Nilpotent product of nilpotent $2\times 2$ matrices is $0$

Question

Let $A,B$ be $2 \times 2$ matrices such that $A^2 = B^2 = (AB)^2 = 0$ does it then follow that $AB =0$?

Answer 1

Yes.

Firstly, your conditions imply $A, B, AB$ are nilpotent. Then you can suppose $A = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$ (if you are acquainted with Jordan normal forms of matrices).

Then suppose $B = \begin{bmatrix} a& b \\ c& d\end{bmatrix}.$ Then compute $AB = \begin{bmatrix}c&d\\ 0&0\end{bmatrix}.$ Since $AB$ is nilpotent, we deduce $c = 0.$ Therefore $B = \begin{bmatrix} a & b \\ 0 & d\end{bmatrix}$ is upper-triangular, so $a = d =0.$