How can we prove the following:
If a matrix $A \in \mathbb{C}^{n\times n} $ is invertible, then $\begin{bmatrix} A & I \\ I & X\\ \end{bmatrix} \in \mathbb{C}^{2n\times 2n}$ has the same rank as $A$ if and only if $X=A^{-1}.$
I would appreciate any hints or ideas to show this.
$\pmatrix{I & 0\cr -I & A\cr} $ is invertible, its inverse being $\pmatrix{I & 0\cr A^{-1} & A^{-1}\cr}$, so the rank of $\pmatrix{A & I\cr I & X}$ is the same as the rank of $$ \pmatrix{I & 0\cr -I & A\cr} \pmatrix{A & I\cr I & X} =\pmatrix{A & I\cr 0 & AX-I}$$ But that would have greater rank than that of $A$ if some entry of $AX-I$ is nonzero.
If $A$ is invertible, then you can reduce $A$ to $I$ by elementary row operations: $EA=I$.
Performing the same operations on $\begin{bmatrix} A & I \\ I & X\\ \end{bmatrix}$ gives $\begin{bmatrix} I & E \\ I & X\\ \end{bmatrix}$ and then $\begin{bmatrix} I & E \\ 0 & X-E\\ \end{bmatrix}$.
If any entry in $X-E$ is nonzero, then the last matrix has rank larger than $n$.
Therefore, $X-E=0$ and so $X=E=A^{-1}$.