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Home , Algebra, Linear algebra

Math 113: Linear Eigenvectors and Eigenvalues

Ilya Sherman

October 31, 2008

1 Finding Eigenvalues and Eigenvectors

Our goal is, given T (V, V ), to !nd bases that are “well-adapted” to T . In particular, it ∈ L would be nice to have ei so that T ei = λiei.

De!nition 1 (Eigenvalue, eigenvector). An eigenvector is a nonzero vector v V so that ∈ T (v) = λv for some λ F . An eigenvalue is any λ F that occurs in this way. ∈ ∈

Last time, we showed that if F = , there is always an eigenvector. In fact, λ0 is an eigenvalue if and only if λ Id A is not an , i.e. det(λ Id A) = 0. Recall that this follows from 0 − 0 − the fact that det(λ Id A) is a in λ of degree n, so any complex root of det(λ Id A) − − is an eigenvalue.

Recall that if F = R, there doesn’t have to be an eigenvector (consider, for example, a in R2). Recall also that repeated roots of the polynomial can result in fewer than n distinct eigenvalues, and hence fewer than n linearly independent eigenvectors. Note that this proof also shows that if F = R and n is odd, then any A (V, V ) has an ∈ L eigenvector (because any polynomial of odd degree has a real root). Axler gives a di"erent proof of this property:

Proposition 1. For F = C, any A (V, V ) has an eigenvector. ∈ L

Proof. Take v = 0. Consider v, Av, A2v, . . . , Anv. These are n + 1 elements in V . Since the # of V is n, there exist a F so that i ∈ n i aiA v = 0, i=0 ! with not all ai = 0.

1 Ilya Sherman Math 113: Eigenvectors and Eigenvalues October 31, 2008

a0 If n = 1, then this boils down to a1Av + a0v = 0, i.e. Av = v, so v is an eigenvector. In − a1 general, we’ll reduce to the linear case by factoring the sum into linear factors. Let p by the polynomial with complex coe"icients with

n i p(z) = aiz . i=0 ! It factors: p(z) = C (z z ), − j where the zj are the roots. Then, we can “substitute”" A for z, as

p(A) = C (A z Id). − j " This is because all the algebraic rules used to verify the original factoring (e.g. the ) remain valid when substituting A for z. For more detail, check out “ applied to operators” in Axler, Chapter 5. Warning: note that is not in general commutative; but fortunately, the only matrices we’re multiplying are A and Id, which commute with one another. Since p(A)v = 0, (A z Id)v = 0, i.e. (A z Id) is not injective, i.e. there exists some j − j − j so that A zj Id is not injective, and hence, zj is an eigenvalue. − # #

Exercise: work out the proof numerically to !nd an eigenvalue of a 2 2 matrix. × Note that the !rst proof we gave of this property gave us all of the eigenvalues: they are exactly the roots of the polynomial, det(λ Id A). In contrast, this proof only tells us one − eigenvalue.

2

Proposition 2. Suppose v1, . . . , vk are eigenvectors of A with eigenvalues λ1, . . . , λk distinct, i.e. λ = λ for i = j. Then, v are linearly independent. In particular, if the characteristic i # j # i polynomial of A has n distinct roots, then there exists a v1, . . . , vn of eigenvectors for V .

Proof. First of all, note that the second part follows immediately from the !rst: If the character- istic polynomial has n distinct roots, λ1, . . . , λn, then all of the λi are eigenvalues. Let v1, . . . , vn be the corresponding eigenvectors. By the !rst assertion, they are linearly independent, so they are a basis. Now, suppose we have some linear

α v + α v + + α v = 0, 1 1 2 2 · · · k k

Page 2 of 3 Ilya Sherman Math 113: Eigenvectors and Eigenvalues October 31, 2008

where not all ai are 0. Apply A:

α λ v + α λ v + + α λ v = 0. 1 1 1 2 2 2 · · · k k k

Subtracting λ1 times the !rst from the second equation gives

α (λ λ )v + + α (λ λ )v = 0. 2 2 − 1 2 · · · k k − 1 k k But, we could choose the original i=1 αivi = 0 to have the smallest possible of nonzero αi (among all possible linear relations). The reasoning above shows that one $ can produce a shorter linear relation (fewer nonzero αi’s), though we need to make sure that rather than just eliminating α , we choose α = 0 to eliminate. This is a contradiction. 1 j # Another way to show this is to repeatedly apply A to

α v + α v + + α v = 0, 1 1 2 2 · · · k k where not all ai are 0. Apply A repeatedly:

α v + α v + + α v = 0 1 1 2 2 · · · k k α λ v + α λ v + + α λ v = 0 1 1 1 2 2 2 · · · k k k α λ2v + α λ2v + + α λ2v = 0 1 1 1 2 2 2 · · · k k k . . α λmv + α λmv + + α λmv = 0 applying A m times 1 1 1 2 2 2 · · · k k k . .

2 k 1 k In an earlier , we showed that (1, λi, λi , . . . , λi − ) are linearly independent vectors in F for each 1 i k (while studying polynomial interpolation). Hence, these vectors are a basis k ≤ ≤ 2 2 k 1 k 1 for F . This implies that (1, 1, 1, . . . , 1), (λ1, . . . , λk), (λ1, . . . , λk), . . . , (λ1− , . . . , λk− ) (because in matrices, row is equal to column rank). Hence, there exist βi so that

k 1 k 1 (1, 1, . . . , 1)β + (λ , . . . , λ )β + + (λ − , . . . , λ − )β = (1, 0, 0, 0, . . . , 0). 1 1 k 2 · · · 1 k k Taking a of , we see that a v = 0 a = 0; similarly all a v = 1 1 ⇒ 1 i i 0 a = 0. ⇒ i The thing to learn from this proof is that considering matrix powers A, A2, . . . is a useful way of studying eigenvalues and eigenvectors.

Note that the condition that the characteristic polynomial has n distinct roots is usually satis- !ed, in the sense that if we choose a matrix at random, most of the time this is true.

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