I was thinking about the problem that says:
If $A$ and $B$ are $3\times 3$ real matrices such that $\operatorname{rank}(AB)=1$, then $\operatorname{rank}(BA)$ can not be which of the following?
(a) $0$
(b) $1$
(c) $2$
(d) $3$.
My attempt: I have chosen suitable $3 \times 3 $ matrices for $A$ and $B$ keeping in mind that $\operatorname{rank}(AB)=1$. Say for example if I take $A$ and $B$ to be $A = \begin{pmatrix} 1 &2 &0 \\ 0 & 0 &0 \\ 0 & 0 &0 \end{pmatrix}$ and $B = \begin{pmatrix} -2 &1 &0 \\ 1 & 0 &0 \\ 0 & 0 &0 \end{pmatrix}$
respectively, then I see $\operatorname{rank}(AB) = \operatorname{rank}(BA) = 1$. So, option (b) can not be correct. Do I have to keep choosing the matrices and then observe which of the option holds good. Is this kind of approach right to tackle the problem? I am looking for a direct way (e.g. application to some theorem) which can give me the result. I have also noticed that $AB$ and $BA$ are similar matrices as we see that $A^{-1}(AB)A=BA$. Is this observation going to help me in any way? Thanks in advance for your time.