I've stumbled across the above question. Usually, I don't spend my time on such problems, though it's a nice question in my opinion.

My idea:

Case $p=2$ is easy, follows from $x^2 \equiv 2 \;\;\; (\text{mod} \;\; 4)$ has no integer solution.

Assume the contrary and $p$ is an odd prime number. Since, $A+A^t$ is symmetric, then is has only real eigenvalues, so if $\det(xI-(A+A^t)) \in \mathbb{Z}[x]$ is the characteristic polynomial, it factors $(x-\lambda_1) \cdots (x-\lambda_p)$ where $\lambda_i$'s are real numbers. By rational root test, if any of $\lambda_i$ is rational, then it will be an integer. Therefore, if we show that non of them is irrational (hence, all $\lambda_i$ are integers), we may then conclude as follows;

$\sum_i \lambda_i=-tr(A+A^t)=2k$ where $k \in \mathbb{Z},$ and $\lambda_1 \cdots \lambda_p=-p.$ Now, $p$ is odd, then wlog,

  1. $\lambda_p=-p$ and $\lambda_1=\cdots=\lambda_{2t}=-1$ and $\lambda_{2t}=\cdots=\lambda_{p-1}=1,$ but $\sum_i \lambda_i=-2t+p-1-2t-p$ which is odd,

  2. $\lambda_p=p$ and $\lambda_1= \cdots=\lambda_{2t+1}=-1$ and $\lambda_{2t+1}=\cdots=\lambda_{p-1}=1,$ but again $\sum_i \lambda_i=-2t-1+p-1-2t-1+p$ which is odd,

hence contradiction.

Now, we're left to show that (in my words)

Lemma: Given $p$ integers and irrational numbers, where $p$ is an odd prime number. We have that

$$\sum_i\lambda_i=2k \in \mathbb{Z}, $$

$$\sum_{i<j}\lambda_i\lambda_j \in \mathbb{Z},$$




$$\lambda_1\cdots \lambda_p=-p \in \mathbb{Z}.$$

Then, non of the $\lambda_i$ is irrational.

An idea to prove the lamma: Since any symmetric polynomial in $p$ variables $\lambda_i$ with integer coefficient can be expressed in terms of elementary symmetric polynomials above, then any symmetric (polynomial with integer coefficient) expression of $\lambda_i$'s will be an integer.

Is it possible to conclude from here that non of the $\lambda_i$ is irrational?

P.S. I'd appreciate any other ideas for solving it.


Or straight from a definition of the determinant: Let $A$ be a symmetric square integer matrix with odd dimension and even entries on the diagonal. Then

$$ \det(A) = \sum_{\sigma}\textrm{sgn}(\sigma)\prod_iA_{i\sigma(i)}. $$

Now $\sigma$ and $\sigma^{-1}$ have equal sign and

$$ \prod_iA_{i\sigma^{-1}(i)} = \prod_iA_{\sigma(i)i} = \prod_iA_{i\sigma(i)}. $$

So if $\sigma \neq \sigma^{-1}$ then their combined contribution to the determinant is even. If $\sigma = \sigma^{-1}$ then $\sigma$ has a fixed point and so its contribution is also even. Therefore $\det(A)$ is even.

  • $\begingroup$ yes, a lot easier way! $\endgroup$ – Ehsan M. Kermani Mar 11 '12 at 22:42

If $p$ is odd, the determinant of the skew-symmetric $p\times p$ matrix $A^t-A$ is zero. But $$ A+A^t\equiv A^t-A\pmod2, $$ so $$ \det(A+A^t)\equiv\det(A^t-A)\equiv 0\pmod2. $$ Thus $\det(A+A^t)$ is an even integer, and hence $\neq p$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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