Greatest common divisor of Characteristic Polynomials

Question

I am stuck with the following proof about matrices:

Let $A\in \textbf{M}_{n}(k),B\in \textbf{M}_{m}(k)$ and $M\in \textbf{M}_{n\times m}(k)$ such that $AM=MB$. Prove that the degree of gcd{$ P_{A} ,P_{B} $} is larger or equal to $r=rankM$. (with $P_{M}$ I mean characteristic polynomial)

I know that if $v$ is an eigenvector of $B$ with eigenvalue $\lambda$, then $AMv=MBv=M\lambda v= \lambda Mv$, so if $Mv$ is nonzero i could say that $\lambda$ is an eigenvalue of A, though I'm not sure if this information will help in any way to solve the problem. Any help would be great!

Answer 1

$AM=MB$ gives $f(A)M=Mf(B)$ for any polynomial $\in k[x]$, and $M\ker(f(B))\subset \ker(f(A))$.

We can assume that $k$ is algebraically closed.

Let $d=\max(n,m)$.

The multiplicity of $\lambda$ in $\det(B-xI)$ is $\dim(\ker( (B-\lambda I)^d)$.

Moreover $k^m=\sum_\lambda \ker( (B-\lambda I)^d)$.

Whence $M k^m=\sum_\lambda M\ker( (B-\lambda I)^d)$.

This space has dimension $rank(M)$ and $\lambda$ has multiplicity $\dim\ker( (A-\lambda I)^d)\ge \dim(M\ker( (B-\lambda I)^d))$ in $\det(A-xI)$.