If v is an eigenvector of A and c(x) is the characteristic polynomial, then c(A)v = 0?

For any polynomial p(x) = a0+a1x+...+ak x^k and any square matrix A, the matrix p(A) is defined as p(A) = a0 I +a1A+...+ak A^k. Show that if v is any eigenvector of A and c(x) is the characteristic polynomial of A, then c(A)v = 0. Deduce that if A is diagonalisable then c(A) is the zero matrix.

If v is an eigenvector of A and c(x) is the characteristic polynomial, then c(A)v = 0?
We can start with two lemmas:





Lemma 1: If v is an eigenvector of A with eigenvalue b, then A^n v = b^n v for all n %26gt; 0





This is trivially true for n = 0 since A^0 = I and Iv = v = b^0 v





It's also true for n = 1 by the definition of eigenvector, A^1 v = Av = bv = b^1 v





And if you assume it's true for n, then it's true for n = 1 since:





A^(n + 1)v


= AA^n v


= Ab^n v (by assumption)


= b^n Av


= b^n bv


= b^(n+1) v





So by induction, the lemma holds.





Lemma 2: For any polynomial p, and A, b, and v as above, p(A)v = p(b)v





Suppose that p(x) = c0 + c1 x + c2 x^2 + c3 x^3 ...





Then p(A) = c0 + c1 A + c2 A^2 + c3 A^3 ...





And p(A) v = (c0 + c1 A + c2 A^2 + c3 A^3 ...) v


= c0v + c1 Av + c2 A^2v + c3 A^3v ...


= c0v + c1 b v + c2 b^2 v + c3 b^3 v ... (by lemma 1)


= (c0 + c1 b + c2 b^2 + c3 b^3 ...) v


= p(b) v





Which is the required result.





From lemma 2, it follows immediately that if we substitute the characteristic polynomial c for p, then:





c(A)v = c(b)v = 0





Since the eigenvalues are zeros of the characteristic polynomial.





And if A is diagonalizable, that means that there is a set of eigenvectors for it {v0, v1, v2, v3, ...} which form a basis for the space. By the above, c(A) is going to be zero for all of these vectors, and since any other vector in the space can be represented as a linear combination of {v0, v1, v2, v3, ...}, and c(A) is linear, this means c(A) x = 0 for all vectors x in the space. This means c(A) is identically zero.
Reply:If v is an eigenvector and Av = av (where a is a constant, possibly complex), then A^2v = AAv = Aav = aAv = a^2v, and so on. That is, A^n v = a^n v. Hence for any polynomial P,


P(A)v = P(a)v. Note that c(a)=0, since a is an eigenvalue of A.


Thus c(A)v = c(a)v = 0.





If A is diagonable, then C^n (n dimensional complex space) has a basis of eigenvectors of A. So for any z in C^n, we can write


z = sum_k z_k v_k, where the z_k are complex constants and the v_k are the eigenvectors of A. Say Av_k = a_k v_k.





Thus c(A)z = c(A) sum z_k v_k = sum z_k c(A) v_k


= sum z_k . 0 = 0, since each v_k is an eigenvector and we're applying the result we just proved.





We say that A satisfies its characteristic equation, and we've just proved it in the diagonable case. You'll soon learn that it holds for any matrix.