How can we expand a matrix while maintaining the entries of its inverse?

Let $A$ be an invertible $n \times n$ matrix. Fix some $m > n$. An invertible $m \times m$ matrix $B$ inverts with $A$ if the principal submatrix consisting of the first $n$ rows and cols of $B^{-1}$ is equal to $A^{-1}$.

Given $A$ and $m$, I am trying to completely characterize the set of matrices that invert with $A$. How can I do this?

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Judging by the title, don't you also want $B$ to have its uppermost principal submatrix equal to $A$? – Erick Wong Feb 5 '13 at 3:18
The title is ambiguous, but I don't necessarily need that to be true. – GMB Feb 5 '13 at 7:24

Suppose $B=\left[ \begin{matrix}P&Q\\R&S\end{matrix}\right]$ is invertible, where the size of $P$ is equal to the size of $A$.

The leading principal submatrix of $X^{-1}$ that conforms to the size of $A$ is given by $Y=\left(P-QS^{-1}R\right)^{-1}$ if $S$ is invertible, or $Y=P^{-1}+P^{-1}Q(S-RP^{-1}Q)^{-1}RP^{-1}$ if $P$ is invertible (where $P-QS^{-1}R$ is the Schur complement of $S$ in $B$ and $S-RP^{-1}Q$ is the Schur complement of $P$ in $B$). So, the condition you need is $Y=A^{-1}$.

The case where both $P$ and $S$ are singular needs further investigation. It seems like the leading principal submatrix of $X^{-1}$ is not invertible in this case, but I am not sure.

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Exactly what I was looking for. Thank you. – GMB Feb 5 '13 at 17:18

Let's take a simple case. Let $A=(1)$, $m=2$. We want all the $2\times2$ matrices $B=\pmatrix{a&b\cr c&d\cr}$ such that $$\pmatrix{a&b\cr c&d\cr}^{-1}=\pmatrix{1&*\cr*&*\cr}$$ where we don't care what the stars stand for. Well, we know a formula for the inverse of a $2\times2$ matrix, and it tells us the answer to this problem is all $B$ such that $ad-bc=d\ne0$.

I doubt that there will be a useful answer for bigger matrices, but, who knows?

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If we add the semi-plausible condition that $B$ contains $A$ in a similar manner (or just $a=1$ in this example), then the criterion becomes $bc=0$ (and $d\ne 0$), which has a much nicer ring to it. – Erick Wong Feb 5 '13 at 3:27

I don't know, whether this gives really a solution in the way that you sought for. But what about the formalism of the LDU-decomposition? Assume the LDU-factors of A as $A = L_A \cdot D_A \cdot U_A$ where $L_A$ is lower triangular having units on the diagonal , $D_A$ is diagonal, $U_A$ upper triangular also having units on the diagonal. Then $A^{-1} = U_A^{-1} \cdot D_A^{-1} \cdot L_A^{-1}$ Assume the analogue decomposition of $B$ and of $B^{-1}$ as $B = L_B \cdot D_B \cdot U_B$ and $B^{-1} = U_B^{-1} \cdot D_B^{-1} \cdot L_B^{-1}$ . Then the matrix $A^{-1}$ is determined by the dot product of the nxm - submatrix of $U_B^{-1}$ with $D_B^{-1}$ and the mxn - submatrix of $L_B^{-1}$.

With this we can describe an infinite set of solutions in that submatrices (because m>n and we have many degrees of freedom), and because the matrices are triangular and diagonal, it might be possible to describe the non-inverses depending on that initial setting with not too much hazzle. Then in the final step the matrix B itself has to be described by the dot-product of the non-inverse LDU-factors.
I expect, that for instance with Pari/GP one can manage this with a parametrized solution for instance for n=2 and m=3 as an example of the general feasibility of a solution.

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