Suppose a matrix $A \in \mathbb{C^{n\times n}}$ is invertible. Then: $M =\begin{pmatrix}A&I\\I&X\end{pmatrix}$ $ \in \mathbb{C^{2n\times 2n}}$ has the same rank as $A$ if and only if $X = A^{-1}$ .
What happens if $A \in \mathbb{C^{n\times n}}$ is singular?
Any hint please.
This is not a full answer, but we have that $\det(M)=\det(AX-I)=\det(XA-I)$, regardless of whether $A$ is or isn't invertible (see, for instance, this link). This means that whenever $AX$ does not have $1$ as an eigevalue, then $M$ will necessarily be invertible and hence full-rank.