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I'm interested in invertible matrices that are built out of invertible sub-blocks. For example, four sub-blocks from $GL_n(F)$ (i.e. the group of $n \times n$ invertible matrices over a field $F$) can be assembled into a $2n \times 2n$ matrix, which may or may not be invertible.

Suppose that a $kn \times kn$ matrix, $M$, composed of $k^2$ invertible sub-blocks is invertible. Is it true that each sub-block of $M^{-1}$ is invertible?

I think that it is true (although I am happy to be shown otherwise), but I am having difficulty constructing a general proof.

In the case $k=1$, there is nothing to prove.

For the case $k=2$, we can use block-wise row reduction, as described here.

For $k>2$, I'm stuck. I've tried examining the maps $X \mapsto MX$ and $Y \mapsto M^{-1}Y$ where I suppose that some sub-block $B_{ij}$ of $M^{-1}$ is not invertible. My thought was that this might tell me something about block $B_{ji}$ of $M$, but I can't seem to make any conclusions.

Any suggestions?

Note that I am mainly interested in the case where $F = \mathrm{GF}(2)$.

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I just found a counter-example for the case $k=3$ using $2 \times 2$ invertible sub-blocks over $\mathrm{GF}(2)$. I will post it below. So, no, in general $M^{-1}$ cannot be constructed in the same way that $M$ is. –  user61836 Feb 11 '13 at 20:49

2 Answers 2

Let $M=\left(\begin{array}{lll} 1 & 1 & 2 \\ 1 & 1 & 3 \\ 2 & 3 & 1 \end{array}\right)$

If I understand your question right, this is a counterexample as the upper left $2\times2$ block is not invertible (over $\mathbb{R}$, say). This would be for $k=3$ and $n=1$.

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Well this sort of works, I just realized you wanted sublocks of $M^{-1}$ to be invertible. But for this example, $M^{-1}$ has a zero entry, so that block is not invertible. –  Trevor Feb 11 '13 at 14:56
    
Your example doesn't apply. We start by fixing the size of the sub-blocks, say $2 \times 2$. To build a $4 \times 4$ matrix, $M$, you put four sub-blocks together in a grid. This bigger matrix $M$ may or may not be invertible. Suppose it is. I want to know if the four sub-blocks of $M^{-1}$ are invertible (i.e. can we similarly build $M^{-1}$ by combining four $2 \times 2$ invertible sub-blocks?). –  user61836 Feb 11 '13 at 15:33
up vote 0 down vote accepted

I've just found a counter-example using Maple.

Let the sub-blocks be the set of invertible $2 \times 2$ matrices over $\mathrm{GF}(2)$ (note that there are 6 different sub-blocks). We use nine sub-blocks to build the following matrix: $$ M=\begin{bmatrix} 1 & 1 & 1 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 & 1 & 0 \end{bmatrix} $$ $M$ is invertible; here is its inverse: $$ M^{-1}=\begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 0 & 1 \\ \end{bmatrix} $$ When $M^{-1}$ is considered as a 3-by-3 array of nine sub-blocks, there are a number of sub-blocks that are not invertible (e.g. the top left corner). So $M^{-1}$ cannot be constructed in the same way that $M$ was.

Thanks for reading :-)

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