If $A$ is an $m$ by $n$ matrix, show that there is an $n$ by $n$ nonzero matrix $B$ such that $AB=0$ iff rank($A$) < $n$

Question

Question is: If $A$ is an $m$ by $n$ matrix, show that there is an $n$ by $n$ nonzero matrix $B$ such that $AB=0$ iff rank($A$) < $n$.

Another (similar) problem is: If $A$ is an $m$ by $n$ matrix, show that there is an $m$ by $m$ nonzero matrix $B$ such that $BA=0$ iff rank($A$) < $m$.

I was looking at some argument involving image and kernel, since these questions are from the chapter dealing them. I thought there would be some use for the rank-nullity theorem, but I cannot easily figure out the solution.

Answer 1

It is easy to argue that if $A$ has full rank, then $AB = 0 \iff B = 0$ and similarly, $BA = 0 \iff B = 0$. In particular: $ABx = 0$ for every vector $x$ means that every $(Bx)$ is in the kernel of $A$. Similarly, in order to have $x^TBA = [(A^T)(B^Tx)]^T$, every $B^Tx$ should be in the kernel of $A^T$. Both of these kernels are necessarily trivial.

The converse wherein we guarantee that such a $B$ exists, is tricky. For the $AB = 0$: it suffices to choose an $n \times 1$ vector $x$ such that $x \neq 0$, but $Ax = 0$ (i.e. a non-trivial element of the kernel). It follows that for any second non-zero vector $y$ (you can take $y = x$ if you'd like), we have $$ A(xy^T) = 0 $$ verify that $xy^T$ is of the correct size. In fact, it is common to take "nice vectors" for $y$ such as $y = (1,0,\dots,0)^T$ or $y = (1,1,\dots,1)^T$.

For the second: note that $$ BA = 0 \iff (BA)^T = 0 \iff A^TB^T = 0 $$ so it suffices to apply the previous argument.