Second Year Mathematics Revision...Second Year Mathematics Revision Linear Algebra - Part 2 Rui Tong UNSW Mathematics Society Rui Tong Second Year Mathematics Revision Eigenvalues

Second Year Mathematics RevisionLinear Algebra - Part 2

Rui Tong

UNSW Mathematics Society

Rui Tong Second Year Mathematics Revision

Eigenvalues and Eigenvectors

Today’s plan

1 Eigenvalues and EigenvectorsSingular Value Decomposition (MATH2601)

2 The Jordan Canonical FormFinding Jordan FormsThe Cayley-Hamilton Theorem

3 Matrix ExponentialsComputing Matrix ExponentialsApplication to Systems of Differential Equations


Eigenvalues and Eigenvectors Singular Value Decomposition (MATH2601)

Singular Values (MATH2601 only section)

Definition 1: Singular Values

A singular value of a m × n matrix A is the square root of aneigenvalue of A∗A.

Recall: A∗A denotes the adjoint of A.

Definition 2: Singular Value Decomposition

A SVD for an m × n matrix A is of the form A = UΣV ∗ where

U is an m ×m unitary matrix.

V is an n × n unitary matrix.

Σ has entries

σii > 0. (These are determined by the singular values.)σij = 0 for all i 6= j .



Nice properties of A∗A

Lemma 1: Properties of A∗A

1 All eigenvalues of A∗A are real and non-negative (even if A hascomplex entries!)

2 ker(A∗A) = ker(A)

3 rank(A∗A) = rank(A)

The first one is pretty much why everything works.



SVD Algorithm

Algorithm 1: Finding a SVD

1 Find all eigenvalues λi of A∗A and write in descending order.Also find their associated eigenvectors of unit length vi .

2 Find an orthonormal set of eigenvectors for A∗A.

Automatically occurs when all eigenvalues are distinct, whichwill usually be the case. Otherwise require Gram-Schmidt forany eigenspace with dimension strictly greater than 1..

3 Compute ui = 1√λiAvi for each non-zero eigenvalue.

4 State U and V from the vectors you found, and Σ from thesingular values.

Lemma 2: Used to speed up step 1

A∗A and AA∗ share the same non-zero eigenvalues.

If rank(A) = r , then A∗A has r non-zero eigenvalues. All othereigenvalues are 0.



SVD Example

Example 1: MATH2601 2017 Q2 c)

For the matrix A =

(3 1 1−1 3 1

)1 Find the eigenvalues of AA∗.

2 Explain why the eigenvalues in part 1 are also eigenvalues ofA∗A, and state any other eigenvalues of A∗A.

3 Find all eigenvectors of A∗A.

4 Find a singular value decomposition for A.



SVD Example

Part 1: We compute

AA∗ =

(3 1 1−1 3 1

)3 −11 31 1

=

(11 11 11

).

This matrix has two eigenvalues, and they sum to tr(AA∗) = 22 andmultiply to det(AA∗) = 120. By inspection, λ1 = 12 and λ2 = 10.



SVD Example

Part 2: Quoted word for word from the answers...

”We know that A∗A and AA∗ have the same nonzero eigenvalues,so 12 and 10 are eigenvalues of A∗A.

Also, all eigenvalues of A∗A are real and nonnegative, so its thirdeigenvalue is 0.”



SVD Example

Part 3: We compute

A∗A =

3 −11 31 1

( 3 1 1−1 3 1

)=

10 0 20 10 42 4 2

For λ = 12:

A∗A− 12I =

−2 0 20 −2 42 4 −10

.

Looking at row 1, arbitrarily set first component to 1, and then thethird component is 1. Equating row 2, the second component is 2.

∴ v1 = t

121

.



SVD Example

Part 3: We compute

A∗A =

3 −11 31 1

( 3 1 1−1 3 1

)=

10 0 20 10 42 4 2

For λ = 10:

A∗A− 10I =

0 0 20 0 42 4 −8

.

Rows 1 and 2 force the third component to be 0. Looking at row 3,it is easier to set the second component to 1, and then the firstcomponent will be −2.

∴ v2 = t

−210

.



SVD Example

Part 3: We compute

A∗A =

3 −11 31 1

( 3 1 1−1 3 1

)=

10 0 20 10 42 4 2

For λ = 0, looking at A∗A itself, there really are many possibilities

we can go about it. But I follow the answers, which arbitrarily setthe first component to −1. See if you can then show that

v3 = t

−1−25

.

Note: In each case, t ∈ R.



SVD Example

Part 4: In each case, choose the value of t that normalises theeigenvectors:

v1 =1√6

121

(λ1 = 12)

v2 =1√5

−210

(λ2 = 10)

v3 =1√30

−1−25

(λ3 = 0)



SVD Example

Part 4: Compute ui = 1√λiAvi for each non-zero eigenvector:

u1 =1√12

(3 1 1−1 3 1

)1√6

121

=1√2

(11

)

u2 =1√10

(3 1 1−1 3 1

)1√5

−210

=1√2

(−11

)



SVD Example

Part 4: We conclude that a SVD for A is A = UΣV ∗, where

U =1√2

(1 −11 1

)Σ =

(√12 0 0

0√

10 0

)

V =

1√6− 2√

5− 1√

302√6

1√5− 2√

301√6

0 5√30



Remark

For A = UΣV ∗, where A ∈ Mn×n(C):

The columns of U =(u1 . . . um

)are called the left singular

vectors.

The columns of V =(v1 . . . vn

)are called the right

singular vectors.

Word of advice: Write some of these numbers very quickly! SVDsare instructive when you know the method, but it always takesforever to do.



Reduced SVD

I don’t see these examined, but I should still mention them.

1 Obtain Σ̂ by removing any zero columns in Σ

2 Obtain V̂ by removing the corresponding columns in V .

3 Then, A = UΣ̂V̂ ∗.

For the earlier example:

Σ̂ =

(√12 0

0√

10

)

V̂ =

1√6− 2√

52√6

1√5

1√6

0


The Jordan Canonical Form

Today’s plan





The Jordan Canonical Form Finding Jordan Forms

Jordan Blocks

Definition 3: Jordan blocks

The k × k Jordan block for λ is the matrix

Jk(λ) =

λ 1 0 . . . 00 λ 1 . . . 00 0 λ . . . 0...

......

. . ....

0 0 0 . . . 10 0 0 . . . λ

∈ Mk×k(C).

That is, put λ on every entry along the main diagonal, and a 1immediately above each λ wherever possible.



Jordan Blocks

Quick examples:

J3(−4) =

−4 1 00 −4 10 0 −4

J4(0) =

0 1 0 00 0 1 00 0 0 10 0 0 0



Powers of Jordan Forms

Find the pattern.

J1(λ)n =(λn)

J2(λ)n =

(λn

(n1

)λn−1

0 λn

)

J3(λ)n =

λn (n1

)λn−1

(n2

)λn−2

0 λn(n

1

)λn−1

0 0 λn

J4(λ)n =

λn

(n1

)λn−1

(n2

)λn−2

(n3

)λn−3

0 λn(n

1

)λn−1

(n2

)λn−2

0 0 λn(n

1

)λn−1

0 0 0 λn



Powers of Jordan Forms

Lemma 3: Computing powers of Jordan forms

1 Start with λn on every diagonal entry.

2 Put(n

1

)λn−1 wherever you can immediately above λn

3 Put(n

2

)λn−2 wherever you can immediately above

(n1

)λn−1

4 Keep doing this, increasing the binomial coefficient anddecreasing the power on λ.

Note: Not quite the above. If you ever bump into(nn

), that’s the

last diagonal you fill. Just put 0’s everywhere else above.



Matrix Direct Sums

Definition 4: Direct sums of matrices

The direct sum of matrices A1,A2, . . . ,An is the matrix formed byputting these matrices on the diagonals and zeroes everywhere else.

A1 ⊕ A2 ⊕ · · · ⊕ An =

A1 0 . . . 00 A2 . . . 0...

.... . .

...0 0 . . . An

In MATH2501 and MATH2601, we only worry about this withJordan blocks.



Matrix Direct Sums

Quick example:

J2(3)⊕ J4(5) =

3 1 0 0 0 00 3 0 0 0 00 0 5 1 0 00 0 0 5 1 00 0 0 0 5 10 0 0 0 0 5



Matrix Direct Sums

Lemma 4: Powers on direct sums of Jordan blocks

The power of a Jordan form is the direct sum of powers on eachindividual block. I.e.,

(J1 ⊕ · · · ⊕ Jm)n = Jn1 ⊕ · · · ⊕ Jnm.

Example:

[J2(3)⊕ J1(2)]n =

3n(n

1

)3n−1 0

0 3n 00 0 2n



The Generalised Eigenvector

Definition 5: Generalised Eigenvector

A generalised eigenvector corresponding to eigenvalue λ is a

non-zero vector v satisfying the property (A− λI )kv = 0 , for

some k ≥ 1.

This differs from the (usual) eigenvector in the sense that thosemust satisfy (A− λI )v = 0, i.e. we must take k = 1.





Let C =

9 7 −3−2 2 12 5 2

and v =

1−2−3

.

Show that for the matrix C , v is a generalised eigenvectorcorresponding to λ = 5.




We compute that

C − 5I =

4 7 −3−2 −3 12 5 −3

and continuously left-multiplying to v,

(C − 5I )v =

−111

(C − 5I )2v =

000

so v ∈ GE5.



Generalised Eigenspaces

Definition 6: Generalised Eigenspace

The generalised eigenspace of λ, denoted GEλ, is the set of allgeneralised eigenvectors corresponding to λ.

GEλ = {v ∈ Cn | (A− λI )kv = 0 for some k ≥ 1}

Lemma 4: Alternate representation of GEλ

GEλ = ker(A− λI ) ∪ ker(A− λI )2 ∪ ker(A− λI )3 ∪ . . .



Computing Jordan Forms

Definition 7: Jordan matrix

A Jordan matrix J is a direct sum of Jordan blocks.

Lemma 5: Uniqueness

Every matrix A has one unique Jordan matrix, up to somepermutation (arrangement) of the Jordan blocks.




Theorem 1: Useful properties in computing Jordan forms

Let dim ker(A− λI )k , i.e. nullity(A− λI )k = dk . Set d0 = 0. Then

1 ker(A− λI ) ⊆ ker(A− λI )2 ⊆ . . .2 d0 ≤ d1 ≤ d2 ≤ d3 ≤ . . .3 d1 − d0 ≥ d2 − d1 ≥ d3 − d2 ≥ . . .

That is to say, the nullities must not decrease, but the difference innullities must not INcrease.

Remark: Multiplicity

As a corollary, the algebraic multiplicity of an eigenvalue λ equals todimGEλ. This allows us to not compute (A− λI )k forever - westop when nullity(A− λI )k = AM.




We use Jordan chains to find the matrices P and J, such thatA = PJP−1. For an eigenvalue λ with algebraic multiplicity k , weneed to start with some vector v1 such that on multiplication, wehave

v1A−λI−−−→ v2

A−λI−−−→ . . .A−λI−−−→ vk

A−λI−−−→ 0.

We then include (note the reverse order!)(vk . . . v2 v1

)to P . This corresponds to one Jordan block Jk(λ) in the direct sumfor the Jordan matrix J of A.




(Or maybe your lecturer taught things the other way around.) Weconsider this chain

vkA−λI−−−→ vk−1

A−λI−−−→ . . .A−λI−−−→ v1

A−λI−−−→ 0.

and we include this to P instead.(v1 . . . vk−1 vk

)We still use the Jordan block Jk(λ).



Computing Jordan Forms: Example 1


Let C =

9 7 −3−2 2 12 5 2

and v =

1−2−3

.

1 Calculate (C − 5I )v and (C − 5I )2v. (Done earlier)

2 Without using any matrix calculations, write down all theeigenvalues of C and their algebraic and geometricmultiplicities. Give reasons for your answers.

3 (Not originally in the exam:) Find an invertible matrix P and aJordan matrix J such that C = PJP−1.




Part 2: The trace is usually helpful, because it is the sum of theeigenvalues.

tr(C ) = 9 + 2 + 2 = 13

From part 1, 5 is an eigenvalue of C with algebraic multiplicity atleast 2. The third eigenvalue λ3 satisfies

5 + 5 + λ3 = 13 =⇒ λ3 = 3.

Which is, of course, the only remaining eigenvalue and hence musthave AM = 1. So we have:

Eigenvalue 5: AM = 2, GM = 1

Eigenvalue 3: AM = 1, GM = 1

Note: I haven’t justified the GM’s! Try doing that yourself!




Part 3: Row reducing C − 3I ,

C − 3I =

6 7 −3−2 −1 12 5 −1

∼

0 −12 0−2 −1 10 4 0

∼

−2 −1 10 1 00 0 0

so we can take a corresponding eigenvector v3 =

102

.




So our chains are: 1−2−3

C−5I−−−→

−111

C−5I−−−→ 0

102

C−3I−−−→ 0

and hence A = PJP−1 where

J =

5 1 00 5 00 0 3

and P =

−1 1 11 −2 01 −3 2



Computing Jordan Forms: Example 2 (time permitting...)

Example 4: MATH2601 2017 Q3 a)

Let A =

3 1 −22 6 −72 2 −2

. We are given that

GE2 = span

1

32

,

011

and GE3 = span

1

42

.

1 Find the Jordan chain for λ = 2 starting with

011

.

2 Without any calculation, write down the geometric multiplicityof λ = 2, giving reasons for your answer.

3 Find a Jordan form J and invertible matrix P for A, such thatA = PJP−1.




Part 1: 011

A−2I−−−→

−1−3−2

A−2I−−−→

000

.

Part 2: The algebraic multiplicity is 2, since we have anotherdistinct eigenvalue, so GM ≤ 2. But GM 6= 2 since we have a chainof length 2, so GM = 1.




Part 3: A = PJP−1 where

P =

−1 0 1−3 1 4−2 1 2

J =

2 1 00 2 00 0 3





4 Find v ∈ GE2 and w ∈ GE3 such that v + w =

210

Theorem 2: Cn and the generalised eigenspaces

The direct sum of generalised eigenspaces of any A ∈ Mn×n spanCn.

Hence we just need to express

210

as a linear combination of132

,

011

and

142

.




You can have fun with the row reduction... I’ll just state the finalanswer: 2

10

= 3

132

− 4

011

−1

42

=

352

︸︷︷︸

v

+

−1−4−2

︸︷︷︸

w



Remark: Similarity Invariants

Theorem 3: Jordan forms are the complete similarity invariant

Two matrices A and B are similar, i.e. A = PBP−1 for someinvertible matrix P, if and only if they have the same Jordan forms.(At least, to within a different arrangement of direct sums.)



Jordan forms given nullities

The Jordan matrix J can sometimes be found with less informationif we don’t need to find P.

Example 5: MATH2601 2016 Q4 b)

Let B be a 10× 10 matrix and let λ be a scalar. Suppose it isknown that

nullity(B − λI ) = 5,

nullity(B − λI )2 = 8,

nullity(B − λI )3 = 9.

Find all possible Jordan forms of B.

Idea: Our Jordan chains can be drawn on an onion diagram.




There are 5 eigenvectors in ker(B − λI ). The idea is that there are8− 5 = 3 more generalised eigenvectors in ker(B − λI )2. This isbecause we know that ker(B − λI ) ⊆ ker(B − λI )2.

Similarly, there is another 9− 8 = 1 in ker(B − λI )3.




We’ve addressed 9 of the 10 eigenvalues for B. There is only onemore left to go.

Case 1: The tenth eigenvalue is NOT λ.Then it must be some other value µ 6= λ. It can only correspond toone eigenvector, so we include J1(µ) to the direct sum.

The Jordan chains for λ have lengths 3, 2, 2, 1 and 1, so therefore(up to some permutation),

J = J3(λ)⊕ J2(λ)⊕ J2(λ)⊕ J1(λ)⊕ J1(λ)⊕ J1(µ).

(again, up to some permutation of the Jordan blocks).




We’ve addressed 9 of the 10 eigenvalues for B. There is only onemore left to go.

Case 2: The tenth eigenvalue IS also λ.Problem: We cannot add it in ker(B − λI ), ker(B − λI )2 orker(B − λI )3 without screwing up the nullities!

Recall that the difference is nullities is non-increasing. This meansthat the last generalised eigenvector must be in ker(B − λI )4. Ouroriginal chain of length 3 also becomes chain of length 4. So we get

J = J4(λ)⊕ J2(λ)⊕ J2(λ)⊕ J1(λ)⊕ J1(λ)

(again, up to some permutation of the Jordan blocks).




Remark: Why ker(B − λI )4? For completeness sake, here’s a quickcontradiction.

Suppose, say, the remaining generalised eigenvector was inker(B − λI )5 but not in ker(B − λI )4. Then ker(B − λI )4 must infact be equal to ker(B − λI )3, so d4 = d3, i.e. d4 − d3 = 0. Yetd5 − d4 = 1. Therefore d5 − d4 > d4 − d3, which cannot happen.



Invalid nullities

The property d1 − d0 ≥ d2 − d1 ≥ d3 − d2 ≥ . . . helps determinethings that are impossible.

Example 6: David Angell’s MATH2601 notes

Let A be a matrix with eigenvalue λ. Explain why this is notpossible:

nullity(A− λI ) = 5,

nullity(A− λI )2 = 8,



nullity(A− λI )k = 12 for all k > 4.

Answer: d4 − d3 = 3 > 1 = d3 − d2, which can’t happen.



From Jordan forms back to nullities

Example 7: Peter Brown’s MATH2501 notes

If A is similar to J = J2(−4)⊕ J2(−4)⊕ J2(−4)⊕ J3(5)⊕ J1(5).find

nullity(A + 4I )k and nullity(A− 5I )k

for all k ≥ 1.

Solution: Go backwards!




We know the lengths of the chains...




So we see that:

nullity(A + 4I ) = 3

nullity(A + 4I )k = 6 for all k ≥ 2

nullity(A− 5I ) = 2

nullity(A− 5I )2 = 3

nullity(A− 5I )k = 4 for all k ≥ 3



Remark: T -invariance (MATH2601)

I’ve seen questions on this pop up in tutorials and exams, so I’ll givean example involving proving this result. I won’t have time to goover it in class though.

Definition 8: Invariance under T

A subspace U of V is said to be invariant under a transformation Tif T (U) ⊆ U.

Example 8: MATH2601 2018 Q3 b)

Let V be a vector space, let S and T be linear transformations fromV to V , and write W = ker(S − T ). Show that if ST = TS thenW is invariant under T .




Let V be a vector space and let S and T be linear transformationsfrom V to V . Let W = ker(S − T ) and suppose that ST = TS .

Let v ∈ T (W ). Then v = T (w) for some w ∈W .

Goal: Show that v ∈W = ker(S − T ), i.e. (S − T )(v) = 0 .

Then,

(S − T )(v) = S(v)− T (v)

= S(T (w))− T (T (w))

= T (S(w))− T (T (w))

since ST = TS .




Further, since T is linear,

(S − T )(v) = T (S(w)− T (w))

= T ((S − T )(w)) .

But since w ∈W = ker(S − T ), we know that (S − T )(w) = 0.Hence

(S − T )(v) = T (0)

= 0.

Therefore v ∈W , so T (W ) ⊆W and henceW is invariant under T .


The Jordan Canonical Form The Cayley-Hamilton Theorem

The Companion Matrix (MATH2501)

The companion matrix allows us to go backwards from acharacteristic polynomial to a matrix. (Or at least, one such matrix.)

Definition 9: Companion matrix

Consider the polynomial

f (λ) = λn + an−1λn−1 + an−2λ

n−2 + · · ·+ a1λ+ a0.

A matrix C whose characteristic polynomial is f (λ) is

C =

0 0 . . . 0 0 −a0

1 0 . . . 0 0 −a1

0 1 . . . 0 0 −a2...

. . ....

0 0 . . . 1 0 −an−2

0 0 . . . 0 1 −an−1



The Companion Matrix (MATH2501)

Example: The companion matrix corresponding top(λ) = λ3 − 3λ2 + 2λ− 5 is

C =

0 0 51 0 −20 1 3

.

Here, we set a0 = −5, a1 = −2 and a2 = 3.



Recursively finding matrix powers

Theorem 4: The Cayley-Hamilton Theorem

Let A be an n × n matrix and f (z) be its characteristic polynomial.

Then f (A) = 0 , the zero matrix.


1 Verify the Cayley-Hamilton Theorem for A =

(1 34 −2

)2 Use the Cayley-Hamilton Theorem to express A4 and A−1 in

terms of A and I , where I is the 2× 2 identity matrix.




Part 1: Begin by computing

cpA(z) =

∣∣∣∣1− z 34 −2− z

∣∣∣∣= (z − 1)(z + 2)− 12

= z2 + z − 14

Then observe that

cpA(A) =

(13 −3−4 16

)+

(1 34 −2

)− 14

(1 00 1

)=

(0 00 0

)where we note that the constant term gets multiplied to theidentity matrix.




Part 2: Using the Cayley-Hamilton Theorem, we know thatA2 = −A + 14I . Hence

A3 = −A2 + 14A

= −(−A + 14I ) + 14A

= 15A− 14I

A4 = 15A2 − 14A

= 15(−A + 14I )− 14A

= −29A + 210I

Also A = −I + 14A−1, so A−1 = 114A + 1

14 I .



Minimal polynomials (MATH2501)

Note: This has been taken out of the higher syllabus.

Definition 10: Minimal polynomial of a matrix

Let A be an n × n matrix. The minimal polynomial m of A is thepolynomial:

of smallest degree possible

and monic (i.e. the leading coefficient is 1)

such that m(A) = 0.

Lemma 6: Minimal polynomials and characteristic polynomials

The minimal polynomial is a factor of the characteristic polynomial.(Not really useful for computations, but it can be a nice sanitycheck.)

We won’t delve much into the theory, we just illustratehow to find it.




Theorem 5: Explicit form for the minimal polynomial

Let A be an n × n matrix and denote the distinct eigenvalues of Aas λ1, λ2, . . . , λr .

For the i-th eigenvalue λi , let bi be the size of the largest Jordanblock corresponding to λi .

Then the minimal polynomial of A is

m(z) = (z − λ1)b1(z − λ2)b2 . . . (z − λr )br .





The Jordan form of A ∈ M15×15 is

J5(2)⊕ J2(2)⊕ J3(−2)⊕ J3(−2)⊕ J2(−2).

What is its minimal polynomial?

The largest block for λ = 2 has size 5, and the largest block forλ = −2 has size 3. Therefore

m(z) = (z − 2)5(z + 2)3.





The matrix A =

3 5 −4−2 −4 4−1 −3 4

has characteristic polynomial

cpA(z) = z(z − 1)(z − 2).

What is its minimal polynomial?

The characteristic polynomial shows that the 3× 3 matrix A hasthree distinct eigenvalues, so it must be diagonalisable. HenceJ = J1(0)⊕ J1(1)⊕ J1(2), so for this matrix,

m(z) = cpA(z) = z(z − 1)(z − 2).


Matrix Exponentials

Today’s plan





Matrix Exponentials Computing Matrix Exponentials

Matrix Exponential

Definition 11: Exponential of a matrix

The matrix exponential exp(tA) is defined as

exp(tA) = etA =∞∑k=0

tk

k!Ak .



Computing matrix exponentials

We will illustrate the ideas...

Lemma 7: Properties of matrix exponentials

1 If A = PBP−1, then exp(A) = P exp(B)P−1.

2 If A = A1 ⊕ · · · ⊕ An, then exp(A) = exp(A1)⊕ · · · ⊕ exp(An)

3 exp (tJk(λ)) = eλt

1 t t2

2! . . . tk−1

(k−1)!

0 1 t

0 0 1. . .

.... . .

0 0 0 . . . t0 0 0 . . . 1

Also nice to note is that if AB = BA, thenexp(A) exp(B) = exp(A + B).




Example for a Jordan block:

exp (tJ5(2)) = e2t

1 t t2

2!t3

3!t4

4!

0 1 t t2

2!t3

3!

0 0 1 t t2

2!0 0 0 1 t0 0 0 0 1

We’re really filling the matrix in with terms from the power series ofet , but then leaving a usual exponential in front.




Example 12: Not really an example...

Consider C =

9 7 −3−2 2 12 5 2

from earlier. We want exp(tC ).

We have

P =

−1 1 11 −2 01 −3 2

and J =

5 1 00 5 00 0 3

.




The earlier results show that we can do powers of Jordan blocksone at a time. So we obtain

exp(J) =

e5t te5t 00 e5t 00 0 e3t

and hence

exp(C ) = P

e5t te5t 00 e5t 00 0 e3t

P−1



A huge pain, as you can see.

So you probably won’t be asked to do that in an exam. But youmay be asked something else.


Matrix Exponentials Application to Systems of Differential Equations

The ‘Columns’ technique

Theorem 6: Matrix Exponential times Generalised Eigenvector

If we have the Jordan chain

v1A−λI−−−→ v2

A−λI−−−→ v3A−λI−−−→ . . .

A−λI−−−→ vk

then

exp(tA)v1 = eλt(v1 + tv2 +

t2

2v3 + · · ·+ tk−1

(k − 1)!vk

)



The ‘Columns’ technique

This does come with a caveat in that v1 must be a generalisedeigenvector corresponding to λ.

(Otherwise, we have to decompose it into a sum of generalisedeigenvectors first.)



Solving Homogeneous systems of DEs

More often than not, we just need to compute exp(tA)v for somevector v, instead of the actual matrix exponential itself.

Theorem 7: Solution to a homogeneous system

The solution to dydt = Ay with initial condition y(0) = c is

y = exp(tA)c.





Recall for C =

9 7 −3−2 2 12 5 2

and v =

1−2−3

we have the chain

1−2−3

C−5I−−−→

−111

C−5I−−−→ 0.

Use this to solve the initial value problem y′ = Cy, y(0) = v.




The solution is

y = exp(tA)

1−2−3

.

So using the columns method,

y = e5t

1−2−3

+ t

−111

= e5t

1− t−2 + t−3 + t



The more general case (if time permits)

In general, if we can decompose c into a sum of generalisedeigenvectors, we work our way around this issue.


For A =

3 1 −22 6 −72 2 −2

, solve y′ = Ay with initial condition

y(0) =

210

, given that

210

=

352

︸︷︷︸∈GE2

+

−1−4−2

︸︷︷︸∈GE3

.



The more general case (if time permits)

Construct the chains:352

A−2I−−−→

4128

A−2I−−−→ 0

−1−4−2

A−3I−−−→ 0

Our solution will thus be

y = etA

352

+ etA

−1−4−2

= e2t

3 + 4t5 + 12t2 + 8t

+ e3t

−1−4−2

.



Solving Non-homogeneous systems of DEs

Lemma 8: Solution to non-homogeneous systems

The general solution to y′ = Ay can be expressed as y = yH + yPwhere

yH is the general solution to y′ = Ay

yP is any particular solution to y′ = Ay + b

Lemma 9: Variation of parameters

We can approach the particular solution by subbing y = etAz.

Lemma 10: Derivative of matrix exponential

d

dtetA = AetA



Variation of parameters

In general, upon substituting in y = etAz intody

dt= Ay + b, we

have

d(etAz)

dt= AetAz + b

etAz′ + AetAz = AetAz + b

etAz′ = b

z′ = e−tAb

So what we can do is:

1 Find z′, probably using the columns technique again.

2 Integrate out to find z.

3 Recompute y = etAz for our particular solution.





Find a particular solution of y′ = Cy + te5tw, where

C =

9 7 −3−2 2 12 5 2

and w =

−82−4

, given that w is a

generalised eigenvector of C .

Subbing y = etCz gives

CetCz + etCz′ = CetCz + te5tw

z′ = te5te−tCw




We need to construct a Jordan chain starting at w first:−82−4

C−5I−−−→

−666

C−5I−−−→

000

so therefore

etCw = e5t

−82−4

+ t

−666

.But observe how we want the negative exponent e−tC ! This meanswhat we’re really interested in is

e−tCw = e−5t

−82−4

− t

−666




Therefore

z′ = te5te−tCw = t

−82−4

− t2

−666

so upon integrating,

z =t2

2

−82−4

− t3

3

−666

+ c

Question: How to deal with that constant of integration?




In general, you can only deal with it when you know what y(0) is,i.e. you have an initial value for the original systems of DEs. Whenthat’s the case, you let z(0) = y(0) to find it.

Here we don’t, so we just proceed as usual.

yP = etCz =t2

2etC

−82−4

− t3

3etC

−666

+ etCc.




To finish this off, we can recycle our Jordan chain from earlier:−82−4

C−5I−−−→

−666

C−5I−−−→

000

.

This gives us

yP =t2

2e5t

−82−4

− t

−666

− t3

3e5t

−666

+ etCc

The remainder is trivial and is left as an exercise to the audience.




To finish this off, we can recycle our Jordan chain from earlier:−82−4

C−5I−−−→

−666

C−5I−−−→

000

.

This gives us

yP =t2

2e5t

−82−4

− t

−666

− t3

3e5t

−666

+ etCc

The remainder is trivial and is left as an exercise to the audience.




Note: The harsh reality is that if we knew what c was, that wouldpotentially be another Jordan chain we need to deal with. Fingerscrossed you don’t have to do that in your exam.


Final remark: One single eigenvalue

When you’re told that the matrix A only has one eigenvalue, youcan take care of things more easily. The following comments assumen = 3, but the analogy can be adapted for all n × n matrices.

Use the trace to find that eigenvalue λ.

You automatically know that GEλ = C3, so it’s less difficult toconstruct a Jordan chain. Find ker(A− λI ), and onlyker(A− λI )2 if you don’t already have two eigenvectors.

Then, just pick a third vector out of thin air, not linearlyindepedent to the other two. Construct a chain using thatvector.

Done!

You’ll see this in all the examples in your tutorials...


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