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Let $A_{20 \times 20}$ be a real matrix such that:
$\ \ \ \bullet$ $a_{ii}=0$ for $1 \le i \le 20$
$\ \ \ \bullet$ $a_{ij} \in \{-1;1\}$ for $1 \le i,j \le 20$ and $ i \neq j$
Prove that $A$ is nonsingular.

Anyone can help me find out an useful idea for this problem.

HLong
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2 Answers2

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HINT:

$$A^2 = I_{20}\ (\!\! \! \! \!\mod 2\,)$$

orangeskid
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  • Can you explain how to use this? – HLong Oct 25 '14 at 09:32
  • You have $\det A \equiv 1\ (! ! !\mod 2,)$ so a non zero integer. – orangeskid Oct 25 '14 at 09:37
  • But why we can work over $\mathbb Z_2$ instead of $\mathbb R$ – HLong Oct 25 '14 at 09:41
  • Because your matrix has integer entries. You want to show that the determinant is non zero. It's enough to show it's an odd integer. Still it's good that you notice that just because it's invertible mod $2$ it's not guaranteed to be invertible in the integers. But it works over $\mathbb{R}$ since it's a field, and $\det\ne 0$ is enough. – orangeskid Oct 25 '14 at 10:21
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    Take a look at the accepted answer here, it might help to understand why working modulo $2$ works. You can try to prove that in the case $2 \times 2$, it is instructive. – Ivo Terek Oct 25 '14 at 17:48
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I think you are aware that

$$\det A=\sum_{\sigma}\prod_{i=1}^{20} a_{i\sigma(i)}$$

The number of $\sigma$ such that $\forall i:\sigma(i)\ne i$ is $!20$ (http://en.wikipedia.org/wiki/Derangement) which is an odd number (it can be proven by induction that $!n=n-1(\operatorname{mod}2)$). therefore

$$\det A=1(\operatorname{mod}2)$$ so $\det A\ne0$. Direct proof I think.

anonymous67
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