Is every noninvertible matrix over a field a zero divisor?
Related to this: What are sufficient conditions for a matrix to be a zero divisor over a noncommutative ring?
Abstract Algebra – Is Every Noninvertible Matrix a Zero Divisor?
abstract-algebramatrices
Best Answer
New and improved: If $A$ is singular we can get $AB=BA=0$ with no more work.
Original Yes. If $A$ is a singular square matrix then there exists a non-zero vector $v$ with $Av=0$. So if $B$ is the square matrix that has every column equal to $v$ then $AB=0$.
Better There is a non-zero "column vector" $v$ with $Av=0$. There is also a non-zero "row vector" $w$ with $wA=0$. Let $B=(b_{jk})$, where $b_{jk}=v_jw_k$. Then every row of $B$ is a multiple of $w$ and every column of $B$ is a multiple of $v$, so $BA=AB=0$.