Why $\operatorname{rank} AB≠\operatorname{rank} BA$?

Question

Let A,B be two square matrices, say of size n, then why $\operatorname{rank} AB≠\operatorname{rank} BA$? Just need valid reasoning.I just know that above questions has answer 'No'.

Answer 1

Just for completeness sake, I will provide an example of $A$ and $B$ such that $AB = 0$ but $BA\neq 0$, as suggested by @richrow in the comments above:

For $A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 1 & 1 \end{pmatrix}$, we have $AB = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ but $BA = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 1 & 0 \end{pmatrix}$.