# Determinant of a Special Symmetric Matrix

If $A$ is a symmatric matrix of odd order with integer entries and the diagonal entries $0$ then $A$ has determinant value even.

I can prove the result if I can show that the eigenvalues of $A$ are integers,but I am unable to do that. Thanks for any help.

• Note: $\textrm{eig} \begin{pmatrix} 0 & 2 & 1 \\ 2 & 0 & 1 \\ 1 & 1 & 0\end{pmatrix} = \{2.732\ldots, -2, -0.732\ldots\}$. – Emily May 10 '14 at 19:15
• it is enough to have the diagonal entries even – Will Jagy May 10 '14 at 20:59
• Related to but not duplicate of What is the determinant of a symmetric n×n matrix with all diagonals be 0 and all others are non-negative integers? – user1551 May 10 '14 at 21:09
• @user1551 An approach : A is congruent modulo 2 to a skew-symmetric matrix and an antisymmetric matrix of odd size has a zero determinant. – user146010 May 13 '14 at 20:32
• What is your argument if the eigenvalues are integers? (The eigenvalues of an integer matrix are definitely algebraic integers, so perhaps the argument can be generalized...) – Ben Blum-Smith May 16 '14 at 17:05

As suggested by @Grigory M I post this as an answer :

A is congruent modulo 2 to a skew-symmetric matrix and an antisymmetric matrix of odd size has a zero determinant.

• Very nice! Note (if anyone else was confused by this) that if a matrix is antisymmetric "mod 2" that does not imply it has determinant zero mod 2 (for obvious reasons). We must use the fact that the resulting matrix is actually antisymmetric over $\mathbb{Z}$. – 6005 May 16 '14 at 21:49

Hint. $\det(A)=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^na_{i\sigma(i)}$. Partition $S_n$ into three mutually disjoint subsets. The first subset contains those permutations that have some fixed points. The permutations in the second subset are inverses of the permutations in the third one.

• Minor typo: you forgot the term for the sign of the permutation. – 6005 May 10 '14 at 21:58
• @Goos Ha ha, of course, you are right. It's now fixed. Thanks. – user1551 May 10 '14 at 22:00

To expand on user1551's hint, using the formula $$\det A = \sum_{\sigma \in S_n} {\operatorname{sgn} \sigma} \prod_{i=1}^n a_{i, \sigma(i)}$$

• For any permutation $\sigma \in S_n$, the inverse permutation $\sigma^{-1}$ has the same sign as $\sigma$. Furthermore, since the matrix is symmetric, we can conclude that $\sigma$ and $\sigma^{-1}$ each contribute the same value to the determinant, if they are distinct.

• Permutations such that $\sigma = \sigma^{-1}$ are a product of disjoint transpositions. Since $n$ is odd, what can we conclude about the contribution of such permutations to the determinant?

More generally, let $R$ be a commutative ring with unity and with characteristic $2$. If $A\in M_{2k+1}(R)$ is symmetric or skew-symmetric with $0$ diagonal, then $\det(A)=0$. I gave a sketch of a proof in Odd-dimensional skew-symmetric matrix is singular, even in a field of characteristic 2. Curiously user1551 gave also an answer. Note that we may discuss about the definition of a skew-symmetric matrix in characteristic $2$ ; the notion has not any interest if we do not require that the diagonal must be $0$. Thus it is convenient to add this condition in the definition.

Thus assume that $R$ is any commutative ring with unity (with any characteristic) and$A=[a_{i,j}]\in M_{2k+1}(R)$ with $a_{i,j}=-a_{j,i},a_{i,i}=0$. We work in the ring of polynomials $\mathbb{Z}[(a_{i,j})_{i,j}]$ in the indeterminates $(a_{i,j})$. This ring has no zero-divisors, is factorial and its characteristic is $0$. Clearly $\det(A)\in\mathbb{Z}[(a_{i,j})_{i,j}]$ is $0$ (FORMALLY $0$). The determinant that we consider is an evaluation in $R$ and, consequently, is $0$.

Note that $\mathbb{Z}[(a_{i,j})_{i,j}]$ is also an integrally closed ring. This is a key to show that, if $A\in M_{2k}(R)$, then $\det(A)$ is the square of an element of $R$.