Let and be a by and by matrix respectively. Show that
Solution (By Zorn Leung).
For convenience, given any matrix , denote (abuse of notation) to be the linear operator . Plus let and be the null space and range space of operator respectively.
Consider the restriction of on , denoted by .
Observe first since is clearly a subspace of . Also, is a subspace of . Indeed, if , then for some . Hence implying . Therefore .
By dimension theorem,
, which is the conclusion.
Corollary. , as above. Then .
Dimension theorem implies
Using the previous result,
. The result follows.