Eighan values and diagonalization

6.1 Eigenvalues and Diagonalization

DefinitionsA is n x n. λ is an eigenvalue of A if

AX = λX has non zero solutions X (called eigenvectors)

If λ is an eigenvalue of A, the set Eλ = Eλ(A) = {X | X in ℜn, AX = λX}is a vector space called the eigenspace associated w/ λ

(i.e. Eλ is all eigenvectors corresponding to λ & 0 vector)λ is eigenvalue if Eλ has at least one non-zero vector.Can also write AX = λX as (λIn - A)X = 0

Example

Show that λ = -3 is an eigenvalue of A, and find the

eigenspace E-3. A =

5 8 16

4 1 8

−4 −4 −11

Write out (λIn - A)X = 0 and solve.

Get: X = s

−1

1

0

+ t

−2

0

1

So it is an eigenvalue since there is a non-zero solution. Eigenspace is:

E−3 = span

−1

1

0

,

−2

0

1

Discussion

Now we have (λIn - A)X = 0, and λ is an eigenvalue iff there exists a nonzero solution X.

Recall that a matrix U is invertible iff UX = 0 implies X = 0.

So, since we are looking for a nonzero solution above,

(λIn-A) cannot be invertible for λ to be an eigenvalue.

So det (λIn-A) =0.

Definition

The characteristic polynomial of the n x n matrix A is:cA(x) = det(xI - A)

Theorem 1

A (n x n). The eigenvalues of A are the real roots of the characteristic polynomial of A --the real numbers λ satisfying:

cA(λ) = det(λIn - A) = 0The eigenspace Eλ = {X | (λI - A)X = 0} consists of all solutions to a system of n linear equations in n variables.

The eigenvectors corresponding to λ are the nonzero vectors in the eigenspace.

Summary

So there are two issues: finding eigenvalues, and finding eigenspaces (and eigenvectors).

Finding the eigenvalues can be difficult - won’t do much here.

Spend more time dealing with eigenspaces.

Example

Find the characteristic polynomial, eigenvalues, and eigenspaces of A:

A =1 −2 3

2 6 −6

1 2 −1

Set up cA(x) = det (xI - A)

Eigenvalues will be the roots of the polynomial as those will give us where det is 0.

Then use those λ to find eigenspace: X such that (λ I-A)X=0

ExampleIf A is a triangular matrix, show that the eigenvalues of A are the entries on the main diagonal.

Proof: cA(x) = det (xI - A) = det ( a triangular matrix) = product of entries on main diagonal of (xI - A).

The matrix showing entries on main diagonal is:

x − a

11

x − a22

...

x − ann

det = (x-a11)(x-a22)…(x-ann)

So eigenvalues are{a11,a22,…,ann}

ExampleShow that A and AT have the same characteristic polynomial and thus the same eigenvalues.

Proof: From chapter 3, we know that a matrix and its transpose will have the same determinant.

cA

T (x) = det(xI − A T ) = det((xI) T − A)T

= det(xI − A)T = det(xI − A) = cA(x)

Theorem 2If A is a real symmetric matrix, each root of the characteristic polynomial cA(x) is real. (to be proven later)

Show this is true for a (2 x 2):

A =a b

b c

cA (x) = detx − a −b

−b x − c

= (x − a)(x − c) − b2

= x 2 − x(a + c) + (ac − b2 )

Recall that we can determine the nature of the roots from the discriminant: (b2-4ac) = (a+c)2-4(ac+b2) = a2+c2+2ac-4ac+4b2

=a2-2ac+c2+4b2 = (a-c)2 + 4b2 which is always pos so real roots.

Similar MatricesA, B (n x n) are similar (we say A~B) if B = P-1AP

holds for some invertible matrix.

P is not unique.

ExampleFind P-1AP in the following case, then compute An.

P =1 5

1 2

, A =

6 −5

2 −1

We are able to find a similar matrix B.

Then P-1AP=B.

So A = PBP-1

So A2=(PBP-1)(PBP-1)=PB2P-1

Generally An=PBnP-1

Life is made easy is B is diagonal since we just raise entries to n.

Interesting FactSimilar Matrices will have the same determinant.

Proof:

P-1AP = D

det(D) = det (P-1AP) = (detP-1)(detA)(detP) = (1/detP)(detA)(det P) = det A.�

ExampleShow that A and B are not similar.

A =1 2

2 1

,B =

1 1

−1 1

Just need to show that they do not have the same determinant.

TraceThe trace of a square matrix A (tr A) is the sum of the entries on the main diagonal of A.

Theorem 3

A,B (n x n), k is a scalar:

1. tr(A + B) = tr A + tr B and tr(kA) = k tr A

2. tr (AB) = tr (BA)

Proof:

1. (homework)

2. AB =

a1j

bj 1

j =1

n

∑

a2 jb j2j=1

n

∑...

anjb

jnj =1

n

∑

tr(AB) = aijb jij =1

n

∑

i =1

n

∑ = b jiaiji =1

n

∑

j =1

n

∑ = tr(BA)

Theorem 4

If A~B, they have the same determinant, the same rank, the same trace, the same characteristic polynomial, and the same eigenvalues. (similarity invariants)

Proof: Already shown that they have the same determinant.

Rank: Have B = P-1AP

rank (B) = rank (P-1AP) = rank(AP)=rankA since P is invertible (and using cor 4 of thm 4 in 5.5)

tr B = tr (P-1AP) = tr[(AP)P-1] = tr (A) (uses thm 3)

Theorem 4 - cont

Characteristic polynomial

cB(x) = det (xI - B) = det(xI - P-1AP)=det(P-1xIP - P-1AP)

(since xI = P-1xIP -- since xI is diagonal)

= det [P-1(xI - A)P]=(1/detP)(det(xI-A))(det P) = det(xI-A) = cA(x)

Eigenvalues: all matrices with the same characteristic poly will have the same eigenvalues since the eigenvalues are the roots of the characteristic polynomial.

�

Fact

The invariants do not imply similarity.

Ex. I =1 0

0 1

, A =

1 2

0 1

Have same det,tr,rank,characteristic poly, eigenvalues, but are not similar since P-1IP = I A≠

Theorem 5

A,B,C (n x n). Then:

1. A~A for all A.

2. If A ~ B, then B~A

3. If A ~ B and B ~ C, then A~C.

Proof of 2 (others follow):

A~B ⇒ B = P-1AP

Let Q = P-1, then B = QAQ-1, so A= Q-1BQ

Which means that B ~ A.

Use of thm 5

Proving similarity is not always easy. But if we can find a simple (often diagonal) matrix to which both A and B are both similar, then: A~D and B~D means D~B by (2)

and A~B then by (3)