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I am completely stuck in the following linear algebra problem.

Consider a finite group $H$ and $N\times N$-matrices $M_{g,h}$ with entries in $\mathbb{Z}$ for all $g,h\in H$. Assume $\sum_{h\in H} M_{g,h}=D$ for all $g\in H$, where $D$ is some $N\times N$-matrix with $\det(D)\neq 0$. Furthermore $M_{g,h}=M_{1,g^{-1}h}$ for all $g,h\in H$.

The QUESTION is wether the determinant of the $N|H|\times N|H|$-block matrix $M:=(M_{g,h})_{g,h\in H}$ is not zero.

For further reduction one may assume that the entries of the matrices are in the set $-1,0,1$, that $M_{g,h,i,j}\neq 0$, $1\leq i,j\leq N$, implies $M_{g,h',i,j}=0$ for all $h'\neq h$ and not more than three $M_{g,h}\neq 0$ for fixed $g$.

The answer is certainly positive if $H$ is abelian, but unfortunately I cannot assume commutativity.

I would be happy if someone knew a place where this kind of problem was dealt with before or could point out a solution. Of course I would appreciate a counterexample, although less satisfactory.

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Have you tried the case where $H$ has a normal subgroup $G$, then use the Schur complement formula to reduce the question to the case of $G$.? The simplest situation is when $(H:G)=2$. A $2$-group must be a nice situation to carry out. –  Denis Serre Mar 22 '11 at 15:57

1 Answer 1

up vote 3 down vote accepted

Even if $H$ is Abelian (see MO question), the answer is No. Here is a counter-example where $H=\mathbb Z_2$ and $N=2$. We have $M=\begin{pmatrix} A & B \\\\ B & A \end{pmatrix}$, where $D=A+B$. The assumption is $\det(A+B)\ne0$. One verifies easily that $\det M=0$ if and only if $\det (A-B)=0$. Let us choose $$A=\begin{pmatrix} 1 & 1 \\\\ 1 & 0 \end{pmatrix},\qquad B=\begin{pmatrix} 0 & 0 \\\\ 0 & -1 \end{pmatrix}.$$ We do have $\det(A+B)\ne0$, but $\det(A-B)=0$, and therefore $\det M=0$. Yet the extra assumptions about the entries are satisfied too.

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This is definitely true. So I should wonder why I overlooked such easy examples... Thank you very much. –  Abel Stolz Mar 23 '11 at 8:27

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