TITLE: $A$ 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?
QUESTION [7 upvotes]: 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.

REPLY [4 votes]: This question has lots of good answers.But I just want to share a very short answer.It may already be discussed in the previous answers,so really sorry to answer again.
$AB$ is a $3\times 3$ matrix,rank$(AB)=1\ne3\Rightarrow AB$ is not of full rank$\Rightarrow \operatorname{det}(AB)=0\Rightarrow \operatorname{det}(BA)=0\Rightarrow BA$ is not of full rank$\Rightarrow$rank$(BA)\ne3$