$\mathbf {The \ Problem \ is}:$ If $\operatorname p(x)$ is a polynomial over an algebraically closed field $\mathbb F[x]$ and let $A$ be an $n×n$ square matrix, then it is known that every eigenvalue of $\operatorname p(A)$ is of the form $\operatorname {p(\lambda)}$ for some eigenvalue $\lambda$ of $A$ .
If the algebraic and geometric multiplicity of $\lambda$ be known, what can be inferred about that of $\operatorname p(\lambda) ?$
Can something interesting be told about the eigenvectors of $\operatorname p(A)$ knowing that of $A$ ???
$\mathbf {My \ approach}$ : I tried some examples and the algebraic and geometric multiplicity of $\operatorname p(A)$ turned out to be equal respectively to that of $A$ , so I tried to prove $\operatorname {det(xI-A) = det(xI- \operatorname p(A))}$ , but I couldn't and I'm not sure whether this is true or not .
$\mathbf {Bonus} :$ If $\operatorname p(A)$ is diagonalizable over $\mathbb F[x]$, can anything be said about diagonalizability of that of $A$ ???
Best Answer
The starting statement must be that since $\Bbb F$ is a closed field all the matrices are triangolable, so for every matrix $A$ it exists a basis such that in that coordinates $A$ is a triangolar matrix with the eigenvalues on the diagonal, counted with algebraic multiplicity.
Then when we evaluate a polinomial $p$ in a matrix $A=[a]_{ij}$ we obtain that $p(A)=[b]_{ij}$ is so that $\forall i\;\; [b]_{ii}=p([a]_{ii})$ (you can easily prove this property demonstrating it for simple powers by induction and concluding by linearity). So you see that for all $\lambda$ eigen value of $A$ then $p(\lambda)$ is eigen value of $p(A)$.
But nothing can be concluded about the multiplicities without knowing more informations on $p$: for example thanks to Hamilton-Cayley Theorem we know that if $q$ is the characteristic polinomial of $A$ then $q(A)=0$, while if $f$ is a polinomial so that $f(\lambda)\ne f(\mu)$ for all $\lambda, \mu$ different eigen values of $A$, then algebraic multiplicity of $f(\lambda)$ for $f(A)$ is the same of $\lambda$ for $A$.
About the bonus question the answer is the same: in general $A$ has not to be diagonalizable, infact for every $A$ not diagonal $q(A)=0$ that is diagonal ($q$ is always the characteristic polinomial of $A$). I hope I answered your question, maybe if you have more details I will be happy to approach them with you!