Let $n$ be a positive odd integer and let $A$ be a symmetric $n\times n$ matrix of integer entries such that $a_{ii}=0,i=1,2.....n$. Show that the determinant of $A$ is even.

I tried using definition of determinant. Also I can't use induction since the problem is for odd integers. Please give some thoughts on how to solve it.

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    $\begingroup$ You can use induction for odd integers, but I doubt it would work here. (You prove for $n=1$ then prove if true for $n$ that it is true for $n+2$.) $\endgroup$ – Thomas Andrews Nov 6 '14 at 15:46

The basic trick is to note that if $\sigma\in S_n$, the complete permutation group on $n$ elements, and $\sigma^2=1$, then there is a fixed point of $\sigma$. That's only true if $n$ is odd.

So this means that in the definition:

$$\det A = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) a_{1\sigma(1)}\cdots a_{n\sigma(n)}$$

the terms with $\sigma^2=1$ contribute zero, since $a_{ii}=0$, and the term from other $\sigma$ can be paired with th term from $\sigma^{-1}$. Those terms are equal by the symmetry of the matrix and the sign of a permutation being equal to the sign of its inverse.

(Technically, you don't need to know the signs are the same, since if they are different, they still contribute an even number together, $0$.)

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  • $\begingroup$ how to prove the fact that u have used in the beginning $\endgroup$ – Learnmore Nov 6 '14 at 16:06
  • $\begingroup$ If there were no $i$ such that $\sigma(i)=i$ then you can partition $\{1,2,\dots,n\}$ into pairs $\{i,\sigma(i)\}$. In group theory terminology, the "orbits" of $\sigma$ each must have $1$ or $2$ elements, and when $n$ is odd, there must be at least one orbit with one element. $\endgroup$ – Thomas Andrews Nov 6 '14 at 16:08
  • $\begingroup$ Try out some examples. What does it mean for $\sigma^2=1$? $\endgroup$ – Thomas Andrews Nov 6 '14 at 16:10
  • $\begingroup$ upto the terms are equal by symmetry is clear but "sign of.. inverse "can u please explain $\endgroup$ – Learnmore Nov 6 '14 at 16:24

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