Does the inverse of matrix A exist if we define the first k columns of A and set the last (n-k) rows of A^-1 as orthognal space of the k columns? I am new to the platform and currently undergoing a problem.
Consider the matrices $A \in \mathbb{R}^{n \times k}$, $B\in \mathbb{R}^{n-k \times n}$ both of full rank, where $im(A)=\ker(B)$. In other words,the kernel of B describes the image of A and thus we have connections like $im(A)\oplus \ker(B)^\perp$ and since $\ker(B)=im(B^T)^\perp$ the vectors $(a_1, \dots, a_k, b^1, \dots, b^{n-k})$ [with the columns of A and rows of B] are linearly independent (full rank and orthognonality). The matrices are also called Gale Dual.
Assume we would want to expand $A$ to a new matrix $M \in \mathbb{R}^{n\times n}$ , where we set the first k columns as A and leave the last $n-k$ free for now. Additionally, define a matrix $\tilde{M} \in \mathbb{R}^{n\times n}$ , where we set the last $n-k$ rows as B and leave the first k rows arbitrary. Then we have something like:
\begin{equation}
M=
\begin{bmatrix}
   \vert & \vert & \vert & * & & * \\
    \vert   & \vert  & \vert  & * && *  \\
    a_1   & \dots  & a_k  & * &\dots & * \\
    \vert   & \vert  & \vert  & * && * \\
    \vert   & \vert  & \vert  & * && *  \\
\end{bmatrix}
\end{equation}
\begin{equation}
\tilde{M}=
\begin{bmatrix}
  * & * &*&*&* \\ &&\dots&& \\  * & * &*&*&* \\ 
  \text{---} \hspace{-0.2cm} &\text{---} \hspace{-0.2cm} & b_1 & \hspace{-0.2cm} \text{---} &\text{---} \hspace{-0.2cm} \\
  \text{---} \hspace{-0.2cm} &\text{---} \hspace{-0.2cm} & \dots & \hspace{-0.2cm} \text{---} &\text{---} \hspace{-0.2cm} \\
  \text{---} \hspace{-0.2cm} &\text{---} \hspace{-0.2cm} & b_{n-k} & \hspace{-0.2cm} \text{---} &\text{---} \hspace{-0.2cm} \\
\end{bmatrix}
\end{equation}
My question is: Can we set the last $n-k$ columns of $M$ and the first $k$ rows of $\tilde{M}$, s.t.
\begin{equation}
M^{-1}=\tilde{M}  ?
\end{equation}
I would assume so. I tried to interpret the unknown matrix entries in $M$ and $\tilde{M}$ as variables that we derive from the linear equation system
\begin{equation}
M\tilde{M}=\mathit{I}_n
\end{equation}
(where the variables are in the last columns of M and the first rows of \tilde{M}).
But unfortunately this gets technical when I then try to show that the new linear equation system has a solution...
Is there another way? An abstract argument is more than fine. Only need the existence.
Thanks for your help!
Greatly appreciated
 A: The answer is yes. First, fill in the columns of $M$. In particular, take
$$
M = \pmatrix{A_1 & A_2},
$$
where $A_1 = A$ and $A_2$ is a full column-rank matrix satisfying $B A_2 = I$. Such an $A_2$ must exist because $B$ has full row-rank; one example of such a matrix is the pseudoinverse $A_2 = B^+$. To see that $M$ is invertible, it suffices to note that $A_1,A_2$ each have linearly independent columns and that the images of $A_1$ and $A_2$ do not intersect. To see that the images do not intersect, note that the image of $A_2$ does not have any intersection with the kernel of $B$; otherwise, $BA_2$ would have a non-trivial kernel, but $BA_2 = I$.
Write $\tilde M = \left[ \begin{smallmatrix}B_1\\B_2\end{smallmatrix}\right]$, where $B_2 = B$. $\tilde M$ is the inverse of $M$ if and only if $\tilde M M = I$, which if we expand the product $\tilde M M$ using block-matrix multiplication gives us the required conditions
$$
B_1 A_1 = I, \quad B_1 A_2 = 0, \quad B_2A_1 = 0, \quad B_2A_2 = I.
$$
The equation $B_2A_1 = 0$ (i.e. $BA = 0$) holds automatically, and $B_2A_2 = I$ follows from our construction of $A_2$. Thus, it suffices to select $B_1$ such that $B_1A_1 = I$ and $B_1 A_2 = 0$. We can more simply write this equation as
$$
B_1M = \pmatrix{I&0} \implies B_1 = \pmatrix{I&0}M^{-1}.
$$
In other words, now that we have constructed $M$, we can simply take $B_1$ to be the top-portion of $M^{-1}$, and the necessary relations for the $A_i$ and $B_i$ will follow.
