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If $A \in \mathbb{R}_{n\times n}$ has only integer entries, show that the only rational characteristic values of A are integers.

I have absolutely no idea how to approach this problem.

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Please make the body of the post self-contained. You don't expect a book to start at the spine, do you? The subject line is not part of the post, it's an indexing feature. –  Arturo Magidin Jul 15 '12 at 20:40

2 Answers 2

If the entries of $A$ are integers, then the characteristic polynomial is the determinant of a matrix with coefficients in $\mathbb{Z}[t]$, and so is a polynomial with integer coefficients. Moreover, the characteristic polynomial is monic.

What do you know about rational roots of polynomials with integer coefficients? There's even a "Test" named after them...

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There is a direct argument too, as I mentioned in a recent post on a related question. Suppose that $Av = \lambda v$ where $v$ is a non-zero vector with rational entries and $\lambda = \frac{r}{s}$, where $r$ and $s$ are relatively prime integers. We may multiply $v$ by an integer and assume that all its entries are integers. Furthermore, we may suppose that the gcd of the entries of $v$ is $1$ (otherwise divide through by the gcd). Then note that $1$ may be written as an integer combination of the entries of $v.$ Now we have $sAv = rv.$ Now every entry of the vector on the left is an integer multiple of $s$. On the other hand, $r$ may be written as an integer combination of the entries of the vector on the right. Hence $s$ divides $r.$ Since ${\rm gcd}(r,s) = 1,$ we must have $s = \pm 1.$ Hence $\lambda$ is an integer.

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