Let $f$ be an endomorphism of a finite-dimensional vector space $V$.
Definition. A Jordan–Chevalley decomposition of $f$ is a decompositon $f = d + n$ where $d$ and $n$ are endomorphisms of $V$ such that $d$ is diagonalizable, $n$ is nilpotent, and $d$ and $n$ commute with each other.
It is known that $f$ admits a Jordan–Chevalley decomposition if and only if it admits a Jordan normal form. More explicitely, if $\mathcal{B}$ is a basis of $V$ such that the representing matrix $[f]_{\mathcal{B}}$ is in Jordan normal form, then $[d]_{\mathcal{B}}$ is the diagonal part of $[f]_{\mathcal{B}}$ and $[n]_{\mathcal{B}}$ is the strictly upper triangular part of $[f]_{\mathcal{B}}$.
The Jordan–Chevalley decomposition of $f$ is unique, and the Jordan normal form is unique up to permutation of the Jordan blocks. In the past, I’ve seen students claim (without proof) that the uniqueness of the Jordan–Chevalley decomposition follows from the uniqueness of the Jordan normal form. This has lead me to the following question:
Question. Is there a connection between the uniqueness of the Jordan–Chevalley decomposition and the uniqueness of the Jordan normal form?