Eigenvector Facts

KEY FACTS ABOUT EIGENVECTORS Let A be an n × n matrix. Let f (k) be the characteristic polynomial det(A − k · Id). In this note, we will establish the following facts: Big Fact 1. For a real number λ, there is a (nonzero) eigenvector of A with eigenvalue λ if and only if f (λ) = 0. Big Fact 2. If A has r linearly independent eigenvectors v1 , v2 , . . . , vr , all with eigenvalue λ, then (k − λ)r divides f (k). Given a number λ, the geometric multiplicity of λ is defined to be the dimension of the space of vectors ~v for which A~v = λ~v . The algebraic multiplicity is the exponent of k − λ which occurs in the factorization of f (k). So Big Fact 1 says that the algebraic multiplicity is positive if and only if the geometric multiplicity is positive. Big Fact 2 says that algebraic multiplicity is greater than or equal to geometric multiplicity. Big Fact 3. Let λ1 , λ2 , . . . , λr be distinct zeroes of f , with ~v1 , ~v2 , . . . , ~vr corresponding eigenvectors. Then ~v1 , ~v2 , . . . , ~vr are linearly independent. We deduce some consequences about when there is a basis of eigenvectors. Big Fact 4. If there is a basis of eigenvectors of A, then f (k) factors as (λ1 −k)(λ2 −k) · · · (λn −k). If k has such a factorization, and all the λi are distinct, then there is a basis of eigenvectors. Big Fact 1 For a nonzero vector ~v , we have A~v = λ~v if and only if (A − λ · Id)~v = 0, if and only if ~v is in the kernel of A − λ · Id. So there is a nonzero vector ~v with A~v = λ~v if and only if the kernel of A − λ · Id is nonzero. By our previous discussion of determinants, this happens precisely when det(A − λ · Id) = 0. Big Fact 2 Take the linearly independent vectors ~v1 , ~v2 , . . . , ~vr and find n − r more vectors so that ~v1 , ~v2 , . . . , ~vr , ~vr+1 , . . . , ~vn is a basis of Rn . (I should justify that you can always do this, but I’ll gloss over that.) For 1 ≤ i ≤ r, we have A~vi = λ~vi . For r + 1 ≤ i ≤ n, we can write n X A~vi = cij ~vj j=1 for some coefficients cij . We can organize these facts into a matrix. For concreteness, we take r = 2 and n = 5.      λ c31 c41 c51   | | | | | | | | | |     λ c32 c42 c52      ~v1 ~v2 ~v3 ~v4 ~v5   ~ v ~ v ~ v ~ v ~ v c c c = A · 1 2 3 4 5 33 43 53          | | | | | | | | | | c34 c44 c54 c35 c45 c55 The blank entries in the matrix are zero. Let   | | | | |    S = ~v1 ~v2 ~v3 ~v4 ~v5   | | | | | So our equation above says  λ c31  λ c32  c33 S   c34 c35 c41 c42 c43 c44 c45  c51 c52   c53   = AS. c54  c55 Since ~v1 , ~v2 , . . . , ~vr , ~vr+1 , . . . , ~vn is a basis, the matrix S is invertible and we can write   λ c31 c41 c51  λ c32 c42 c52    −1 c33 c43 c53  S   S = A.  c34 c44 c54  c35 c45 c55 So  λ c31 c41  λ c32 c42   c33 c43 det(A − k · Id) = det(A − kSS −1 ) = det S    c34 c44 c35 c45  λ−k c31  λ − k c 32  c −k = (det S) · det  33   c34 c35   c51  c52    −1  c53   − k · Id S   c54 c55  c41 c51 c42 c52   −1 c43 c53   · (det S )  c44 − k c54 c45 c55 − k The det S and det S −1 factors cancel so the characteristic polynomial is   λ−k c31 c41 c51  λ−k c32 c42 c52     c33 − k c43 c53  det    c34 c44 − k c54  c35 c45 c55 − k Using the zeroes in the left r columns, we see that every term of this determinant is divisible by (λ − k)r . Big Fact 3 The proof of Big Fact 3 is a bit challenging, and I imagine I won’t cover it in class. But it is a nice result! Our proof is by induction on r. The base case r = 1 is trivial. To build confidence, let’s also check r = 2. We have A~v1 = λ1~v1 and A~v2 = λ2~v2 . If ~v1 were proportional to ~v2 , then A would stretch both ~v1 and ~v2 by the same factor. But it stretches ~v1 by λ1 and ~v2 by λ2 , and we said that λ1 6= λ2 . So, in fact, ~v1 and ~v are not proportional after all, which is what we wanted. Let’s now do the case of larger r. Suppose we already know that ~v1 , ~v2 , . . . , ~vr−1 are linearly independent. Suppose that we have a nontrivial linear relation c1~v1 + c2~v2 + · · · cr−1~vr−1 + cr~vr = 0. (∗) Multiplying equation (∗) by A, we get c1 A~v1 + c2 A~v2 + · · · cr−1 A~vr−1 + cr A~vr = 0 or c1 λ1~v1 + c2 λ2~v2 + · · · cr−1 λr−1~vr−1 + cr λr~vr = 0 Now subtract off λr times equation (∗): (c1 λ1~v1 + c2 λ2~v2 + · · · cr−1 λr−1~vr−1 + cr λr~vr ) − λr (c1~v1 + c2~v2 + · · · cr−1~vr−1 + cr~vr ) = 0 or c1 (λ1 − λr )~v1 + c2 (λ2 − λr )~v2 + · · · + cr−1 (λ1 − λr−1 )~vr−1 = 0. (†) Equation (†) is a linear relation among ~v1 , ~v2 , . . . , ~vr−1 . Our induction assumption was that we had already showed this couldn’t happen. At this point, we have basically reached a contradiction, and the rest is just mopping up. The only way we might not have a contradiction is if all the coefficients in (†) were zero. In other words, if c1 (λ1 − λr ) = c2 (λ2 − λr ) = · · · = cr−1 (λ1 − λr ) = 0. Since λr is not equal to any of the other λ’s, this would mean c1 = c2 = · · · = cr−1 = 0. Plugging into equation (∗), we get cr~vr = 0. But ~vr is not zero, so this means cr = 0. We have now shown that all the coefficients in equation (∗) are 0. In other words, the only linear relation between the ~vi is a trivial relation. Now, we have a contradiction, and the proof is complete. Big Fact 4 First, suppose that there is a basis ~v1 , ~v2 , . . . , ~vn of eigenvectors. Then, as Big Fact 4, we deduce that     λ1 |   | | | | | λ2 | |     ~v1 ~v2 ~v3 ~v4 ~v5     λ ~ v ~ v ~ v 3   =A· 1 2 3 |  | | | | |  λ4 | | λ5 in the argument for  | |  ~v4 ~v5   | | As in that argument, we deduce that  λ1 − k  λ2 − k  λ3 − k det(A − k · Id) = det    λ4 − k    .   λ5 − k The right hand side is clearly (λ1 − k)(λ2 − k) · · · (λn − k). We now make the reverse argument. Suppose that f (k) = (λ1 − k)(λ2 − k) · · · (λn − k) and that the λi are distinct. From Big Fact 1, for each λi , there is an eigenvector ~vi . By Big Fact 3, they are linearly independent. Since there are n of them, and they are linearly independent, they are a basis. So what if f has repeated roots? If f (k) does factor as (λ1 − k)(λ2 − k) · · · (λn − k), but with some repeated λi , then you can’t predict whether or not A has a basis of eigenvectors until you actually do the computation and see. Consider     2 1 2 0 A= B= . 0 2 0 2 Then A and B both have characteristic polynomial (2 − k)2 . The matrix A only has a one dimensional space of eigenvectors, so we can’t find a basis of eigenvectors for A. For B, every vector is an eigenvector, so any basis of R2 is a basis of eigenvectors.