Proof for the case: BA=0 but B=/0 where A is a singular matrix?

Question

How it is possible to prove that for some B, BA=0 but B=/0 where A is a singular matrix and bot A and B are n x n matrices? I found such matrices like

A=

1 1

1 1

and

B=

1 -1

1 -1

which hold the case.

However, I could not end with a proof without making assumptions on 'n', where I assumed n=2 in my examples. I know that, since A is not invertible, Ax=0 has infinitely many solutions for x in R.

Answer 1

If A is singular you can transform it to the Row Reduced Echelon form where you have rows of zeros at the bottom of your matrix.

Now you can easily find a non-zero matrix B where $AB=0$

Answer 2

Since $A$ is singular, $A^T$ is singular too. So, there is a non-zero vector $v$ such that $A^Tv=0$. But this means that $v^TA=0$. So, let $B$ be the $n\times n$ matrix such that each of its lines is $v^T$ and then $B\neq0$ and $BA=0$.