In other words, CAix0 = 0 for all i = 0 : n − 1, or, by theCayley-Hamilton theorem, for all i ≥ 0.

Now the output corresponding to the initial condition x(0) = x0 is y(t) = CeAtx0 ≡ 0. Butsince x0 cannot be determined from y(·) = 0, the system is unobservable. Due to the apparentcontradiction, we conclude that observability of (C,A) implies rank(O) = n. ¥

Remark 5.3.4. The remarks about controllability we listed can be modified for the case of observ-ability by replacing A by AT and B by CT . Among others, we make two remarks:

(i) Similar to a point being reachable, we say a point x0 is observable if the output trajectoryof the system with the initial condition x0 is sufficient to determine the initial condition x0.The set of all observable states is obtained as Im(O).

78

(ii) If (C, A) is observable, the initial condition x0 that results in the output y(t) can be obtainedas

x0 = WO(T )−1

∫ T

0eAT tCT y(t) dt.

This is obtained by pre-multiplying (5.3.2) by eAT tCT to obtain

eAT tCT y(t) = eAT tCT CeAtx0.

Integration of the equation above from 0 to T yields∫ T

0eAT tCT y(t) dt = WO(T )x0.

(iii) Suppose A is diagonalizable. Then, A can be decomposed as A = SΛS−1. Now consider thefollowing:

y(t) = CeAtx0

= CSeΛtS−1x0

=n∑

i=1

Cvieλitxi0 , x

4= S−1x

Now assume Cvi = 0 for some i. Then, all information about the initial xi disappears andthe system is unobservable.

(iv) In line with the controllability gramian, we define the observability gramian as

LO4=

∫ ∞

0eAT tCT CeAt dt.

When A is stable, the pair (C, A) is observable if and only if

AT LO + LOA + CT C = 0.

(iv) The set of observable states is given by Im(O).

In the same vein as in controllability/stabilizability, we have the weaker condition, detectability:

Definition 5.3.5 (Detectability). The pair (C,A) is said to be detectable if there exists anL ∈ IRn×p such that A + LC is stable.

The characterization of detectability is parallel to that of stabilizability also.

Theorem 5.3.6. The following statements are equivalent:

(i) (C, A) is detectable.

79

(ii) rank([

A− λIC

])= n for all Re(λ) ≥ 0.

(iii) If Ax = λx for some nonzero x ∈ Cn and Re(λ) ≥ 0, then Cx 6= 0.

(iv) There exists an L ∈ IRn×p such that A + LC is stable.

(v) (AT , CT ) is stabilizable.

5.4 State-Space and Transfer Function Descriptions

5.4.1 State-Space to Transfer Functions

Given a systemx = Ax + Buy = Cx + Du,

obtaining a transfer function description of the system is straightforward. Simply, take the Laplacetransform of the first equation and solve for X(s), then plug it into the second. We obtain

Y (s) = [C(sI −A)−1B + D]U(s)4= G(s)U(s).

We have not assumed anything about the input/output dimensions. Hence, in the most generalcase, G(s) is a transfer matrix. Moreover, it is clear that G(s) is rational in s, that is, each entryin G(s) is a ratio of polynomials in s.

In SISO systems, the zeros and the poles of the system are defined as the roots of the polynomialsin the numerator and denominator of the transfer function, respectively. In the case of MIMOsystems, the definitions of poles and zeros are more complicated. Here, we will discuss poles only.The transfer function G(s) can be expressed as

G(s) =C [cofactor(sI −A)]T B

det(sI −A)+ D

=C [cofactor(sI −A)]T B + D det(sI −A)

det(sI −A)4=

N(s)d(s)

With these definitions, the order of N(s) is at most n (when is it equal to n?) and the order of d(s)is n. We say G(s) is

(i) Proper iforder(N(s)) ≤ order(d(s))

(ii) Strictly proper iforder(N(s)) < order(d(s)).

As such, the eigenvalues of A are the roots of d(s). But there is the crucial fact that a cancellationis possible between N(s) and d(s). That is, we may reduce the transfer function to

G(s) =N(s)d(s)

=N(s)d(s)

,

80

where the order of d(s) is strictly less than n. When all such cancellations are carried out, the polesof G(s) are the roots of d(s), not those of d(s).

Hence, every pole of G(s) is an eigenvalue of A but not every eigenvalue of A is a pole of G(s).In fact, we will see very soon that the ones that disappear are those corresponding to uncontrollableand/or unobservable modes. The converse is also true: The uncontrollable and/or unobservablemodes of the state-space system disappear in the transfer function description.

Multivariable Poles and Zeros

5.4.2 Transfer Functions to State-Space

We begin by the definition of a realization.

Definition 5.4.1. Given a proper, rational transfer function G(s), the state-space system

x = Ax + Bu

y = Cx + Du

is said to be a realization (or a state-space realization) of G if

C(sI −A)−1B + D = G(s).

It is easy to show that the realization of a transfer function is never unique. That is, differentstate-space systems may have the same transfer functions. For instance, note that transfer functionsremain unchanged under coordinate transformations. That is, the systems

x = Ax + Bu

y = Cx + Duand

x = TAT−1x + TBu

y = CT−1x + Du

have the same transfer functions, namely G(s) = C(sI −A)−1B + D.But coordinate transformations are not the only reason for the nonuniqueness of state-space

realizations. State-space systems with different dynamic orders can have the same transfer functionsalso.

Example 5.4.2. Consider the system

x =[ −1 1

0 −2

]x +

[01

]u

y =[

0 1]x.

The transfer function can be easily computed to be

G(s) =1

s + 2.

But among others, the system

x = −2x + u

y = x

also has the same transfer function.

81

At this point, it is worth noting that the second-order system above is controllable, but notobservable. We will soon see that we can describe the I/O behavior of a state-space function byanother one of smaller dimensions if and only if it is uncontrollable and/or unobservable. We willcall a realization of G(s) that is controllable and observable a minimal realization.

Suppose the p×m transfer function matrix from u to y is in the irreducible1 form

G(s) =N(s)d(s)

.

where d(s) is the least common multiple of the denominators of the entries of G(s), specified as

d(s) = sn + an−1sn−1 + · · ·+ a1s + a0

andN(s) = Nn−1s

n−1 + Nn−2sn−2 + · · ·+ N1s + N0.

There are two basic state-space forms, namely the controllable canonical and observable canon-ical forms, that result in the transfer function G(s).

Controllable Canonical Form

x =

0 Im 0 · · · 00 0 Im · · · 0...

......

. . ....

0 0 0 · · · Im

−a0Im −a1Im −a2Im · · · −an−1Im

x +

00...0

Im

u (5.4.1a)

y =[

N0 N1 N2 · · · Nn−1

]x (5.4.1b)

Observable Canonical Form

x =

0 0 · · · 0 −a0Ip

Ip 0 · · · 0 −a1Ip

0 Ip · · · 0 −a2Ip...

.... . .

......

0 0 · · · Ip −an−1Ip

x +

N0

N1

N2...

Nn−1

u (5.4.2a)

y =[

0 0 · · · 0 Ip

]x (5.4.2b)

1That is, no further cancellations between N(s) and d(s) are possible.

82

Remark 5.4.3. We will see in the next chapter that it is important to be able to put any givenstate-space system into the controllable or observable canonical form via an appropriate coordinatetransformation. We will study the two cases separately:

(i) Transformation into the controllable canonical form: Possible if and only if (A,B) is control-lable. We can use x = Tx, where T = (CM)−1, where

C 4=

[b Ab · · · An−1b

],

where b is a nontrivial linear combination of the columns of B such that rank(C) = n and

M4=

a1 a2 · · · an−1 1a2 a3 · · · 1 0...

......

an−1 1...

1 0 · · · · · · 0

,

where the characteristic polynomial of A is

pA(λ) = λn + an−1λn−1 + · · ·+ a1λ + a0.

(ii) Transformation into the observable canonical form: Possible if and only if (C,A) is observ-able. The necessary transformation can be obtained by noting the equivalence of the observ-ability of (C, A) and the controllability of (AT , CT ) (Problem 5.21).

5.5 Canonical Decomposition

We may be interested in decomposing a given state-space system into different parts that are

(i) Controllable and observable

(ii) Uncontrollable and observable

(iii) Controllable and unobservable

(iv) Uncontrollable and unobservable

In order to achieve this decomposition, we first discuss decomposition of a given state-spacesystem into (i) controllable and uncontrollable parts and (ii) observable and unobservable parts.We do this in two theorems:

Theorem 5.5.1. Given (5.1.1), assume rank(C) = kC ≤ n. Then, there exists a coordinatetransformation

x4= TCx =

[xC

xC

]

83

such that[ ˙xC

˙xC

]=

[AC A12

0 AC

]x +

[BC

0

]u (5.5.1a)

y =[

CC CC

]x + Du, (5.5.1b)

where AC ∈ IRkC×kC and (AC , BC) is controllable. Moreover,

G(s) = C(sI −A)−1B + D = CC(sI − AC)−1BC + D.

Remark 5.5.2. The theorem is begging for interpretations:

(i) When kC = n, the system is controllable. Then, AC simply does not exist.

(ii) When kC < n, the modes of A are the modes of AC combined with those of AC . The modesof AC are clearly uncontrollable, since they are unaffected by the control input u. Dependingon whether AC is stable or not, the system may or may not be detectable. Nothing can besaid about observability and detectability, however.

(iii) The coordinate transformation matrix TC is determined as follows: Let q1, · · · , qkCbe the

linearly independent columns of C. Now choose n− kC vectors qkC+1, · · · , qn such that

Q4=

[q1 · · · qkC

qkC+1 · · · qn

]

is invertible. Then, the required transformation matrix can be given by TC = Q−1.

(iv) When kC < n, i.e., the system is uncontrollable, there exists a state-space system of smallerdynamic order that gives the same transfer function. That is,

G =[

A B

C D

]=

[AC BC

CC D

].

Now the decomposition into observable/unobservable parts:

Theorem 5.5.3. Given (5.1.1), assume rank(O) = kO ≤ n. Then, there exists a coordinatetransformation

x4= TOx =

[xO

xO

]

such that[ ˙xO

˙xO

]=

[AO 0A21 AO

]x +

[BO

BO

]u (5.5.2a)

y =[

CO 0]x + Du, (5.5.2b)

where AO ∈ IRkO×kO and (CO, AO) is observable. Moreover,

G(s) = C(sI −A)−1B + D = CO(sI − AO)−1BO + D.

84

When the two theorems above are combined, we obtain the promised decomposition into fourdifferent parts:

Theorem 5.5.4. Given (5.1.1), there exists a coordinate transformation x = Tx such that

˙xCO˙xCO˙xCO˙xCO

=

ACO 0 A13 0A21 ACO A23 A24

0 0 ACO 00 0 A43 ACO

xCO

xCO

xCO

xCO

+

BCO

BCO

00

u (5.5.3a)

y =[

CCO 0 CCO 0]

xCO

xCO

xCO

xCO

+ Du (5.5.3b)

where ([ACO 0A21 ACO

],

[BCO

BCO

])is controllable

and ([CCO CCO

],

[ACO A13

0 ACO

])is observable.

Moreover,G(s) = C(sI −A)−1B + D = CCO(sI − ACO)−1BCO + D.

5.5.1 Balanced Realizations

5.6 The Input/Output System in Discrete-time

Now that we know how to solve x = Ax + Bu we can convert this continuous-time system to adiscrete-time system, described by the corresponding difference equation

x(k + 1) = Adx(k) + Bdu(k).

Assume that we are sampling the continuous-time system every T seconds. Assume furtherthat between samples numbered k and k + 1, the control input u is constant, i.e., u(t) = u(k)∀t ∈ [kT, (k + 1)T ).

From the solution of x = Ax + Bu, we have

x[(k + 1)T ] = eAT x(kT ) +∫ (k+1)T

kTeA((k+1)T−τ)Bu(τ)dτ.

(From now on we drop T dependence in x for the sake of simplicity.) Since u(t) is assumed to beconstant between kT and (k + 1)T , we can take it out of the integral sign as u(k). This gives

x(k + 1) = eAT x(k) +

(∫ (k+1)T

kTeA((k+1)T−τ)dτ

)Bu(k).

85

For the simplification of the integral, let us use a new variable α4= (k + 1)T − τ . It follows that

dα = −dτ and {τ = (k + 1)T ⇒ α = 0

τ = kT ⇒ α = T.

Therefore,

x(k + 1) = eAT x(k) +(∫ 0

TeAα(−dα)

)Bu(k),

or equivalently

x(k + 1) = eAT x(k) +(∫ T

0eAαdα

)Bu(k).

Thus, x(k + 1) = Adx(k) + Bdu(k), where Ad4= eAt and Bd

4=

(∫ T0 eAαdα

)B.

5.7 MATLAB Commands

(i) The transformations between state-space and transfer function descriptions are done by thecommands

ss2tf(A,B,C,D) and tf2ss(num,den)

5.8 Exercises

Problem 5.1. Consider the systems

G1 =[

A1 B1

C1 D1

]and G2 =

[A2 B2

C2 D2

].

Assuming that the input and output dimensions are compatible in all of the following cases, find(a) the equivalent transfer function in terms of G1(s) and G2(s) and (b) the compound systemmatrices when G1 and G2 are

(a) Connected in series:

G2 G1- - -

(b) Connected in parallel:

G1

G2

-?6

-

-

++

(c) Connected in a feedback loop:

86

G1

G2

-

¾

-6

-+−

Discuss how the eigenvalues of the “augmented” system are related to those of A1 and A2.

Problem 5.2. Given a two-input system

G =[

A B

C D

],

suppose the two inputs have actuator dynamics (i.e., from input commands v1 and v2 to actualcontrol inputs to the plant u1 and u2.)

C1 =[

AC1 BC1

CC1 DC1

]and C2 =

[AC2 BC2

CC2 DC2

].

Find expressions for the combined system from the input commands v1 and v2 to the output y.

Problem 5.3. Given the system

G =[

A B

C D

]

Find a state-space description for the inverse of the system above, assuming m = p (i.e., the systemis square) and D is invertible. Hint: Work with the transfer function from u to y.

Problem 5.4. For a system

G =[

A B

C D

],

we define the conjugate2 system of G as

G∼(s)4= GT (−s) =

[ −AT −CT

BT DT

]=

[ −AT CT

−BT DT

].

(i) Find a condition so that G∼(s)G(s) is always invertible.

(ii) Assuming the condition in (i) is satisfied, find a state-space representation for the left-inverse

of G, that is, G−L 4= (G∼G)−1G∼.

(iii) Is the realization for G−L minimal? If not, find a minimal realization.

(iv) Repeat for G(s)G∼(s) and a right-inverse.

Problem 5.5. Verify that the expression in (5.1.6) is a solution of (5.1.1). You need to show twothings: (i) x(0) = x0 and (ii) x = Ax + Bu.Hint: Leibnitz’ Rule: If

I(α) =∫ h(α)

g(α)f(s, α) ds,

2Note that the conjugate is not defined as G∗(s). Why not?

87

then,dI

dα= f(s, α)|s=h(α)

dh

dα− f(s, α)|s=g(α)

dg

dα+

∫ h(α)

g(α)

∂f

∂α(s, α) ds

Problem 5.6. Consider the system

x =[

0 12 1

]x +

[01

]u, x0 =

[34

].

Compute x(t) when u(t) = 0 and u(t) = 2.

Problem 5.7. Verify that the expression for the control input u in (5.2.2) does drive x0 to xf intime T .

Problem 5.8. Show that the control input that drives the state x0 to xf in T seconds is the sameas that which drives the origin to the final state xf − eAT x0 in T seconds.

Problem 5.9. Solve for u(t) to be applied for t ∈ [0, T ] so that (5.2.3) holds.

Problem 5.10. Consider the following linear model of an aircraft

x =

−10 0 −10 00 −0.7 9 00 −1 −0.7 01 0 0 0

x +

20 2.80 −3.130 00 0

[u1

u2

]

where the state variables are the so-called roll and yaw rates and slip and roll angles and the controlinputs u1 and u2 are aileron and rudder angles, respectively. Answer the following:

(i) Is the aircraft controllable?

(ii) Is the aircraft controllable when the rudder fails (u2 = 0)?

(iii) Is the aircraft controllable when the aileron fails (u1 = 0)?

Problem 5.11. The answers to the previous problem should turn out like this: (i) Yes, and one of(ii) and (iii) Yes. This brings us to the following question: If the system is controllable with onlyone input, why use two? Discuss in terms of the control input required to drive the origin to somefinal state, xf . Hint: Given invertible, symmetric matrices A and B,

A ≥ B ⇐⇒ A−1 ≤ B−1,

where A ≥ B means A−B ≥ 0.

Problem 5.12. Assume (A,B) is controllable.

(i) Show that (TAT−1, TB) is controllable for any non-singular T .

(ii) Show that (A + BF, B) for any F of compatible dimensions. (It follows, of course, that(A + BBT , B) is also controllable.)

(iii) Show that (A + αI, B) is controllable. What about stabilizability?

88

Hint: Use rank(C) = n for (i) and the x∗A = λx∗ ⇒ x∗B 6= 0 argument for (ii) and (iii). In (ii),assume (A+Bf, B) is not controllable first, even though (A,B) is. This will lead to a contradiction.

Problem 5.13. Compute the controllability gramian for the system

x =[

0 1−1 −1

]x +

[10

]u.

Find the smallest possible norm of the control input that can drive the origin to the state xf =[11

].

Problem 5.14. Prove that the control needed to drive the origin to the state xf in T seconds satisfies

∫ T

0‖u(t)‖2

2 dt = xTf Wc(T )−1xf .

Problem 5.15. Show that if T2 ≥ T1, then, for any x ∈ IRn,

xT WC(T2)−1x ≤ xT WC(T1)−1x.

Problem 5.16. Consider the system

x =

0 2 −13 0 10 0 2

x

Check whether the system observable when

(i) y =[

0 −2 10 0 1

].

(ii) y = x2. What is the subspace of observable states?

(iii) y = x1. What is the subspace of observable states?

Problem 5.17. Consider a cart-and-pendulum system:

mc

m-x

θ

-F

L

The equations of motion of the system are:

mL2θ −mgL sin θ + mL cos θx = 0

(mC + m)x + mL cos θθ −mL sin θθ2 = F.

A similar system is the one where the cart is restrained with a spring:

89

mck

θ

-x

-F

m

L

(i) What are the equations of motion for this system?

(ii) Linearize both sets of nonlinear equations and obtain linear state-space models for the twosystems.

(iii) Check controllability of the two models.

(iv) It is proposed that we use the outputy = θ.

Are the systems observable?

(v) How would you modify the output so as to obtain an observable system?

Problem 5.18. Find state-space descriptions for the systems below:

(i)1s

(ii)1s2

(iii)ω

s2 + ω2

(iv)s

s2 + ω2

Problem 5.19. Find state-space descriptions for the systems below:

(i) G(s) =[

1s

s + 2(s + 1)2

]

(ii) G(s) =

s + 1s + 2

s + 2(s + 1)2

Problem 5.20. Consider a differential equation

y′′′ + a2y′′ + a1y

′ + a0y = b2u′′ + b1u

′ + b0u.

Obtain a state-space description for the system.Problem 5.21. Given an observable state-space system, find the necessary coordinate transformationmatrix that is required to put it in the observable canonical form.Problem 5.22. Prove that the transfer function obtained from the state-space representation (5.1.1)will always be proper and strictly proper if D = 0. Can a nonproper rational function be expressedas (5.1.1)?

90

Part III

Synthesis

91

Chapter 6

State Feedback Control Design

In this chapter, we discuss the problem of stabilization via feedback control. Our focal point isthe determination of feedback control laws in the cases where the measured output is (i) the entirestate vector x, i.e., y = x and (ii) a linear combination of some of the states and the control input.Feedback control in the special case y = x is called full-state feedback (FSF) control and the generalcase is given the generic name output feedback (OF) control.

We distinguish between static and dynamic feedback control laws as follows: A static feedbackcontrol law is one in the form

u = Ky,

whereas a dynamic control law is in the form

xc = Acxc + Bcy

u = Ccxc + Dcy,

where we do not make any specification for y. In both cases, the control law can be represented asa block as

¾y

¾ C(s)u

where C(s) = K in the case of a static control law and C(s) = Cc(sI − Ac)−1Bc + Dc in the caseof a dynamic control law. The feedback control problem then becomes one of finding a matrix K,or matrices Ac, Bc, Cc, Dc, such that the closed-loop system is stable.

For reasons that will become clear later on, there is no need to design a dynamic controller wheny = x. That is, –except for very special cases which are beyond the scope of this class– whatevercan be done with a dynamic full-state feedback controller can also be done by a static one. On theother hand, even though static output feedback control is much easier to implement, determinationof a static output feedback gain matrix is known to be an extremely difficult problem to give areasonable, general solution. Therefore, we focus on two cases here:

(i) Static full-state feedback control1

1From this point on, we will drop the word “static” and just say full-state feedback control.

92

(ii) Dynamic output feedback (DOF) control.

In both cases, we will first give traditional design methods (namely exact pole placement inFSF and observer design in DOF) and then discuss how the same problems can be solved by linearmatrix inequalities (LMI’s).

6.1 Full-State Feedback Control

We now assume C = I and D = 0, so that y = x. A full-state feedback control law is a relation ofthe form

u = Kx,

where K ∈ IRm×n is the so-called “gain matrix”. The full-state feedback stabilization problem isone of finding an appropriate K matrix such that the closed-loop system

x = (A + BK)x

is stable.

6.1.1 Exact Pole Placement

Suppose we want to choose the full state feedback gain K so that the closed-loop eigenvalues(i.e., the eigenvalues of A + BK) are at the prescribed values µ1, µ1, . . . , µn. That is, the desiredcharacteristic polynomial is

pA+BK(λ) =n∏

i=1

(λ− µi) = λn + dn−1λn−1 + · · ·+ d1λ + d0.

We begin with the case of a single control input and we assume that the plant is in the control-lable canonical form, namely,

x =

0 1 0 · · · 00 0 1 · · · 0...

......

. . ....

0 0 0 · · · 1−a0 −a1 −a2 · · · −an−1

x +

00...01

u

Then, we want the closed-loop system to be in the form

x =

0 1 0 · · · 00 0 1 · · · 0...

......

. . ....

0 0 0 · · · 1−d0 −d1 −d2 · · · −dn−1

x

This is equivalent to requiring −a + K = −d, or

K = a− d,

93

wherea4=

[a0 a1 · · · an−1

]and d

4=

[d0 d1 · · · dn−1

].

Example 6.1.1. Consider the simple mechanical system in Figure 6.1. The equation of motion for

m-

-

F

z

Figure 6.1: F = ma.

this system is simplyF = mz.

If we take m = 1, we can express the dynamics in the state-space form

x =[

0 10 0

]x +

[01

]F.

The characteristic polynomial for the A matrix is pA(λ) = λ2 and the system description is clearlyin the controllable canonical form.

Now suppose we want to design a full-state feedback controller in order to place the closed-loopeigenvalues at µ1 = µ2 = −1. The desired closed-loop characteristic polynomial will then be

pA+BK(λ) = (λ + 1)2 = λ2 + 2λ + 1.

Hence, the desired gain K is

K = a− d

=[

0 0]− [

1 2]

=[ −1 −2

].

Hence, the closed-loop system dynamics become

x =[

0 1−1 −2

]x.

Remark 6.1.2. Now suppose the system has only one input, but that it is not in the controllablecanonical form. As long as (A,B) is controllable, using the coordinate transformation given in ...,we can work with the equivalent system

˙x = Ax + Bu,

where x4= Tx, A

4= TAT−1 and B

4= TB. Note that the resulting feedback control law will be in

the form u = Kx for some K. When expressed in terms of the original state vector x, we haveu = Kx, where K

4= KT .

Remark 6.1.3. If the system has more than one input, say, m, then the FSF gain matrix K has atotal of mn independent variables.

94

6.2 Feedback Control Design in Discrete-time

6.2.1 Full-state Feedback

Theorem 6.2.1. [ −X XAT + F T BT

AX + BF −X

]< 0. (6.2.1)

6.3 MATLAB Commands

6.4 Exercises

Problem 6.1. Assume (A,B) is controllable. Show that x(t) can be driven to xd and kept constantthere if and only if −Axd ∈ Im(B). Assume this is so and compute the control force u that willaccomplish this.

Problem 6.2. Give a simple formula for the state-feedback control gain that places the closed-loop eigenvalues of the system x = Ax + Bu at desired locations (assume u is scalar and (A,B)controllable). The formula must include the transformation matrix that will put the system in thecontrollable canonical form, the coefficients of the desired and the actual characteristic polynomialsstored in vectors and the control gain obtained for the transformed system.

Problem 6.3. Consider the aircraft model discussed previously:

x =

−10 0 −10 00 −0.7 9 00 −1 −.07 01 0 0 0

x +

20 2.80 −3.130 00 0

u.

Find a stabilizing full-state feedback controller. Give all details.

95

Chapter 7

Output Feedback Control Design

We now consider the case when only a part of the state vector is measured and available forfeedback.

x = Ax + Bu (7.0.1a)y = Cx + Du (7.0.1b)

In this case, we design dynamic output feedback (DOF) controllers to stabilize the system. DOFcontrollers that we will design are LTI systems themselves, with their own internal states, whichwe denote by xc, input y and output u. That is,

xc = Acxc + Bcy (7.0.2a)u = Ccxc + Dcy. (7.0.2b)

The reason why we design a controller in this way will become clear once we discuss observerdesign. Furthermore, we will also discuss the conditions for designing a static output feedback(SOF) controller of the form u = Ky after introducing the design procedure via LMI’s.

7.1 Observer Design

The basic idea of observer design is to approximate the state vector x by using the measured outputy and then use the approximate state vector as if it were the actual one to implement a full-statefeedback control law. That is, from y, we first construct x and then use the control law u = Kx,where K is chosen so that the eigenvalues of A + BK are at the desired locations. But we haveto discuss how x can be approximated by x using y. The basic strategy in observer design is toconstruct a new dynamical system with state vector x such that such that

x → x as t →∞.

At first, one might think that a computer model of the original system given as

˙x = Ax + Bu

96

would do the job, but look closely: If we define the error between the two state vectors as

e4= x− x,

then we obtaine = Ae.

Now, ‖e(t)‖ → 0 as t →∞ if and only if A is stable. But if A is stable, there is no need to controlthe system anyway. Therefore, for practical purposes, this approach is useless. We need to definea new dynamical system with state vector x such that the error dynamics we arrive at in the endwill be in the form

e = Aee,

where Ae is stable regardless of the stability of A itself.The problem is actually solved as follows: Define a new dynamical system by

˙x = Ax + Bu− L[y − (Cx + Du)]= Ax + Bu + LC(x− x),

where L ∈ IRn×p is some matrix to be specified soon. Now define e4= x− x. Then,

e = ˙x− x = (A + LC)e.

Now, as long as L is chosen so that ALC is stable, we have

e → 0 as t →∞,

or, equivalently,x → x as t →∞.

Now, using the control law u = Kx, we arrive at the closed-loop system[

x˙x

]=

[A BK

−LC A + LC

] [xx

].

The question now is, “Is the closed-loop system really stable?”. It is not very easy to answerthis question using the representation above. But if we apply a coordinate transformation

[x

x− x

]=

[I 0

−I I

] [xx

]=

[xe

],

we arrive at a different, yet equivalent, state-space transformation[

xe

]=

[A + BK −BK

0 A + LC

] [xe

].

Hence, the eigenvalues of the closed-loop system are the eigenvalues of A+BK and those of A+LC.This fact is referred to as the “separation principle”. The closed-loop system is thus stable if andonly if K and L are chosen such that A + BK and A + LC are stable. (We know how to makeA+BK stable. In order to make A+LC stable, we can use the same method to make AT +CT LT

stable, since the eigenvalues of A + LC and AT + CT LT are the same).

97

Remark 7.1.1. The overall control law is

˙x = (A + BK + LC + LDK)x− Ly

u = Kx,

which is in the form (7.0.2), with

xc = x

Ac = A + BK + LC + LDK

Bc = −L

Cc = K

Dc = 0.

Note also that the order of the controller dynamics is nc = n. Such a controller is called a full-ordercontroller.

Remark 7.1.2 (Lower-order observers). Since we can assume without loss of generality that p ofthe states are available directly for feedback, we do not have to try to reconstruct them. Hence, itis intuitive that we should need an observer of order n− p at most. In fact, this is so.

We assume without loss of generality that our system is in the form

x = Ax + Bu

y =[

Ip 0]x.

We partition the state vector into measured and unmeasured parts as

xm 4=

x1...

xp

and xum 4

=

xp+1...

xn

Now partition the system matrices accordingly as[

xm

xum

]=

[A11 A12

A21 A22

] [xm

xum

]+

[B1

B2

]u.

Note that in this case, we have y = xm. Now let L ∈ IR(n−p)×p and consider the system

z = (A22 + LA12)z + (B2 + LB1)u + (A21 + LA11)xm − (A22 + LA12)Lxm.

We define an approximation of xum and the error between xum and its approximation via thefollowing:

xum 4= z − Lxm

eum 4= xum − xum.

Then, one can show thateum = (A22 + LA12)eum.

98

Therefore, if L is chosen so that A22 + LA12 is stable, then we have

eum(t) → 0 as t →∞,

or equivalently,xum(t) = z(t)− Lxm(t) → xum(t) as t →∞.

Now that we have an estimate of xum, we can use the following control law to stabilize thesystem:

u =[

Km Kum] [

xm

xum

],

where K4=

[Km Kum

]is chosen such that A + BK is stable.

There are two points left unclarified:

(i) We did not justify our unspoken assumption that (A12, A22) is controllable, so that we canplace the eigenvalues of A22 + LA12 arbitrarily, and

(ii) We did not show that the overall closed-loop system is indeed stabilized by the control lawabove.

These two points are left as exercises (Problem 7.2 and Problem 7.3).Lastly, if the given system output matrix is not in the form

[Ip 0

], then it can be put in that

form via the coordinate transformation

x4= Tx, where T =

[CR

]∈ IRn×n,

where R ∈ IR(n−p)×n is chosen so that T is invertible.

Remark 7.1.3. By a simple manipulation, it can be shown that the resultant control law is of theform

xc = Acxc + Bcy (7.1.1a)u = Ccxc + Dcy, (7.1.1b)

where

xc4= z Ac

4= ...

y = xm Bc4= ...

Cc4= ...

Dc4= ...

99

7.1.1 Design via LMIs

Remark 7.1.4 (On Static Output Feedback Controllers). A static output feedback (SOF) controllaw is one of the form

u = Ky, (7.1.2)

where K ∈ IRm×p is a constant gain matrix. Even though the SOF problem is very easy to stateand implement, no general procedure exists for finding such a K matrix. This difficulty arises fromthe fact that the search for such a K matrix cannot be expressed as an LMI (except for very specialcases): Consider the problem of checking the stability of the closed-loop system

x = (A + BKC)x.

The system is stable if and only if there exists a P = P T > 0 such that

(A + BKC)T P + P (A + BKC) < 0.

But there is no change of variables (that does not introduce extra constraints) that leads to anexpression on the left-hand-side of the inequality above that is linear in the unknowns.

For instance, one might be tempted to use the approach we took in the FSF problem (seethe proof of Theorem ??): Pre- and post-multiply the inequality by X = P−1 and then define

F4= KCX (it was F

4= KX in Theorem ??). We then need to solve for X = XT > 0 and an F

such thatAX + XAT + BF + F T BT < 0

and the matrix equationF = KCX

is solvable for some K. But this last algebraic requirement cannot be incorporated into the in-equality preceding it in any reasonable way. Hence, the overall problem cannot be cast as an LMI.In fact, the SOF problem is known to be “NP-hard”, or, in non-specialists’ language, “very hard”in terms of “complexity”.Remark 7.1.5 (On PID Controllers). The classical three-term PID controller is a controller withthe transfer function

C(s) = KP + KDs +KI

s.

Assuming KD = 0, this can be given a state-space representation as

C =[

0 KI

1 KP

]

However, if KD 6= 0, C(s) becomes improper. Hence, no state-space representation can be found ifderivative action is present in the controller.Remark 7.1.6 (On Decentralized Control). When the number of inputs is equal to the number ofoutputs, it is common practice to form several decoupled scalar feedback loops for each input/outputpair. The resulting decentralized feedback control system is depicted in the block diagram below:Unfortunately, the search for Ci(s)’s that satisfy closed-loop stability cannot be cast as an easilysolvable problem except for special cases. In fact, the mathematical nature of the problem is verysimilar to the one encountered in static output feedback. On the other hand, due to the simplicityof its implementation, decentralized feedback control is widespread in industrial applications.

100

G

¾C1

-

C2 ¾

-

¾Cm

.

.

.

-

Figure 7.1: Decentralized feedback control law.

7.2 Exercises

Problem 7.1. Consider a spring-mass system (no damper) with k = 1, m = 1. Assume there is aforce f acting on the mass. Write down the equation of motion and put it in the state-space formusing the position and the velocity of the mass as state variables. Assume that only the position isavailable in the measured output. Design control laws that stabilize the system using

(a) A state feedback controller; feedback eigenvalues at −2,−2.

(b) A full-order observer-based controller; feedback eigenvalues at −2,−2, observer eigenvaluesat −2,−2.

(c) A lower-order observer-based controller; feedback eigenvalues at −2,−2, observer eigenvalueat −2.

Simulate the system using SIMULINK in all three cases using the initial condition x0 =[

11

].

Note: If you happen to use the place command in MATLAB, be careful. It gives −K as the outputand not K. If place doesn’t work (as it does for certain problems), use the command acker.

Problem 7.2. Show that if the pair (C,A) is observable, then so is (A12, A22), where A and C aredecomposed compatibly as

A =[

A11 A12

A21 A22

]and C =

[I 0

].

Problem 7.3. Show that the lower-order observer design discussed in Remark 7.1.2 does indeedstabilize the closed-loop system.

Problem 7.4. Draw a block-diagram for a lower-order observer-based controller, assuming the outputof the system is y =

[Ip 0

]x.

101

Bibliography

[1] P.J. Antsaklis and A.N. Michel, “Linear Systems”, McGraw Hill, 1997.

[2] S. Boyd, L. El Ghaoui, E. Feron and V. Balakrishnan, “Linear Matrix Inequalities in Systemand Control Theory”, SIAM Studies in Applied Mathematics, 1994.

[3] Chen, “Linear Systems - Theory and Design”

[4] Gahinet, Chilali, ...

[5] R. A. Horn and C. A. Johnson, “Matrix Analysis”, Cambridge University Press, 1985.

[6] R. A. Horn and C. A. Johnson, “Topics in Matrix Analysis”, Cambridge University Press,19...

[7] A. Isidori, “Nonlinear Control Systems”, Springer, 1995.

[8] T. Kailath, “Linear Systems”, Prentice-Hall Information and System Science Series, 1980.

[9] D. G. Luenberger, “Optimization by Vector Space Methods”, John Wiley & Sons., Series inDecision and Control, 1997.

[10] A. W. Naylor and G. R. Sell, “Linear Operator Theory in Engineering and Science”, Springer,1982

102

Appendix A

Solutions to Problems

Solution to Problem 1.1. The fact

α

00...0

= 0

holds for any α ∈ C. Hence, the zero vector is linearly dependent.

Solution to Problem 1.2. The dimension of (C, IR) is 2. An obvious basis for it is {1, j}. Taking anappropriate real combination of 1 and j will create any point in C.

The dimension of (C,C), however, is 1. To see this, note that any nonzero z ∈ C constitutesa basis for C. This is because span(z) = C, i.e., for any w ∈ C, there exists an α ∈ C such thatw = αz.

Solution to Problem 1.3. In all three cases, the conditions except for the triangle inequality aretrivial. Let’s focus on the triangle inequality then. In the case of the l1-norm,

‖x + y‖1 =n∑

i=1

|xi + yi| ≤n∑

i=1

(|xi|+ |yi|) =n∑

i=1

|xi|+n∑

i=1

|yi| = ‖x‖1 + ‖y‖1.

For the l∞-norm,‖x + y‖∞ = maxi |xi + yi|

Let the index that gives maxi |xi + yi| be i∗. Then,

‖x + y‖∞ = |xi∗ + yi∗ | ≤ |xi∗ |+ |yi∗ | ≤ maxi|xi|+ max

i|yi| = ‖x‖∞ + ‖y‖∞.

The case of the l2-norm is more involved. Let us first prove the statement given in the hint. Letz = a + bi. Then,

Re(z) = a ≤ |a| =√

a2 ≤√

a2 + b2 =√

zz = |z|.

1

Now,

‖x + y‖22 =< x + y, x + y >

=< x, x > + < x, y > + < y, x > + < y, y >

=< x, x > + < x, y > +< x, y >+ < y, y >

=< x, x > +2Re < x, y > + < y, y >

≤< x, x > +2| < x, y > |+ < y, y >

≤< x, x > +2 < x, x >1/2< y, y >1/2 + < y, y > (Cauchy-Schwarz)

= (< x, x >1/2 + < y, y >1/2)2

= (‖x‖2 + ‖y‖2)2.

And therefore, ‖x + y‖2 ≤ ‖x‖2 + ‖y‖2.

Solution to Problem 1.4. Obviously, the 2−norm unit ball is the unit circle. The statement ‖x‖∞ ≤1 is equivalent to max{|x1|, |x2|} ≤ 1 in the case of x ∈ IR2, which in turn is equivalent to |x1| ≤ 1and ‖x2| ≤ 1. Thus, the ∞−norm unit ball is the square region centered at the origin. As for the1−norm unit ball, consider the following:

|x|+ |y| =

x + y if x ≥ 0 and y ≥ 0x− y if x ≥ 0 and y < 0−x + y if x < 0 and y ≥ 0−x− y if x < 0 and y < 0

When we plot the region |x| + |y| ≤ 1 for each case, the intersection becomes a diamond-shapedregion centered at the origin. All three regions are shown in Figure A.1.

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Figure A.1: Unit balls for 1-, 2- and ∞-norms. The square (dashed line) is the ∞-norm unit ball,the circle is the 2-norm unit ball and the diamond (dotted line) is the 1-norm unit ball. All threecross the axes at ±1.

2

Solution to Problem 1.5. By the Cauchy-Schwarz inequality, we have

−1 ≤ < x, y >

‖x‖‖y‖ ≤ 1

for any nonzero x, y ∈ X . Hence, the value cos−1 <x,y>‖x‖‖y‖ is well-defined if we make the restriction

θ ∈ [0, π].In IR2, consider Figure A.2.

6

-

±

~:

x

y

z = y − x

θ

Figure A.2: Derivation of the “cosine law”.

Then,‖x‖‖y‖ cos θ =< x, y >

and‖z‖2 =< x− y, x− y >= ‖x‖2 − 2 < x, y > +‖y‖2.

Using the expression for < x, y > in the previous equation, we obtain

‖z‖2 = ‖x‖2 + ‖y‖2 − 2‖x‖‖y‖ cos θ,

the so-called cosine law for triangles.

Solution to Problem 1.6. For any x, y ∈ X , we have

x = x− y + y so that ‖x‖ ≤ ‖x− y‖+ ‖y‖.Equivalently,

‖x‖ − ‖y‖ ≤ ‖x− y‖.Repeat for y = y − x + x to conclude

−‖x− y‖ ≤ ‖x‖ − ‖y‖ ≤ ‖x− y‖,or equivalently,

|‖x‖ − ‖y‖| ≤ ‖x− y‖.

Solution to Problem 1.7. Since for every subspace S of X , if x ∈ S, then αx ∈ S for every α ∈ IF,choosing an arbitrary x ∈ S and setting α = 0, we conclude that every subspace of X contains theorigin of X .

3

Solution to Problem 1.8. For any x, y ∈ X , we have

‖x + y‖2 + ‖x− y‖2 =< x + y, x + y > + < x− y, x− y >

= ‖x‖2+ < x, y > + < y, x > +‖y‖2 + ‖x‖2− < x, y > − < y, x > +‖y‖2

= 2‖x‖2 + 2‖y‖2.

Solution to Problem 1.9. Given x, y ∈ X and x ⊥ y,

‖x + y‖2 =< x + y, x + y >

= ‖x‖2+ < x, y > + < y, x > +‖y‖2

= ‖x‖2 + ‖y‖2.

Solution to Problem 1.10. No. Because the condition ‖x + y‖2 + ‖x − y‖2 = 2‖x‖2 + 2‖y‖2 isviolated. A counterexample in C2 is

x =[

j0

]and x =

[1j

].

Try it.

Solution to Problem 2.1. Since Φ is upper triangular and all diagonal elements are zero, φij = 0∀i ≥ j, j = 1 : n. Then, for Φ2, we have φ2

ij = 0 ∀i ≥ j − 1, ∀j = 1 : n by inspection. Applyinginduction, we have, for Φn, that φn

ij = 0 ∀i ≥ j − (n− 1), ∀j = 1 : n, that is, Φn = 0.

For instance, let A4=

[0 1 2 30 0 4 50 0 0 60 0 0 0

]. Then, A2 =

[0 0 4 170 0 0 240 0 0 00 0 0 0

], A3 =

[0 0 0 240 0 0 00 0 0 00 0 0 0

]and A4 = 0. Consider

also the matrix (Ji − λiI) in the proof of the Cayley-Hamilton theorem.

Solution to Problem 2.6. If rank(A) = n, then AL = A, AR = In is a suitable choice. If rank(A) <minn, then we can form a matrix AL from the linearly independent columns of A. Then, we canform AR from the combinations that produce the matrix A. In this case, rank(AL) = rank(AR) =rank(A).

More explicitly, consider the case where rank(A) = k < n. Denote the columns of A as ai,i = 1 : n and assume (without loss of generality) the first k columns be linearly independent. Then,

A =[

a1 · · · ak ak+1 · · · an

]=

[a1 · · · ak

]

1 · · · 0 bk+1,1 · · · bn,1...

. . ....

......

...0 · · · 1 bk+1,k · · · bn,k

4= ALAR,

where

ak+j =k∑

i=1

aibk+j,i ∀j = 1 : n− k.

Clearly, rank(AL) = rank(AR) = rank(A) = k.

4

Solution to Problem 2.7. We can write

B + BAB = B(I + AB) = (I + BA)B.

Now pre-multiply by (I + BA)−1 and post-multiply by (I + AB)−1. We obtain

(I + AB)−1B = B(I + BA)−1.

Solution to Problem 2.8. Note that

rank([

I 0A−1 I

])= rank

([I A−1

0 I

])= 2n.

Then, by Sylvester’s rank inequality,

rank([

A II B

])= rank

([A 00 B −A−1

])= rank(A) + rank(B −A−1).

Since rank(A) = n and rank(M) = 0 if and only if M = 0, we have

rank([

A II B

])= n ⇐⇒ A = B−1

Solution to Problem 2.9. The jth diagonal element of AT A is given by

n∑

i=1

a2ij .

When all such diagonal elements are added up, we have the given formula.

Solution to Problem 2.11. Let p(λ) be a polynomial specified by

p(λ) =n∑

i=0

aiλi

where the coefficients ai are real. Then, we have

p(λ) =n∑

i=0

aiλi =n∑

i=0

aiλi =n∑

i=0

aiλi =n∑

i=0

aiλi.

We have used to basic properties of complex numbers:

z1z2 = z1z2 and z1 + z2 = z1 + z2 ∀z1, z2 ∈ C.

5

Solution to Problem 2.12. (i) If A is diagonal, then

det(sI −A) =n∏

i=1

(s− aii).

(ii) Same as the diagonal case.

(iii) Consider the eigenvalue-eigenvector equation:

Ax = λx.

Add αx to both sides to obtain

(A + αI)x = (λ + α)x.

Hence, if λ is an eigenvalue of A, λ+α is an eigenvalue of A+αI. Note that the correspondingeigenvector is unchanged.

(iv) If A is invertible, Ax 6= 0 for any x 6= 0. Then, the eigenvalue-eigenvector equation

Ax = λx

is satisfied for nonzero λ’s only. Pre-multiply the equation by A−1 and multiply by λ−1 toobtain

λ−1x = A−1x,

implying that if A is invertible and if λ is an eigenvalue of A, then λ−1 is an eigenvalue ofA−1.

(v) Recall that A is invertible if and only if Ax = 0 implies x = 0. Since the eigenvectors of anymatrix are nonzero by definition, if 0 is an eigenvalue of A, then

Ax = 0x = 0,

which implies that there exists a nonzero x such that Ax = 0, i.e., A is singular. Conversely,if A is singular, there exists a nonzero x such that

Ax = 0 = 0x,

meaning that 0 is an eigenvalue of A.

(vi) This is very similar to proving that al the eigenvalues of a hermitian matrix are real.

Ax = λx ⇒ x∗Ax = λx∗x⇐⇒ (x∗Ax)∗ = (λx∗x)∗

⇐⇒ x∗A∗x = λx∗x⇐⇒ − x∗Ax = λx∗x⇐⇒ − λx∗x = λx∗x⇐⇒ − λ = λ

⇐⇒ Re(λ) = 0.

6

(vii) Let p(x)4= akx

k + ak−1xk−1 + · · ·+ a1x + a0 and let Ae = λe. In this case,

p(A)e = akAke + ak−1A

k−1e + · · ·+ a1Ae + a0Ie

= akλke + ak−1λ

k−1e + · · ·+ a1λe + a0e

= p(λ)e.

Here, of course, we have used the fact that if λ ∈ σ(A), then λk ∈ σ(Ak). To see this, considerthe following:

Akx = Ak−1Ax = Ak−1λx = λAk−2Ax = λ2Ak−2x = · · · = λk−1Ax = λkx.

Solution to Problem 2.13. Using the poly command, we obtain the characteristic polynomial of Aas

pA(λ) = λ3 + 3λ2 − 92λ + 24.

Then, by the deconv command,λ17 = q(λ)pA(λ) + r(λ),

wherer(λ) = −6.1794E16λ2 + 5.1943E17 λ− 1.3250E17.

Since pA(λ) = 0, we have

A17 = −6.1794E16A2 + 5.1943E17A− 1.3250E17 I3

=

6.3411E17 −8.1494E17 2.0526E181.7804E18 −2.2934E18 5.7621E18

−3.7829E18 4.9704E18 −1.2223E19

.

Solution to Problem 2.14.

Solution to Problem 2.15. Let A = A∗ ∈ Cn×n and consider two distinct eigenvalues λi and λj .Then,

Axi = λixi and Axj = λjxj

for some nonzero xi, xj ∈ Cn. Now pre-multiply the two equations by x∗j and x∗i respectively.Taking the adjoint of the second one, the difference of the two equations gives

x∗jAxi − x∗jA∗xi = x∗jAxi − x∗jAxi = (λi − λj)x∗jxi = 0.

But since λi 6= λj , xi and xj are orthogonal.

Solution to Problem 2.16. (i) False. Counterexample: A =[

0 10 0

]6= 0, but σ(A) = {0, 0}.

(ii) False. Counterexample: A =[

5 −1−1 5

]. The off-diagonal elements are negative, yet A > 0.

7

(ii’) True, since A ≥ 0 implies that all the diagonal elements are nonnegative (see the next solutionalso).

(iii) False. Counterexample: A =[

1 22 1

]. All elements are positive, yet A � 0.

(iii’) False. A ≥ 0 implies Aii ≥ 0, but not vice versa. Counterexample: A =[

1 44 1

]. The

diagonal elements are positive, yet A � 0.

Solution to Problem 2.17. This will be more of a verbal proof: By definition of positive definiteness,x∗Ax > 0 ∀x ∈ Cn, x 6= 0. A principal submatrix is formed by deleting a number of rows andcolumns from A symmetrically. In order to prove that a principal submatrix is positive definite,set the entries in x that correspond to the rows (hence the columns) that are deleted in A to zero.Now, taking all possible combinations of this new ”subspace” of x’s, it follows, by definition, thatthe given principal submatrix is positive definite.

Solution to Problem 2.18. Consider the condition MT M ≥ 0, or equivalently,

xT MT Mx ≥ 0 ∀x ∈ IRn, x 6= 0.

Define y4= Mx. The condition then becomes yT y ≥ 0, which is always true. For strict inequality,

we need y = Mx 6= 0 for all x ∈ IRn, x 6= 0. This is equivalent to rank(M) = n, that is M is fullcolumn-rank.

The condition MMT ≥ 0 is handled similarly by replacing M by MT .

Solution to Problem 2.19. The matrix[

a bb c

]is positive definite if and only if both of its eigenvalues

are real. The eigenvalues are the roots of

det(

λI −[

a bb c

])= λ2 − (a + c)λ + ac− b2.

The eigenvalues can then be expressed as

λ1,2 =a + c±

√(a + c)2 − 4(ac− b2)

2

Then, λ1,2 > 0 if and only ifa + c > 0 and ac− b2 > 0.

But since the second condition implies that a and c have the same sign, it is sufficient to replacea + c > 0 by c > 0 (or a > 0) only. Hence, the matrix is positive definite if and only if

c > 0 and ac− b2 > 0.

Note that an application of the Schur complement formula leads to the equivalent conditions

c > 0 and a− b2

c> 0

8

or

a > 0 and c− b2

a> 0.

Solution to Problem 2.20. From the Rayleigh-Ritz inequality, we have

λmin(A)‖x‖22 ≤ x∗Ax ≤ λmax(A)‖x‖2

2 ∀x ∈ Cn,

which implies

λmax(A) = maxx6=0

x∗Ax

x∗xand λmin(A) = min

x 6=0

x∗Ax

x∗x.

(These max and min values are attained for the x’s that are the eigenvectors corresponding to λmin

and λmax). It follows that

λmax(A + B) = maxx6=0

x∗(A + B)xx∗x

= maxx 6=0

[x∗Ax

x∗x+

x∗Bx

x∗x

]

≤ maxx 6=0

x∗Ax

x∗x+ max

x 6=0

x∗Bx

x∗x= λmax(A) + λmax(B).

The λmin-inequality is proved in the exact same way.

Solution to Problem 2.21. If P > 0, all of its eigenvalues are positive. Moreover,

det(P ) =n∏

i=1

λi and trace(P ) =n∑

i=1

λi.

Since the geometric mean of n positive scalars is less than or equal to their arithmetic mean, wehave

(detP )1/n ≤ trace(P )n

.

Solution to Problem 2.22. Since ‖I‖ = ‖I2‖ ≤ ‖I‖2, we have ‖I‖ ≥ 1.

Solution to Problem 2.23.

Solution to Problem 2.24.

Solution to Problem ??. We apply congruence transformations and the Schur complement formularepeatedly.

(i) Pre- and post-multiply both sides by X and then apply the Schur complement formula.

X−1F T FX−1 < X−1 ⇐⇒ F T F < X ⇐⇒[ −X F T

F −I

]< 0.

9

(ii) Apply the Schur complement formula once:

AX + XAT + BX−1BT < 0 ⇐⇒[

AX + XAT BBT −I

]< 0.

(iii) Pre- and post-multiply by Y and then apply the Schur complement formula:

CT X−1C < Y −1 ⇐⇒ Y CT X−1CY < Y ⇐⇒[ −Y Y CT

CY −X

]< 0.

Solution to Problem 3.1. Let x4=

hvθ

. Then,

x =

0 1 00 −1/τ2 σ0 0 −1/τ1

x +

001

u.

Solution to Problem 3.3. Simply carry out the multiplications given in (3.2.7).

Solution to Problem 3.4. Since T is now time-varying,

z = T x + T x

= T x + T (Ax + Bu)

= (T + TA)T−1z + TBu

4= Az + B

and

y = CT−1z + Du.

4= Cz + Du

Solution to Problem 3.5. (i) Since the number of foxes leads to a decrease in the number ofrabbits, ar < 0. Similarly, since more rabbits result in an increasing fox population, af > 0.Lastly, since rabbits eat carrots to feed themselves and then multiply, b > 0.

(ii) In state space, the predator-prey equations can be expressed as[

fr

]=

[0 af

ar 0

] [fr

]+

[b0

]c.

10

When compared with the spring-mass equations,[

xx

]=

[0 1

−k/m 0

] [xx

]+

[0

1/m

]F,

and noting the signs of the entries, the similarity is obvious. Hence, the fox population isanalogous to x, the rabbit population, to x and the carrot amount to the applied force F . Itis also clear that in the absence of carrots, the rabbit and fox populations will be oscillatoryin nature. (This is clearly counter-intuitive, since if there are no carrots, the rabbits willeventually die out. But we did admit that this is a simplified model.)

(iii)

Solution to Problem 3.6. The solution to x = a(t)x + b(t)u is

x(t) =∫ t

0eR t

τ a(σ) dσb(τ)u(τ) dτ.

Then, by Leibniz’ rule,

x =∫ t

0

∂

∂t

(eR t

τ a(σ) dσb(τ)u(τ))

dτ + b(t)u

=∫ t

0

∂

∂t

(∫ t

τa(σ) dσ

)eR t

τ a(σ) dσb(τ)u(τ) dτ + b(t)u

=∫ t

0a(t)e

R tτ a(σ) dσb(τ)u(τ) dτ + b(t)u

= a(t)∫ t

0eR t

τ a(σ) dσb(τ)u(τ) dτ + b(t)u

= a(t)x(t) + b(t)u(t)

Solution to Problem 3.7. In the absence of any external input, the system is essentially driven byinitial conditions. If the solution x(t) of the system x = f(x) is linear in x0, we can say the systemis linear. Obviously, the system x = Ax is linear since the solution is x(t) = eAtx0.

Solution to Problem 3.8.

Solution to Problem 3.9. In order for the angle to remain constant at π/4, we need xeq =[

π/450

].

The corresponding Teq can be obtained from[

00

]=

[0

gL sin θ + Teq

],

which leads to Teq = − gL sinπ/4.

11

The linearized equations are, then,

˙x =[

0 1gL sinπ/45

]x +

[01

]T ,

where x4= x− xeq and T

4= T − Teq.

Solution to Problem 3.10. The jacobian of f is

0 1 0 0 0 0 0θ2 cos2 φ + φ2 + 2k/r3 0 0 2rθ cos2 φ −2rθ2 cosφ sinφ 2rφ 1/m

0 0 0 1 0 0 0−2rθ

r2− uθ

mr2cos θ −2θ

r0

2r

r+ 2φ tanφ

2θφ

cos2φ− uθ

mrsinφ 2φ tanφ

cosφ

mr0 0 0 0 0 1 0

2rφ

r2− uφ

mr2−2φ

r0 −2θ cosφ sinφ −θ2(cos2 φ− sin2 φ) −2r

r

1mr

.

When evaluated at (xsol, usol), we obtain the indicated A and B matrices.

Solution to Problem 4.1. If A = MJM−1, then

eA = I + MJM−1 +(MJM−1)2

2!+

(MJM−1)3

3!+ · · ·

= M(I + J +J2

2!+

J3

3!+ · · · )M−1

= MeJM−1.

Solution to Problem 4.2. We integrate eAt term by term:∫ t

0eAτ dτ =

∫ t

0

(I + Aτ +

A2τ2

2!+ · · ·

)dτ

=(

Iτ +Aτ2

2!+

A2τ3

3!+ · · ·

)∣∣∣∣t

0

= It +At2

2!+

A2t3

3!+ · · ·

= A−1

(At +

A2t2

2!+

A3t3

3!+ · · ·

)

= A−1

(∓I + At +

A2t2

2!+

A3t3

3!+ · · ·

)

= A−1(eAt − I).

The equality∫ t0 eAτ dτ = (eAt − I)A−1 is obtained by pulling out A−1 to the right in the 4th step

above.

12

Solution to Problem 4.4. Recall that x(t) = eAtx(0) is the solution of x = Ax.

(i) Note that x(t+15) = e(15A)x(t) = Zx(t). Using the command expm in MATLAB, we obtain

Z = e15A =

0.2032 −0.0068 −0.0552 −0.07080.0340 0.0005 −0.0535 0.10690.0173 0.1227 0.0270 0.06160.0815 0.0186 0.1151 0.1298

.

(ii) Similar to (i), x(t) = e20Ax(t− 20), so that x(t− 20) = e−20Ax(t) = Y x(t). Then,

Y = e−20A =

6.2557 3.3818 1.7034 2.2064−2.1630 −2.8107 −14.2950 12.1503−3.3972 17.3931 −1.6257 −2.8004−1.7269 −6.5353 10.7081 2.9736

.

(iii) x0 is the solution to the equationx(10) = e10Ax0.

Then,

x0 = e−10Ax(10) =

3.99611.06503.81141.7021

.

Solution to Problem 4.5. Note that A2 = 0.

(i) Since Ak = 0 for all k ≥ 2,

eAt = I + At =[

1− t t−t 1 + t

].

(ii) Since x(1) = e1Ax(0), we have[

x1(1)2

]=

[0 1

−1 2

] [1

x2(0)

].

Solving for x1(1) and x2(0), we obtain x1(1) = x2(0) = 1.5. Then,

x(2) = eAx(1) =[

22.5

]

13

Solution to Problem 4.6. (i) The eigenvalues of A are obtained from det(λI−A) = det

[λ −ωω λ

]=

λ2 + ω2 = 0. Hence, λ1,2 = ±jω. We know that x(t) = eAtx0, where

eAt = L−1{(sI −A)−1}

= L−1

{1

s2 + ω2

[s ω−ω s

]}

=[

cosωt sinωt− sinωt cosωt

].

(ii) If the Euclidean distance (the 2-norm) of x(t) to the origin does not change, the trajectoriesare circular.

‖x(t)‖24= (x(t)T x(t))1/2

= (xT0 eAT teAtx0)1/2

=(

xT0

[cosωt sinωt

− sinωt cosωt

] [cosωt − sinωtsinωt cosωt

]x0

)1/2

=(

xT0

[1 00 1

]x0

)1/2

= ‖x0‖2.

Therefore, the distance to the origin of any trajectory initiating from x0 is constant at ‖x0‖.(iii) We need to prove x(t)T x(t) = 0 for all t > 0.

x(t)T x(t) = x(t)T AT x(t)

= x(t)T

[0 −ωω 0

]x(t)

=[

ωx2(t) −ωx1(t)]x(t)

= 0.

Solution to Problem 4.7. (i) Find the eigenvalues of A first (let’s abbreviate cos t as ct and sin tas st):

det(λI −A) = det[

λ + 1− 32c2t −1 + 3

2st ct1 + 3

2st ct λ + 1− 32s2t

]

= (λ + 1− 32c2t)(λ + 1− 3

2s2t)− (−1 +

32st ct)(1 +

32st ct)

= λ2 +12λ +

12

14

after using simple trigonometry. Hence,

λ1,2 = −14(1±

√7j),

so that the eigenvalues of A lie in the open LHP at all times.

(ii) Now let’s verify that the given x∗(t) is a solution.

x∗(t) = 12et/2

[ −ctst

]+ et/2

[stct

]

= et/2

[ −12ct + st

12st + ct

].

Let’s separately calculate A(t)x∗(t) also.

A(t)x∗(t) = et/2

[ −1 + 32c2t 1− 3

2stct−1− 3

2stct −1 + 32s2t

] [ −ctst

]

= et/2

[ct− 3

2c3t + st− 32s2tct

ct + 32stc2t− st + 3

2s3t

]

= et/2

[ct− 3

2ct(c2t + s2t) + stct + 3

2st(c2t + s3t)− st

]= x∗(t).

Therefore, the given x∗(t) is indeed a solution.

(iii) Since the given x∗(t) is a solution and since it does not go to zero as t → ∞, the system isnot stable by definition, even though its eigenvalues always have negative real parts.

(iv) The moral of the story is that the statement a system is stable iff all the eigenvalues of A havenegative real parts is valid only for continuous-time LTI systems of the form x(t) = Ax(t),not for time-varying systems like x(t) = A(t)x(t). In the case of a linear time-varying systemx = A(t)x, x(t) = eA(t)tx0 is not a solution anymore. Therefore, none of the things we saidabout how the solution is related to the eigenvalues of A makes sense anymore. In fact, thesolution of x = A(t)x is

x(t) = eR t0 A(τ) dτx0.

Solution to Problem 4.8. The spectrum (set of eigenvalues) of the first matrix is σ(A1) = {0,−1,−2}and σ(A2) = {0, 0,−1}. (Your results may be slightly off, but I constructed these matrices so thatthey have these eigenvalues.) Hence, neither system is stable, but there is a crucial difference be-tween the two. The matrix A1 has a simple eigenvalue at the origin, whereas A2 has a repeatedeigenvalue at the origin. Consequently, there is a constant term in the solution of x(t) for thefirst system, and it neither decays nor blows up, but remains constant as t → ∞. In the secondsystem, however, there is a constant term and a ramp term (time t multiplied by a constant) dueto the multiplicity of the eigenvalue at the origin. Therefore, as t →∞, x(t) does not decay, doesnot remain constant, but blows up instead. Such a system is unstable, whereas the first system isusually referred to as marginally stable.

15

Solution to Problem 4.9. Note that

AT P + PA + 2αP < 0 ⇐⇒ (A + αI)T P + P (A + αI) < 0.

Hence, the inequality is satisfied if and only if all the eigenvalues of A+αI have negative real parts,i.e., λ(A + αI) < 0. But from Problem 2.12, part (iii), this is equivalent to the eigenvalues of Asatisfying λ(A) < −α and we are done.

Solution to Problem 4.10. Define k m4= k/m and c m

4= c/m first and then execute the following

commands:

A=[0 1;-k_m -c_m];B=[0;0];C=eye(2);D=[0;0];sys=ss(A,B,C,D);x0=[1;1];y=initial(sys,x0);plot(y(:,1),y(:,2));

In order to plot different cases on the same graph, use the command hold on. For plotting twographs on the same page (like I have done), use the subplot command.

In the next two figures, you can see how different initial conditions behave. The first plotcorresponds to the case c = 0 and k/m = 1 and the second, to the case k/m = 1 and c/m = 1.The arrows indicate directions in which the states change. This is what is actually called a phaseportrait. You can obtain these plots using the following commands in MATLAB (define k m

4= k/m

and c m4= c/m first):

[X1,X2]=meshgrid(-2:.2:2,-2:.2:2);T=.1;

transmat=expm([0 1;-k_m-c_m]*T);X1T=transmat(1,1)*X1+transmat(1,2)*X2;X2T=transmat(2,1)*X1+transmat(2,2)*X2;quiver(X1,X2,X1T-X1,X2T-X2);grid on;axis([-2 2 -2 2]);

The command meshgrid takes points in the x1 − x2 plane that correspond to different initialconditions in our application. We compute the matrix eAT next and find for each initial condition,the difference between itself and what it becomes after T seconds. The plot as a whole shows howthe system evolves in time. For more details, read the help files for the commands mentioned.

Solution to Problem 5.1. The solution requires only straightforward calculations.

16

(i) Consider the dynamic equation of G2:

x1 = A1x1 + B1u1

= A1x1 + B1(C2x2 + D2u2), since u1 = y2

y1 = C1x1 + D1u1

= C1x1 + D1(C2x2 + D2u2)

Combining with x2 = A2x2 + B2u2 yields[

x1

x2

]=

[A1 B1C2

0 A2

] [x1

x2

]+

[B1D2

B2

]u2

y2 =[

C1 D1C2

] [x1

x2

]+ D1D2u2.

In a compact form, we have

G1G2 =

A1 B1C2 B1D2

0 A2 B2

C1 D1C2 D1D2

.

(ii) Note, first, that the dynamic equations of G1 and G2 are decoupled, so that the dynamicequation of the overall system is

[x1

x2

]=

[A1 00 A2

] [x1

x2

]+

[B1

B2

]u.

As for the output equation,

y = y1 + y2

= C1x1 + D1u + C2x2D2u

=[

C1 C2

] [x1

x2

]+ (D1 + D2)u.

Or,

G1 + G2 =

A1 0 B1

0 A2 B2

C1 C2 D1 + D2

.

(iii) Let the input to the overall system be u and let u1 and u2 be the inputs to G1 and G2

respectively. Define the output of the system as y and let y1 and y2 be the outputs of G1 andG2 respectively. Then, we have y = y1, u2 = y1 and u1 = u− y2. Then,

u2 = y1 = C1x1 + D1u1

= C1x1 + D1(u− y2)= C1x1 + D1[u− (C2x2 + D2u2)],

17

Solving for u2, we haveu2 = R−1

1 (C1x1 −D1C2x2 + D1u),

where R14= I + D1D2. The dynamics for G1 is

x1 = A1x1 + B1u1

= A1x1 + B1(u− y2)= A1x1 + B1u−B1(C2x2 + D2u2)

= A1x1 + B1u−B1C2x2 −B1D2R−11 (C1x1 −D1C2x2 + D1u)

=[

A1 −B1D2R−11 C1 −B1(I −D2R

−11 D1)C2

] [x1

x2

]+ B1(I −D2R

−11 D1)u

=[

A1 −B1D2R−11 C1 −B1R

−12 C2

] [x1

x2

]+ B1R

−12 u,

where R24= I + D2D1. The output equation is

y = y1 = C1x1 + D1u1

= C1x1 + D1(u− y2)= C1x1 + D1(u− C2x2 −D2u2).

Plugging in the expression for u2, we have

y =[

(I −D1D2R−11 )C1 −D1R

−12 C2

] [x1

x2

]+ D1R

−12 u.

Finally, the dynamic equation for G2 is:

x2 = A2x2 + B2u2

= A2x2 + B2R−11 (C1x1 −D1C2x2 + D1u)

[B2R

−11 C1 A2 −B2R

−11 D1C2

] [x1

x2

]+ B2R

−11 D1u.

Hence, the state-space representation for the system mapping u to y is

G =

A1 −B1D2R−11 C1 −B1R

−12 C2 B1R

−12

B2R−11 C1 A2 −B2R

−11 D1C2 B2R

−11 D1

(I −D1D2R−11 )C1 −D1R

−12 C2 D1R

−12

The same result could have been obtained by using transfer function manipulations to obtain

Y (s) = (I + G1(s)G2(s))−1 G1(s)︸ ︷︷ ︸

G(s)

U(s).

Then, a state-space representation could be obtained for G(s), but we would first have to beable to invert a given transfer function.

18

Solution to Problem 5.2. The situation is depicted in the diagram below:

C1

C2

-

-

-

-G -

v1

v2

u1

u2

y

We can define a single system C that maps v4=

[v1

v2

]to u

4=

[u1

u2

]as

C(s) =[

C1(s) 00 C2(s)

].

Then, the state-space realization of C becomes

C =

AC1 0 BC1 00 AC2 0 BC2

CC1 0 DC1 00 CC2 0 DC2

4=

[AC BC

CC DC

].

Now, the overall system mapping v to y can be represented in state space by

GC =

AC 0 BB

BCC A BDC

DCC C DDC

.

Solution to Problem 5.3. The transfer function is G(s) = C(sI − A)−1B + D. Compute G(s)−1

using the matrix inversion lemma:

G(s)−1 = D−1 −D−1C[(sI −A) + BD−1C]−1BD−1

= D−1 + (−D−1C)[sI − (A−BD−1C)]−1BD−1

= D−1 + D−1C[sI − (A−BD−1C)]−1(−BD−1).

Thus, a state-space description of the inverse of G(s) is:

G−1 =[

A−BD−1C BD−1

−D−1C D−1

]=

[A−BD−1C −BD−1

D−1C D−1

].

Solution to Problem 5.4. The left-inverse is defined as

G−L 4= (G∼G)−1G∼

where a state-space representation of G∼(s)4= G is

G∼ =[ −AT −CT

BT DT

]=

[ −AT CT

−BT DT

].

19

Then,

G∼G =

−AT −CT C −CT D

0 A B

BT DT C DT D

.

The inverse of G∼G exists if and only if DT D is invertible (i.e., D is full column-rank) and in thatcase

(G∼G)−1 =

−AT −BT DJBT −CT C −BT DJDT C −BT DJ

−BJBT A−BJDT C BJ

JBT JDT C J

,

where J4= (DT D)−1. The left-inverse (G∼G)−1G∼ is, then,

−AT −BT DJBT −CT C −BT DJDT C −BT DJBT −BT DJDT

−BJBT A−BJDT C BJDT

0 0 −AT −CT

JBT JDT C JBT JDT

.

20