Distinct Eigen Valueslnlkjh

8/12/2019 Distinct Eigen Valueslnlkjh

1/16

Chapter 6

Eigenvalues and Eigenvectors

6.1 Eigenvalues and the Characteristic Equation

Given a n n matrix A, if

Av= v, v = 0 (6.1.1)

where is a scalar and v is a non-zero vector, is called an eigenvalueofA and v an eigenvector. It is important here that an eigenvector should bea non-zero vector. For the zero vector, (6.1.1) is always true for any , andso, it is not very interesting. Let us look at some examples.

Example 15. Let

A=

2 11 2

(6.1.2)

Then, we have:

2 11 2

11

= 3

11

,

2 11 2

11

=

11

. (6.1.3)

We thus see that 3 and 1 are eigenvalues of the matrix A and that thecorresponding eigenvectors are (1, 1)T and (1,1)T. An important thing tonote here that the eigenvector corresponding to 3, say, is not unique. Anynon-zero scalar multiple of (1, 1)T is also an eigenvector corresponding tothe eigenvalue3.

Now consider:

A=

2 1 11 2 1

1 1 2

(6.1.4)

77


2/16

In this case, the eigenvalues are4 and1, and the eigenvectors are:

A

11

1

= 4

11

1

, A

11

0

=

11

0

, A

101

=

101

. (6.1.5)

Note here that there are two linearly independent eigenvectors for the eigen-

value1.

Given a matrix A, how can we find the eigenvalues ofA? We argue asfollows. Equation (6.1.1) can be written as:

Av

Iv= 0, (6.1.6)

where Iis the identity matrix. We thus have:

(A I)v= 0. (6.1.7)

Now, note that v is a non-zero vector. The above equation states that anon-zero vector is being sent to the origin. Therefore, the matrixAIhasa non-trivial kernel, and this implies thatAI is not invertible. Therefore,

det(A I) = 0. (6.1.8)

This gives us an equation for the eigenvalues. The above equation is called

the characteristic equationand det(AI) is called the characteristic poly-nomial. Let us state this as a theorem.

Theorem 14. Given annnmatrixA, is an eigenvalue ofA if and onlyif it satisfies the following characteristic equation:

det(A I) = 0. (6.1.9)

Once the eigenvalues are found, we may find the eigenvectors by solvingthe equation:

(A I)v= 0 (6.1.10)

viewing v as the unknown. This is just a homogeneous linear equation,which we know very well how to solve.

Example 16. Consider the matrix:

A=

1 21 4

. (6.1.11)

MATH 2574H 78 Yoichiro Mori


3/16

The characteristic equation is given by:

det

1 21 4

= (1)(4) + 2 =2 5 + 6 = ( 2)( 3) = 0.

(6.1.12)Therefore, the eigenvalues of this matrix are2 and 3. To obtain the eigen-vector for= 2, we solve the equation:

1 21 2

ab

=

00

(6.1.13)

from which we see that the solution is given by:

ab

=c

21

. (6.1.14)

Thus, one eigenvector corresponding to= 2is(2, 1)T. In a similar fashion,we see that an eigenvector corresponding to = 3 is(1, 1)T.

Consider the following example of a3 3 matrix:

A=

3 0 10 3 0

1 0 2

. (6.1.15)

The characteristic equation is:

det

3 0 20 3 0

1 0 2

= (3 )((3 )(2 ) 1) = 0. (6.1.16)

Solving this equation, we see that the eigenvalues are= 1, 3, 4. The eigen-value for= 1 can be obtained by solving the equation:

2 0 20 3 01 0 1

abc

=

000

. (6.1.17)

An eigenvector for = 1 is (1, 0,1)T. For = 3, an eigenvector is(0, 1, 0)T and for= 4, one eigenvector is(2, 0, 1)T.

Let us examine the characteristic equation is greater detail. First of all,we have the following result.



4/16

Theorem 15. Given an n

n matrix A, the characteristic polynomial

det(A I) is an-degree polynomial in, and has the form:det(A I) = (1)nn + . (6.1.18)

To prove this statement, we first prove the following.

Proposition 19. Consider annmatrixB whose elements are polynomialsin. Assume further that the polynomials are at most degree1 in. Then,det(B) is a polynomial of degree at mostn.

Proof. We proceed by induction on the size n of the matrix. For n = 1, thisis clearly true. Suppose this is true for n 1. The determinant ofB can bewritten as:

det B = b11det(B11) + (1)n+1b1ndet(B1n) (6.1.19)where bij is the ij component of B and Bij is the ij minor of B. Sinceb1k, k= 1, , n are at most degree 1 and det B1k are at most degree n 1by the induction hypothesis, each term in the above sum is at most degreen. This completes the proof.

We are now ready to prove Theorem 15.

Proof of Theroem 15. We use induction on the size of the matrix n. Forn= 1, the statement is clear. Suppose the statement is true for n 1. LetB = A I. We have:

det B = b11det(B11) + (1)n+1b1ndet(B1n) (6.1.20)where we used the same noation as in (6.1.19). The first term in the abovesum is:

b11det(B11) = (a11 )det(A11 I) (6.1.21)where aij is the ij component of the matrix A and Aij is the ij minor.By the induction hypothesis, det(A11 I) is a polynomial in of degreen 1, and the leading order term is (1)n1n1. Therefore, (6.1.21) is apolynomial in of degree n with leading order term (1)nn. Consider theother terms in (6.1.20):

(1)k+1

b1kdet(B1k) = (1)k+1

a1kdet(B1k), k 2. (6.1.22)By Proposition 19, det(B1k) is a polynomial in of degree at most n 1.Since a1k is just a scalar, (6.1.22) is a polynomial in of degree at mostn 1. Since the first term of (6.1.20) is degree n and the rest of the termsare at most degreen 1 in, det B = det(AI) is a n degree polynomialin .



5/16

Given a n

n matrix A, let p() = det(A

I) be its characteristic

polynomial. It is a well-known fact (that we will not prove) that a degreen polynomial equation has n roots in C counting multiplicities. That is tosay, the characteristic equation can be written as:

det(A I) =p() = (1)n( 1)( 2) ( n), (6.1.23)where i mathbbCare the eigenvalues ofA. This implies that a matrixAcan have at most n different eigenvalues. We state this as a theorem.

Theorem 16. A matrixA has at least one eigenvalue inC and at mostndistinct eigenvalues inC.

One complication here is that eigenvalues are not necessarily going to be

real numbers but are in general complex.


A=

2 11 2

(6.1.24)

The characteristic equation and the eigenvalues for this matrix are:

2 4+ 5 = 0, = 2 i. (6.1.25)The eigenvector corresponding to 2 +i is (i, 1)T and the eigenvector corre-sponding to 2 i is(i, 1)T.

We also point out that it is very possible for a n nmatrix to have lessthan n distinct eigenvalue. We have already seen an example of this for thematrix in equation (6.1.4).

6.2 Diagonalization

Suppose we have a nnmatrix A with n linearly independent eigenvectorsv1, , vn. Each eigenvector satisfies:

Avk =kvk. (6.2.1)

We may form a matrix P = (v1,

, vn) so that the column vectors of Pare the eigenvectors ofA. Let

D=

1 0 00 2 0...

... . . .

...0 0 n

(6.2.2)



6/16

The matrixD is thus a diagonal matrix whose diagonals are the eigenvalues

ofA. Equation (6.2.1) can now be written as:

AP =P D. (6.2.3)

Since the column vectors ofPare linearly independent, P has an inverse,and we have:

P1AP =D. (6.2.4)

This is called the diagonalization of a matrix. From the above calculation,we have the following result.

Theorem 17. A nn matrixA can be diagonalized if and only ifA hasnlinearly independent eigenvectors.Proof. If A has n linearly independent eigenvectors, we saw above that Acan be diagonalized. IfA can be diagonalized, there is an invertible matrixP such that P1AP is a diagonal matrix. The column vectors ofP are nlinearly independent eigenvectors ofA.

All examples of matrices we have seen so far in this chapter can bediagonalized.


A=

1 3

2 0

(6.2.5)

Its eigenvalues and eigenvectors are:

A

11

= 2

11

, A

32

= 3

32

. (6.2.6)

The two eigenvectors are linearly independent. Let

P =

1 31 2

. (6.2.7)

We may useP to diagonalizeA as follows:

P1AP =D =

2 00 3

. (6.2.8)

There are cases in which the matrix does not have nlinearly independenteigenvectors. In such a case,Acannot be diagonalized.



7/16


A=

2 10 2

. (6.2.9)

The characteristic equation is(2)2 = 0, and therefore, the only eigenvalueis2. To find the eigenvector corresponding to 2, we must solve:

0 10 0

ab

=

00

. (6.2.10)

An eigenvector is(1, 0)T, and there are no other linearly independent eigen-

vectors.

There is large class of matrices for which diagonalizability is guaranteed.

Proposition 20. Suppose a nn matrix A has n distinct eigenvalues.Then, A hasn linearly independent eigenvectors, each corresponding to dif-ferent eigenvalues. The matrixA is thus diagonalizable.

Proof. Let1, , nbe then distinct eigenvalues ofA. Take one eigenvec-tor vk for each eigenvalue k so that

Avk =kvk. (6.2.11)

Consider the linear combination:

c1v1+ +cnvn= 0. (6.2.12)

We now show that all ck must be equal to 0. This shows thatvk are linearlyindependent, and thusAis diagonalizable. To conclude thatc1is 0, multiplyboth sides of (6.2.12) from the left by the matrix:

(A 2I)(A 3I) (A nI). (6.2.13)

Since (A kI)vk = 0, all terms disappear except for the first term:c1(1 2)(1 3) (1 n)v1= 0. (6.2.14)

Since all the k are distinct and v1= 0, we conclude that c1 = 0. In muchthe same way, we can conclude that all the other ck are also 0.



8/16

6.3 Matrix Powers

One of the reasons why eigenvalues/eigenvectors of diagonalization is souseful is that it allows us to compute matrix powers (or more generallymatrix functions) very easily. The idea is the following. Suppose we havesuccessfully diagonalized a matrixA as follows:

P1AP =D, D=

1 0 00 2 0...

... . . .

...0 0 n

(6.3.1)

We are interested in computing the matrix power Ak. Using the above,

A= P DP1. (6.3.2)

Therefore,Ak =P DP1P DP1 P DP1 =P DkP1 (6.3.3)

since all the intervening P1P =I. The beautiful thing here is that Dk isvery easy to compute. It is just:

Dk =

k1 0 00 k2 0...

... . . .

...

0 0 kn

. (6.3.4)

Example 20. Let us look at matrix powers of the matrix in Example (18).Using the same notation, we have:

Ak =P DkP1 =

1 31 2

2k 00 (3)k

1

5

2 31 1

=1

5

2k+1 (3)k+1 3 2k + (3)k+12k+1 2 (3)k 3 2k + 2 (3)k+1

.

(6.3.5)

6.4 Traces and Determinants

We explore connections between eigenvalues and the determinant.

Proposition 21. LetA be annmatrix and1, , n be the eigenvaluesofA, counting multiplicities. Then,

det(A) =12 n. (6.4.1)



9/16

Proof. Consider the characteristic polynomial:

det(A I) = (1)n( 1)( 2) ( n). (6.4.2)

Set = 0 to obtain the desired result.

The determinant is thus just a product of the eigenvalues.

Another interesting quantity in this connection is the trace. The traceis the sum of the diagonal components of a matrix. For a n n matrix Awhose elements are given by aij,

Tr(A) =a11+a22+ +ann. (6.4.3)

One interesting property of the trace is the following. Given two n nmatrices Aand B ,

Tr(AB) = Tr(BA). (6.4.4)

This can be seen by writing out both sides in terms of the components. Letaij and bij be the components ofA and B respectively. Then,

Tr(AB) =

ni=1

nj=1

aijbji =

nj=1

ni=1

bjiaij = Tr(BA). (6.4.5)

This can be used to establish the following fact.

Proposition 22. The trace of an n n matrix A is the sum of its eigen-values, counting multiplicities.

Proof. We only prove this for matrices withn distinct eigenvalues (althoughin fact, the statement is actually true in general). In this case, A is diago-nalizable, so that

P1AP =D, (6.4.6)

where D is a diagonal matrix with the n distinct eigenvalues ofA along thediagonal. Take the trace on both sides. We have:

Tr(P1AP) = Tr(D) =1+ +n. (6.4.7)

Note that:

Tr(P1AP) = Tr(AP P1) = Tr(A) (6.4.8)

where we used (6.4.4).



10/16

6.5 Classification of 2

2 matrices by similarity

LetA and B be two n n matrices related by:

P1AP =B (6.5.1)

we say thatAandB are similarmatrices. We have seen that, in many cases,we can find a matrix P so that B is a diagonal matrix. However, in somecases, this is not possible. Even if it is possible, it is sometimes the case thatthe eigenvalues are complex conjugates, in which case diagonalization willproduce a complex matrix. Here, we try to see how far we can go, only withreal matrices. This will yield a classification of 2 2 matrices by similarity.

Let the eigenvalues of the 2 2 matrix A be 1 and 2. When 1=2and real, we know that it is diagonalizable with a real matrix P. In thiscase, we have:

P1AP =

1 00 2

. (6.5.2)

The matrixA, from a geometric standpoint, is a transformation that stretchesthe plane in the eigenvector directions by a factor of1 and2 respectively.

When1= 2, there are two situations. It may happen that the matrixA still has two linearly independent eigenvectors v1 and v2. In this case,

P1AP =1I, P = (v1, v2) (6.5.3)

where I is the 2 2 identity matrix. This means that:

A= 1P IP1 =1I (6.5.4)

Therefore, if1 = 2 and A has two linearly independent eigenvectors, A isnecessarily a constant multiple of the identity matrix.

Suppose 1 = 2 and that A has only one linearly independent eigen-vector. Let this vector be v1. We now seek a vector v2 that satisfies:

(A 1I)v2= v1. (6.5.5)

The claim is that there is such a vector v2 and that v1 and v2 are linearlyindependent. To see that the above equation can be solved, we need toshow that v1 is in the image or (A1I). This can be seen as follows. Anyelement w in the image of (A 1I) can be written as:

w= (A 1I)v (6.5.6)



11/16

for some v. By the Cayley-Hamilton theorem, we know that:

(A 1I)2 =O (6.5.7)where O is the 2 2 zero matrix. Therefore,

(A 1I)w= (A 1I)2v= 0. (6.5.8)This means that w is in the kernel of (A 1I). The image ofA 1I isthus included in the kernel ofA 1I. Since the kernel ofA 1I is one-dimensional (Ahas only one linearly independent eigenvector), the image ofA 1Iis also one dimensional. This means that the kernel and image ofA 1I must coincide. Since the eigenvector v1 is in the kernel ofA 1I,this means that (6.5.5) must have a solution. It is also clear that v2 and v1

must be linearly independent. For if not, v2 will be a constant multiple ofv1 and will thus be mapped to 0 byA 1I. We have, therefore:

Av1 = 1v1, Av2 = 1v2+ v1. (6.5.9)

From this, we have:

P1AP =

1 10 1

, P = (v1v2). (6.5.10)

Let us consider matrix powers. Let

B =

1 10 1

(6.5.11)

We can see that B can be written as:

B = 1I+N, N =

0 10 0

. (6.5.12)

Now, using the binomial theorem and noting thatN I=I N, we see that

Bk = (1I+N)k =k1I

k +kk11

Ik1N+k(k 1)

2 k2

1 Ik2N2 + .

(6.5.13)Since N2 = 0, all terms except for the first two terms are just the zeromatrix. therefore,

Bk =k1I+kk11

N =

k1 kk11

0 k1

. (6.5.14)

Now that we know the matrix powers of B, we can compute the matrixpower ofA in (6.5.10) as:

Ak =P BkP1. (6.5.15)



12/16


A=

1 11 3

. (6.5.16)

The characteristic equation is:

2 4+ 4 = ( 2)2 = 0. (6.5.17)

Therefore, = 2 is the only eigenvalue, and the eigenvector correspondingto this value is:

v1=

11

. (6.5.18)

We now must solve:(A 2I)v2 = v1, (6.5.19)

and one solution to this equation is

v2=

10

. (6.5.20)

Therefore, we have:

P1AP =

2 10 2

, P =

1 11 0

. (6.5.21)

The matrix powers ofA can be computed as:

Ak =P

2k k2k1

0 2k

P1 =

2k k2k1 k2k1

k2k1 2k +k2k1

. (6.5.22)

Let us now turn to the case when the 2 2 matrix A has two complexconjugate eigenvalues. In this case, the two eigenvalues are:

= a+ib, = a ib (6.5.23)

where a and b are real numbers. The eigenvector corresponding to can bewritten as:

Av= v, Av= vv = u +iw, v= u iw (6.5.24)

where u and w are real vectors. We see from the above that

Au= 1

2A(v+v) =

1

2(v+v) =au bw. (6.5.25)



13/16

Likewise,

Aw= 12i

A(v v)12i

(v v) =bu +aw. (6.5.26)This shows that:

AP =P B , B =

a bb a

, P = (w, u). (6.5.27)

(You may have realized that we are letting P = (wu) instead ofP = (uw).It is slightly more convenient to let P = (wu)). In fact, the matrix P isinvertible. This can be seen as follows: v and v must be linearly indepen-dent (they correspond to different eigenvalues) and therefore, u and w mustbe linearly independent. Therefore, any matrix with complex conjugate

eigenvalues can be written as:

P1AP =B, B =

a bb a

. (6.5.28)

Recall that we have seen the matrix B before. This is just a composition ofmagnification and rotation. Therefore, any matrix with complex conjugateeigenvalues can be seen as a linear transformation involving magnificationand a rotation.

Now, we may consider the matrix powers of A. Note that B can bewritten as:

B =

a2 +b2

cos() sin()sin() cos()

, cos() = aa2 +b2

, sin() = ba2 +b2

.

(6.5.29)This is just the statement that B is magnification with rotation. Therefore,

Bk = (

a2 +b2)k

cos(k) sin(k)sin(k) cos(k)

. (6.5.30)

Using this, we may compute Ak as:

Ak =P AkP1. (6.5.31)

Example 22. Take the matrix

A=

2 12 0

(6.5.32)

The eigenvalues of this matrix are

= 1 i. (6.5.33)



14/16

The eigenvector corresponding to 1 +i is:

v=

1

1 i

. (6.5.34)

Therefore, if we set:

u=

11

, w=

01

, P = (wu) (6.5.35)

we have:

P1AP =B, B =

1 11 1

(6.5.36)

Now

B =

2

cos(/4) sin(/4)sin(/4) cos(/4)

. (6.5.37)

Thus,

Bk = (

2)k

cos(k/4) sin(k/4)sin(k/4) cos(k/4)

. (6.5.38)

Using this fact, we may computeAk as:

Ak =P BkP1 = (

2)k

cos(k/4) + sin(k/4) sin(k/4)2sin(k/4) cos(k/4) sin(k/4)

.

(6.5.39)

6.6 Exercises

1. Find the eigenvalues and eigenvectors of the following matrices.

3 11 3

,

1 11 1

,

2 14 1

,

2 11 4

,

2 21 0

3 1 11 3 11 1 3

,

1 1 20 1 10 0 2

,

0 1 00 0 1

1 0 0

,

2 1 0 11 2 1 00

1 2 1

1 0 1 2

(6.6.1)

2. Diagonalize the 2 2 matrices in problem 1, if possible.3. Find the matrix powers of the 2 2 matrices in problem 1.



15/16

4. Given a 2

2 matrix A, show that its characteristic equation can be

written as:2 Tr(A)+ det(A) = 0. (6.6.2)

Use this to prove that for 2 2 matrices, the sum of the eigenvaluesgives the trace and the product of eigenvalues gives the determinant.

5. Show that, for 2 2 matrices A and B, AB and B A always have thesame eigenvalues. Hint: Use the result from previous question.

6. Consider a 2 2 matrix of the form:

A=

a bb c

(6.6.3)

wherea,b, c are real numbers and b = 0. Show that the eigenvalues ofthis matrix are always real. Show further that the matrix always hastwo distinct eigenvalues and that the two eigenvectors are orthogonalto each other.

7. Consider two n n matrices A and B that commute. That is to say,AB =BA. Show that ifv is an eigenvector ofA, then Bv is also aneigenvector ofA, provided B v =0.

8. Let be one eigenvalue of an n n matrix A. Consider the set:

All vectors v Rn

such that Av= v. (6.6.4)

Show that this set is a subspace. This subspace is called theeigenspaceof.

9. A n n matrix Asatisfies:

A2 3A+ 2I=O (6.6.5)

where O is the zero matrix. What are the possible eigenvalues ofA?(Hint: Letv be an eigenvector and apply the above expression to v.

10. Find all 2

2 matrices that satisfy A2 = O. Hint: First of all, whatare its eigenvalues?

11. Consider the matrix:

A=

a 1 ab 1 b

(6.6.6)

where 0< a < 1 and 0< b < 1.



16/16

(a) Find the eigenvalues and eigenvectors of this matrix.

(b) Find the matrix power ofA.

(c) Compute:limk

Ak. (6.6.7)


Documents

Distinct Eigen Valueslnlkjh