Let v be an eigenvector associated to the eigenvalue lambda

Let v be an eigenvector associated to the eigenvalue lambda of a matrix A. Let f (x) be any polynomial. Show that v is an eigenvector associated to the eigenvalue f (lambda) of the matrix f (A).

Solution

Let us assume that f is a polynomial with the degree n. Suppose n = 0: then the polynomial is a constant, and certainly v is an eigenvector of f(A) (every vector actually is) with eigenvalue f(). Now suppose that the proposition is true for all polynomials of degree less than or equal to n. Consider f of degree n + 1. If f has no constant term, then f(A) = AQ(f) where Q has degree n. By induction, v is an eigenvector of Q with eigenvalue Q(). Therefore, A Q(A)(x) = A(Q()v) = Q()A(v) = Q()(v) = Q()v. Thus v is an eigenvector of f with eigenvalue Q() = f(). Now, if f has a constant term c, then f c is a polynomial of degree n + 1 with no constant term. By the above argument, v is an eigenvector of f c with eigenvalue (f c)(). We may observe that v is also an eigenvector of c with eigenvalue c (proved for the base case). Now, we know that if v is an eigenvector for both A₁ and A₂, then v is an eigenvector for aA₁ +bA₂ so that v is an eigenvector of the polynomial f c + c = f with eigenvalue (f c)() + c = f()