Let $M, N$ be $2 \times 2$ matrices with entries in $\mathbb{C}$. Suppose the $2 \times 4$ block matrix $\begin{bmatrix} M & N \end{bmatrix}$ has rank 2 (i.e., the rows of this matrix are linearly independent over $\mathbb{C}$).
I would like to show there exists invertible $2 \times 2$ matrices $M^+, N^+ \in \mathbb{C}^{2 \times 2}$ such that the $4 \times 4$ block matrix $$ S = \begin{bmatrix} M & N \\ M^+ & N^+ \end{bmatrix} $$ is invertible.
Furthermore, suppose $$ J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} $$ and $$ MJM^* - NJN^* = 0,$$ where $M^*$ denotes conjugate transpose.
I would like to show $$ \left( S \begin{bmatrix} -J & 0 \\ 0 & J \end{bmatrix} S^*\right)^{-1} = \begin{bmatrix} P & Q \\ R & T \end{bmatrix} $$ has $T = 0$.
Here's what I have so far. Since $\begin{bmatrix} M & N \end{bmatrix}$ has rank $2$, we can certainly form a basis of $\mathbb{C}^4$ by adding two more suitable rows to this set. This would give us an invertible $4 \times 4$ like $S$ above, but I don't know how to show that the desired submatrices are invertible.
As for showing the inverse formula, using the relation $ MJM^* - NJN^* = 0$ I get $$ S \begin{bmatrix} -J & 0 \\ 0 & J \end{bmatrix} S^* = \begin{bmatrix} 0 & Q' \\ R' & T' \end{bmatrix}, $$ but I'm not sure if this will lead to the second conclusion I want.