Given a real polynomial p(x), show that p(λ) is an eigenvalue of p(A)

Question

Consider $A ∈ \mathbb{R}^{n×n}$, and suppose that λ is an eigenvalue of A.

I'm confused. This was all the question told me. Was I not given enough information? How do I approach this? I was thinking you could show the expansion of a real polynomial and plug A in like this $c_kA^k$+$c_{k-1}A^{k-1}+...c_0I_n$ and then doing the same with the eigenvalue but I don't know how to finish?

Attempted proof:

$Ax=λx$

$A^kx=λ^kv$

-->($c_kA^k$+$c_{k-1}A^{k-1}+...c_0I_n$)x

=$c_kA^kx$+$c_{k-1}A^{k-1}x+...c_0I_nx$

=$c_kλ^kx$+$c_{k-1}λ^{k-1}x+...c_0I_nx$

=($c_kλ^k$+$c_{k-1}λ^{k-1}+...c_0I_n$)x

See also https://math.stackexchange.com/questions/492070/how-to-prove-eigenvalues-of-polynomial-of-matrix-a-polynomial-of-eigenvalue (shows that the converse is true over the complex numbers). — Minus One-Twelfth, Feb 15 '20 at 20:56
Yes, that is correct. Purely for clarity, I would be inclined to expand a little by writing $p(t) = \sum_k c_k t^k$ and then $p(A)x = \sum_k c_k A^k x = \sum_k c_k \lambda^k x = p(\lambda) v$. — copper.hat, Feb 15 '20 at 21:05
The last $I_n$ should disappear. You are mixing scalars and matrices :-). — copper.hat, Feb 15 '20 at 21:05

score 4 · Accepted Answer · answered Feb 15 '20 at 21:05

Let $$ p(x) = \sum_{i=0}^n a_i x^i $$ be a polynomial of degree greater than zero.

Let $v$ be an $n \times 1$ eigenvector corresponding to the eigenvalue $\lambda$ of the $n \times n$ matrix $A$. Then we have the equality $$ Av = \lambda v. $$ So $$ A^2 v = A(Av) = A(\lambda v) = \lambda (Av) = \lambda(\lambda v) = \lambda^2 v, $$ $$ A^3 v = A(A^2v) = A(\lambda^2 v) = \lambda^2 (Av) = \lambda^2 (\lambda v) = \lambda^3 v, $$ and so on $$ A^n v = \lambda^n v. $$ Finally, when $\lambda \neq 0$, we also have $$ A^0 v = I_n v = v = 1v = \lambda^0 v. $$ Then $$ p(A)v = \left( \sum_{i=0}^n a_i A^i \right)v = \sum_{i=0}^n a_i A^i v = \sum_{i=0}^n a_i \lambda^i v = \left( \sum_{i=0}^n a_i \lambda^i \right) v = p(\lambda)v. $$

When $\lambda = 0$, then we obtain $$ p(A) v = \left( \sum_{i=0}^n a_i A^i \right)v = \sum_{i=0}^n a_i A^i v = a_0 I_n v = a_0 v = p(\lambda) v. $$

Given a real polynomial p(x), show that p(λ) is an eigenvalue of p(A)

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