I found a problem I got stuck with:
Let $f \in \operatorname{End}(V), V$ be a finite-dimensional $F$-vector space, $F$ algebraically closed.
(a) Let $ a \in F $ and $k \in \mathbb{N}$.
Show: If $(f-a)^{k}=0 $then also $\left(f^{2}-a^{2}\right)^{k}=0 $.
(b) Let $\mu_f$ be the minimal polynomial of $f$ and $\mu_{f^2}$ the minimal polynomial of $f^2$
Show: $\operatorname{deg}\left(\mu_{f^{2}}\right) \leq \operatorname{deg}\left(\mu_{f}\right)$.
I was able to solve (a) but I tried solving (b) since hours now.
I'm sure $\operatorname{deg}\left(\mu_{f^{2}}\right) = \operatorname{deg}\left(\mu_{f}\right)$ holds if $f$ is diagonalizable.
but if $f$ is not diagonalizable, I want to show $\operatorname{deg}\left(\mu_{f^{2}}\right) \leq \operatorname{deg}\left(\mu_{f}\right)$ then.
I know that $\operatorname{deg}\left(\chi_{f^2}\right) =\operatorname{deg}\left(\chi_f\right)$ holds for the characteristic polynomials and $\operatorname{deg}\left(\mu_f\right) + n = \operatorname{deg}\left(\chi_f\right)$ for some $n \in \mathbb{N_0}$.
This is all I got so far:
$\operatorname{deg}\left(\mu_{f^{2}}\right) \leq \operatorname{deg}\left(\chi_{f^{2}}\right) = \operatorname{deg}\left(\chi_{f}\right) = \operatorname{deg}\left(\mu_f\right) + n$ $(*)$
Maybe task $(a)$ can help?
I'm expecting the solution to be some kind of equation like in $(*)$ just with a better estimation.
I was able to find this but I couldn't go on with that.
Thanks for your help!
Best Answer
Let the minimal polynomial of $A$ have degree $r$.
$\mathbf 0= m\big(A\big) = \prod_{k=1}^r\big(A-\lambda_kI\big)$
$\implies \mathbf 0= \prod_{k=1}^r\Big(\big(A-\lambda_kI\big)\big(A+\lambda_kI\big)\Big)=\prod_{k=1}^r\big(A^2-\lambda_k^2 I\big)$
conclude that $A^2$ is annihilated by a degree $r$ polynomial so its minimal polynomial has degree $\leq r$