Find all complex-valued $n \times n$ matrices $A$ such that $\det A = 0$ and $\text{rank}(AB) = \text{rank}(BA)$ for any $n \times n$ complex-valued $B$.
I believe that $A = 0$ is the only answer.
I have been able to prove that, if $A$ is of rank $r$, then any $r$ lines and any $r$ columns are linearly independent. To see this, note that since $A$ is of rank $r$, then A has $r$ linearly independent columns; say that the indexes of these columns are $i_1, i_2, …, i_r$. Then by making B equal to a matrix that has 1 in positions $(i_k,i_k)$ and 0 elsewhere, $AB$ basically "selects" $r$ independent columns from $A$ having all other columns equal to 0, so $\text{rank} AB = r$.
Now, $BA$ selects rows $i_1,i_2,..,i_r$ from $A$. If $A$ were to have $r$ rows that were not linearly independent, then there would be an inversible matrix $M$ which would place these rows in positions $i_1,i_2,…,i_r$. Then
$$\text{rank} (BA) = \text{rank} (BAM) < r,$$
which would contradict our hypothesis. Thus any $r$ rows of $A$ are linearly independent. Running the same argument in reverse, we get that any $r$ lines of $A$ are linearly independent.
Any ideas about how to proceed?
Best Answer
Here is a simple way. Suppose $A\ne 0$ and $\det A=0$. Then its null space $N(A)$ and its range $R(A)$ are both nontrivial. So we can pick any nonzero $v\in N(A)$ and $w\in R(A)$.
Choose any linear map $\mathbf{B}:\mathbb{R}^n\to N(A)\subseteq\mathbb{R}^n$ such that $\mathbf{B}w=v$, and let $B$ be its matrix. Then it is easy to verify that $AB=0$ but $BA\ne 0$.