MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.