Eigenvalues of $p(A$) Let $\lambda$ be an eigenvalue of a matrix $A$. I am trying to show that all the eigenvalues of $p(A)$ are $p(\lambda)$ where $p(x)$ is any polynomial.
I have been able to show that $p(\lambda)$ is an eigenvalue of $p(A)$. But how to show that these will be the only eigenvalues?
For instance, if eigenvalues of $A$ are $1$ and $-1$.
Then I know that 1 is an eigenvalue of $A^2$ but how to show that there is no other eigenvalue?
 A: I assume $A$ is a complex $n \times n$ matrix.
Then $A$ is similar to an upper triangular matrix $D$ whose diagonal coefficients are (in order) $\lambda_1, \dots, \lambda_n$ the eigen values of $A$ (with repetition), so there exist $P$ an invertible (complex) matrix such that $A = P^{-1} D P$.
Then, writing $p = \sum_k \alpha_k X^k$ we have
$$
p(A) = \sum_k \alpha_k (P^{-1} D P)^k = \sum_k \alpha_k P^{-1} D^k P = P^{-1} \left(\sum_k \alpha_k D^k \right) P = P^{-1} p(D) P.
$$
Now it suffices to see that $p(D)$ is upper diagonal and that its diagonal coefficients are $p(\lambda_1), \dots, p(\lambda_n)$. Indeed, since $D$ is upper triangular, $D^k$ is also upper diagonal and its diagonal coefficients are $\lambda_1^k, \dots, \lambda_n^k$, and the result follows by linearity.
Since $p(A)$ is similar to an upper diagonal matrix whose diagonal coefficients are $p(\lambda_1), \dots, p(\lambda_n)$, the spectrum of $p(A)$ is $\{p(\lambda_1), \dots, p(\lambda_n)\}$.
Hope this helps!
A: Let $\alpha$ be an eigenvalue of $p(A)$. Then $W=N(p(A)-\alpha I)$ is nontrivial. Let $g(x)=p(x)-\alpha,$ then $g(x)$ annhilates $A$ when restricted to $W$. So, minimal polynomial of $A$ restricted to $W$, $m_W(x)$ divides  $g(x)$. But $m_W(x)$ divides the minimal polynomial, $m(x)$ of $A$. Hence $m(x)$ and $g(x)$ have a common zero.
Note: $W$ is an $A$ invariant subspace.
