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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.

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2 Answers 2

up vote 3 down vote accepted

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.

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yes, a lot easier way! –  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 $\neq p$.

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