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I wanted to see if there is any connection between the invertibility of a matrix and the invertibility of a particular block of the matrix.

Particularly I want to find out the largest size of a subgroup of $GL_n(\mathbb{F}_q)$ with the property that the first $k \times k$ block of all the matrices in that subgroup is invertible.

I have formalized the problem (trying to do away with matrices) in the following manner.

Let $V$ be a vector space over $\mathbb{F_q}$ of dimension $n$. Let $W$ be a subspace of $V$ so that $V = W \oplus W'$ and dim($W$) $= k$. Let $P$ denote the projection along $W'$ onto $W$ and $G$ be a subgroup of $GL(V)$. Then define $\tilde{G} = \{ T \in G |$ $P\circ T: W \to W$ is invertible $\}$.

1) Under what conditions is $G = \tilde{G}$?

2) What is the largest size of $G$ such that $G=\tilde{G}$?

A few observations:

1) If $G = GL(V)$, then $G \neq \tilde{G}$.

2) I have a group of size $|M_{k \times (n-k)}(\mathbb{F}_q)|$ ($M_{k \times (n-k)}(\mathbb{F}_q)$ is the additive group of $k \times (n-k)$ matrices over $\mathbb{F}_q$) for which $G = \tilde{G}$ holds.

P.S: There was a power cut in the campus suddenly. I will complete the details, if required, at a later moment. Sorry for the inconvenience.

Thank you,


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For your observation 2, consider letting the diagonal blocks be arbitrary invertible matrices, not just the identity matrix. – Jack Schmidt Sep 29 '11 at 19:43
Consider the (parabolic) group $G$ of matrices of the block form $$\left(\begin{array}{cc}A & B\\0 & D\end{array}\right),$$ where $A$ and $D$ are invertible, $B$ arbitrary, $A$ of size $k\times k$, $D$ of size $(n-k)\times (n-k)$. I would be somewhat surprised, if $G$ is not maximal (w.r.t inclusion) among the groups with invertible $k\times k$ blocks in the upper left. I don't know if it is always the largest. Instead of $G$ you can equally well use its transpose. – Jyrki Lahtonen Sep 29 '11 at 20:32
@JackSchmidt: That helps, thank you :) But I still don't know if this is the largest possible group with the required property :( – Isomorphism Sep 29 '11 at 20:35
@JyrkiLahtonen: Isn't your suggestion the same as Jack's suggestion? Am I missing something? Thank you – Isomorphism Sep 29 '11 at 20:37
Oopsie. Yeah, I guess it is. I was busily trying to come up with a proof that $G$ is maximal. Need some sleep... – Jyrki Lahtonen Sep 29 '11 at 20:40

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