If $M$ is a $n\times{n}$ singular matrix with $n>1$, is there always $N≠0$ such thaf $MN=0$?

Question

I read the sentence is true, so I couldn't develop how to prove that, if $M$ is $n\times{n}$ with $n>1$ such that $\det(M)=0$, there is always $N$, also $n\times{n}$, with $N≠0$ such that $MN=0$.

Answer 1

From th egiven, there exists $v\ne 0$ with $Mv=0$. Try $N=vv^T$ (or $vw^T$ for any nontrivial $w$)