Prove that If $f$ and $h$ commute and $h$ is nilpotent, then $f+h$ and $f$ have the same characteristic polynomial

Question

let $V$ be a finite dimensional vector space over $\mathbb{K}$ ($\mathbb{R}$ or $\mathbb{C}$).
and $f$ and $h$ two vector space endomorphisms of $V$, such as that : $h$ is nilpotent and $f \circ h = h\circ f$.
Prove that $f + h$ and $f$ have the same characteristic polynomial.
I first proved that all eigenvalues of $f$ are also eigenvalues of $f+h$, but even if I prove that they have the same spectrum, it wouldn't mean they both have the same characteristic polynomial, so how should I proceed?

Answer 1

Assume that $h^m=0$ for some $m$, and let $N>m$ be so large that $\ker (f-\lambda)^{N-m} = \ker (f-\lambda)^{N-m+1}= \ldots =\ker (f-\lambda)^N$ is already stabilized.

As they commute, we have $$ (f-\lambda + h)^N = \sum_{k=0}^m {N\choose k} h^k (f-\lambda)^{N-k}= \big(\ldots\big)(f-\lambda)^{N-m}. $$ So we have $$ \ker(f-\lambda)^{N} = \ker(f-\lambda)^{N-m} \subseteq \ker(f-\lambda +h)^N $$ so any vector that lies in the null-space of $(f-\lambda)^N$ also lies in the null space of $(f-\lambda+h)^N$. Vice versa, $f$ can be expressed as $(f+h)-h$, so using the same argument, the null-space of $(f-\lambda)^N$ and $(f+h-\lambda)^N$ coincide. So the algebraic multiplicity of $\lambda$ is the same.

Answer 2

You want to show that $\det(f-\lambda I)=\det(f+h-\lambda I)$, provided $f$ commutes with $h$ and $h$ is nilpotent. \begin{align} \det(f-\lambda I-h)&=\det((f-\lambda I)(I-h(f-\lambda)^{-1})) \\ &= \det(f-\lambda I)\det(I-h(f-\lambda)^{-1}). \end{align} So, the conjecture is equivalent to showing $\det(I-k)=1$ for all a nilpotent $k$. And this follows because the Jordan canonical form for $I-k$ is upper triangular with all $1$'s along the diagonal.