If $A$ is nilpotent and $AB = A$, does $B$ have to be identity matrix?

Question

I have seen posts asking under what conditions does $B$ have to be the identity matrix if $AB = A$, but none of them could answer my question. I know that if $A$ is singular, then $A = AB$ doesn't necessarily imply $B = I$. A counter-example would be $A$ being a idempotent matrix (then $B = A$). What if $A$ is nilpotent (so $A$ cannot be idempotent) and nonzero? In that case, can I claim that $B = I$?

Answer 1

If $A$ is nilpotent, $A$ must have $0$ as its eigenvalue, so it is singular.

This means that your second assumption is actually same with your first assumption.

Append :

For these two matrices $A, B$ such that

$$A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix} ,\quad B=\begin{pmatrix} 0 & 1 \\ 0 & 1 \\ \end{pmatrix}. $$

Then you can check that $A$ is not idempotent (In fact, $A$ is nilpotent).

And you can also check that $AB=A$. But $B\neq I$.

Answer 2

How about \begin{align} A &= \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix} & B &= \begin{bmatrix}-1 & 0 \\ 0 & 1\end{bmatrix} \end{align} Then $A$ is nilpotent and not idempotent, $B$ is not the identity, and $AB=A$.

(The inspiration here was the structure of the Lie algebra $\mathrm{sl}(2)$, i.e., the traceless 2x2 matrices.)