Let $f:M_n(F)\rightarrow M_n(F), X\mapsto AX-XA$. If $f$ is diagonalizable, I want to show that $A$ is diagonalizable. I'd prefer to avoid Jordan Blocks. I know that $f$ is diagonalizable if and only if:
- its minimal polynomial is square-free, or
- there exist $d$ linearly independent eigenvectors where $d = \dim M_n(F)$, or
- the characteristic polynomial of $f$ factors into linear terms and each geometric multiplicity equals the corresponding algebraic multiplicity.
Best Answer
If $F$ is a splitting field for $A$ (e.g. when $F$ is algebraically closed), we may prove the statement as follows.
Proof 1. (A simplified version of user8675309’s answer.) Let $\lambda\in F$ be an eigenvalue of $A$ and $v\in F^n$ be a corresponding left eigenvector. Let $B=A-\lambda I$. Define $g:M_n(F)\to M_n(F)$ by $g(X)=BX$. Since $A$ and $B$ commute, so do $f_A$ and $g$. Moreover, for any vector $x\in F^n$, we have $$ f_A(xv^T)=Axv^T-xv^TA=Axv^T-x(\lambda v^T)=Bxv^T=g(xv^T). $$ It follows that $(m(f_A))xv^T=(m(g))xv^T=m(B)xv^T$ for every polynomial $m\in F[x]$. In particular, when $m$ is the minimal polynomial of $f$, we have $m(B)xv^T=0$. Since $x$ is arbitrary and $v$ is nonzero, we must have $m(B)=0$. Hence the minimal polynomial of $B$ divides $m$. However, as $f_A$ is diagonalisable over $F$, $m$ is a product of distinct linear factors. Therefore the minimal polynomial of $B$ is also a product of distinct linear factors. This means $B$ is diagonalisable over $F$. In turn, $A=B+\lambda I$ is diagonalisable over $F$ too.
Proof 2. Since $F$ is a splitting field for $A$, $A$ admits a Jordan-Chevalley decomposition $S+N$. As $S$ is diagonalisable, so is $f_S$. In fact, if $\{v_1,\ldots,v_n\}$ is an eigenbasis of $S$, then $\{v_iv_j^T: i,j\in[n]\}$ will be an eigenbasis of $f_S$. One may also verify that $f_N$ is nilpotent (with $f_N^{2n-1}=0$), $f_S$ commutes with $f_N$, and $f_A=f_S+f_N$. Therefore $f_S+f_N$ is a Jordan-Chevalley decomposition of $f_A$. By assumption, $f_A$ is diagonalisable. Hence $f_N=0$. In turn, $N=0$. Therefore $A=S$ is diagonalisable.