How can I show that $A^{-} A=I_{n} \text { if and only if } \operatorname{rank}(A)=n$?

Question

$A^{-}$ means minus inverse of matrix A.

How can I show that $A^{-}A=I_{n} \text { if and only if } \operatorname{rank}(A)=n$?

I don't know how to start with it, or it's there exist some great theorems can help me figure it out at once.

Any suggestions or hints? Please.

Thanks in advance.

Answer 1

$A^-$ is the generalized inverse of $A$ $$$$Suppose that $A\in\mathbb{F}^{m\times n}$ then $\exists P\in GL_m(\mathbb{F})$ $Q\in GL_n(\mathbb{F})$ s.t. $A=P\begin{bmatrix} I_r&O\\O&O\end{bmatrix}Q$, where $r=\text{rank}(A)$. $A^-=Q^{-1}\begin{bmatrix} I_r&B\\C&D\end{bmatrix}P^{-1}$ for all matrices $B$,$C$,$D$.$$$$ Since rank$(MN)\leq $rank$(M)$, and rank$(MN)\leq $rank$(N)$, it is obvious that $A^{-}A=I_{n} \text { is always correct if and only if } \operatorname{rank}(A)=n$