$C=AB-BA$. If $CA=AC$, then $C$ is not invertible.

Question

Problem

Let $A$, $B \in M_n(\mathbb{C}),$ $C=AB-BA$. If $CA=AC$, then $C$ is not invertible.

To prove a matrix is not invertible, I only came up with $2$ ideas, one is to find a nonzero vector $v$ such that $Cv = 0$, and the other is to prove that $\det C = 0$. However, I have no idea how to use the condition $CA=AC$, so I don't know how to deal with $\det (AB-BA)$.

Is there any other criterion for invertibility of a matrix that is helpful to solve this problem?

I would like to prove $C$ is not invertible, rather than the stronger result that $C$ is nilpotent, so basically this is not a duplicate of another question.

Answer 1

The following proof works over any field whose characteristic is either zero or some finite number that doesn't divide $n$. Suppose the contrary that $C$ is invertible. As $C$ commutes with $A$, its inverse commutes with $A$ too. It follows that $$ I=(AB-BA)C^{-1}=ABC^{-1}-BAC^{-1}=ABC^{-1}-BC^{-1}A. $$ Taking traces on both sides, we get $n=0$, which is a contradiction.

Answer 2

Let $\lambda$ be an eigenvalue of $A$. If we put $A^{\sharp}=A-\lambda I$, then $C=A^{\sharp}B-BA^{\sharp}$ and $C$ commutes with $A^{\sharp}$, so the problem stays the same if we replace $A$ with $A^{\sharp}$. In other words, we can assume without loss that $\lambda =0$, i.e. that $0$ is an eigenvalue for $A$.

We then have a nonincreasing, stationary sequence of subspaces : ${\mathbb C}^n \supseteq {\mathsf{Ker}}(A) \supseteq {\mathsf{Ker}}(A^2) \supseteq \ldots $. Let $r$ be the smallest integer such that ${\mathsf{Ker}}(A^{r})={\mathsf{Ker}}(A^{r+1})$ (denote this subpace by $W$). By construction, $W$ has the property that $(*): w\in W \Leftrightarrow Aw \in W$. Because $C$ commutes wth $A$, we have that $C(W)\subseteq W$. In fact, we also have $B(W)\subseteq W$, but for a slightly different reason : let $w\in W$, and let $j$ be the smallest integer $\geq 0$ such that $B(A^jw)\in W$. Suppose by contradiction that $j\gt 0$. Then we can define $x=A^{j-1}w\in W$, and $Cx\in W$, now $Cx=ABA^{j-1}w-BA^jw$, so $ABA^{j-1}w \in W$, whence $BA^{j-1}w\in W$ which contradicts the definition of $j$. So $W$ is stable by any of $A,B$ and $C$. All computations will henceforth be mande on $W$. In other words, we assume without loss of generality that $A$ is nilpotent with $A^r=0$, $A^{r-1} \neq 0$.

For $k\geq 0$ put $X_k=A^{r-k}BA^k$. Then

$$ X_{k+2}-2X_{k+1}+X_k=A^{r+2-k}(A^2B-2ABA+BA^2)A^k= A^{r+2-k}(AC-CA)A^k=0 \tag{1} $$

It follows by induction on $k$ that $$X_k=X_0+k(X_1-X_0)\tag{2}$$

Taking $r=k$ we deduce $X_1=\frac{X_r+(r-1)X_0}{r}$. Now $X_0$ and $X_r$ are both zero because $A^r=0$. So $X_1=0$, and the sequence $(X_k)$ is identically zero. For $k=r-1$, we deduce $ABA^{r-1}=0$ whence $CA^{r-1}=0$. Now by definition of $r$ there is a $w$ such that $A^{r-1}(w)\neq 0$ ; this vectors shows that $C$ is singular, which finishes the proof.