If $AB+BA=0$ and $B=AX+XB$, then $B$ is nilpotent.

Question

Suppose $A,B,X \in M_n(\mathbb{R})$ and that $AB+BA=0$ and $B=AX+XA$. Prove that $B$ is a nilpotent matrix.

What are your thoughts on this problem, what have you tried? — Najib Idrissi, Aug 09 '14 at 19:40
First I tried to show each eigenvalue is zero, with no success. Then I tried to show $B^n=0$ by multiplying the second equality by B and doing algebraic manipulation, still to no avail. — Q-rious, Aug 09 '14 at 19:46

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

The proof for this one is almost the same.

We have $$AB=-BA$$, thus it follows by multiplying $B$ on the right, that $$AB^2 = -BAB = B^2 A$$ Then by induction, we have $$AB^n=(-1)^nB^nA.$$

Now, following the solution in the link: we have $$B^n = B^{n-1}B = B^{n-1}(AX+XA) = A (-1)^{n-1} B^{n-1}X + B^{n-1} XA.$$

Thus, $$\textrm{Tr} B^{2n} = 0 $$ for all $n$.

This shows that $B^2$ is nilpotent.

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