Minimal Polynomial Degree Inequality for Linear Operator and Its Square

Question

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!

Answer 1

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 $\leq r$

Answer 2

If $f$ satisfies $(f-a)^k=0$ then its minimal polynomial of of the form $(x-a)^{k'}$ for some $k'\leq k$ (I'll come back to this in a minute). Without loss of generality, assume that $k=k'$ and that $(x-a)^k$ is the minimal polynomial.

By part a., you know that $f^2$ is a root of the polynomial $(x-a^2)^k$ and so the minimal polynomial of $f^2$ has degree at most $k$, which is what you wanted to show.

If you want a quick way to verify the claim about the minimal polynomial, put $f$ into Jordan canonical form. The Jordan canonical form is also a root of $(x-a)^k$, from which it is clear that $a$ is the only eigenvalue of $f$. Using the Cayley-Hamilton theorem we see that $a$ is also the only root of the minimal polynomial of $f$

If I wanted to be a bit fancier and not use the algebraically closed hypothesis I would say that the action of $f$ turns $V$ into a $F[x]$-module. Since $F$ is a field then $F[x]$ is a PID. The annihilator of this module consists of all the polynomials for which $f$ is a root and is generated by the minimal polynomial. We know that the annihilator contains $(x-a)^k$ and so this must be a multiple of the minimal polynomial, which must thus be of the form $(x-a)^{k'}$ for $k'\leq k$.

To relate this back to the link you posted in your post notice that if we let $p(x)=(x-a)^k$ and $r(x)=(x-a^2)^k$ then, up to a minus sign, we have $r(x^2)=p(x)p(-x)$

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