Minimal polynomial of f restricted to its image

Question

Let $f:V\to V$ be a $F$-linear map, $V$ an $n$-dimensional vector space over $F$, $\operatorname{rank} E=r$, $W=\operatorname{Im} f$, $\tilde f:=f|_W:W\to W$.

Let $\mu$ be the minimal polynomial of $f$ and $\tilde\mu$ the minimal polynomial of $\tilde f$.

I'm trying to prove that $\mu=\tilde\mu$ or $\mu=X\cdot\tilde\mu$.

I've already proven every non-zero eigenvalue of $f$ to be an eigenvalue of $\tilde f$, thus there must be a $Q\in F[X]$ with $\mu=Q\cdot\tilde\mu$ but I can't seem to understand why $Q$ must either be $1$ or $X$.

Answer 1

Hint: Your argument is sufficient to prove that $Q$ must be $x^k$ for some $k$.

Let $v_1,\dots,v_r$ be a basis for the image, and exend this to a basis $v_{1},\dots,v_n$ for $V$. If we write $f$ with respect to this basis, we have $$ T_f = \pmatrix{T_{\tilde f}&M\\0&0} $$ What does this tell you about the minimal polynomial (or the Jordan Canonical form, if you prefer) of $f$?

Answer 2

If $0$ is not an eigenvector of $f$, then $f$ is surjective, so $W=V$ and $\tilde\mu=\mu$; this case is not very interesting.

So assume $v$ is an eigenvector for eigenvalue $0$. Then since $\mu[f](v)=0$, it must be that $X$ divides $\mu$. Now if for $Q\in K[X]$ the product $QX$ satisfies $(QX)[f]=0$ this means that $Q[f]\circ f=0$ which is equivalent to $Q[f]|_W=0$. The lowest degree monic $Q$ with that property is $Q=\tilde\mu$, from which it follows that $\mu=\tilde\mu\, X$.

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This is in fact the special case $P=X$ of the following useful general fact, similarly proved:

If $P$ is any polynomial dividing the minimal polynomial $\mu$ of $f$, then $\mu/P$ is the minimal polynomial of the restriction of $f$ to the image of $P[f]$.

Often (like in the case above) one knows that some polynomial $P$ divides $\mu$ even before one knows $\mu$, typically because $P$ is the minimal degree monic polynomial for which $P[f]$ kills one or more specific vectors.