I heard the following claim in a youtube video about the rank of a matrix:
If $M$ is an $n \times n$ matrix, and $\mathrm{rank}(M)=n'$ then the set of elements that gets mapped to the origin has dimension $n-n'$. It seems to be an intuitive result (e.g. if $M$ is a $3 \times 3$ matrix with $\mathrm{rank}=2$ it's natural to imagine that some entire line was mapped to the origin.
I believe the above statement can be more formally stated as: If $M$ is an $n \times n$ matrix, and $\mathrm{rank}(M)=n'$ then $M^{-1}(0)=${$ \textbf{v}:\textbf{v}=\textbf{x}_0+\textbf{x}_1t_1+...+\textbf{x}_mt_m$} where $m=n-n'$ for some set {$\textbf{x}_0, ..., \textbf{x}_m$} of linearly independent vectors.
I haven't been able to come up with a proof, so how could the above statement be shown?
Would appreciate any help.