How to prove: if $AB=I$ then $\mathrm{rank}(A)=\mathrm{rank}(B)$ and if $AB=0$ then $\mathrm{rank}(A)+\mathrm{rank}(B)\le n$

Question

If $A$ is an $m\times n$ matrix and $B$ is an $n\times m$ matrix and if $AB=I_m$, how can I prove that $\mathrm{rank}(A)=\mathrm{rank}(B)$ ?

And similarly, if $AB=0$, then $\mathrm{rank}(A)+\mathrm{rank}(B) \le n$.

Answer 1

$\newcommand{\rank}[0]{\text{rank}}$$\newcommand{\nullity}[0]{\text{nullity}}$For the first one, you may use the inequality $$ \rank(A B) \le \min (\rank(A), \rank(B)), $$ which yields $$\tag{ineq} m \le \rank(A), \rank(B) \le \min(m, n). $$

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For the second one, the image of $B$ must be contained in the kernel of $A$. Considering dimensions, this yields $$ \rank(B) \le \nullity(A). $$ Now use rank-nullity for $A$.