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Here is another problem from Golan.

Problem: Let $F$ be a finite field. Show there exists a symmetric $2\times 2$ matrix over $F$ with no eigenvalues in $F$.

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    @BabakSorouh Thank you!2012-06-14

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The solution is necessarily split into two cases, because the theory of quadratic equations has a different appearance in characteristic two as opposed to odd characteristic.

Let $p=\mathrm{char}\, F$. Assume first that $p>2$. Consider the matrix $ M=\pmatrix{a&b\cr b&c\cr}. $ Its characteristic equation is $ \lambda^2-(a+c)\lambda-(ac-b^2)=0.\tag{1} $ The discriminant of this equation is $ D=(a+c)^2-4(ac-b^2)=(a-c)^2+(2b)^2. $ By choosing $a,c,b$ cleverly we see that we can arrange the quantities $a-c$ and $2b$ to have any value that we wish. It is a well-known fact that in a finite field of odd characteristic, any element can be written as a sum of two squares. Therefore we can arrange $D$ to be a non-square proving the claim in this case.

If $p=2$, then equation $(1)$ has roots in $F$, if and only if $tr((ac-b^2)/(a+c)^2)=0.$ By selecting $a$ and $c$ to be any distinct elements of $F$, we can then select $b$ in such a way that this trace condition is not met, and the claim follows in this case also.

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    @Theorem If the characteristic was $ab$ for a,b>1, then the elements $a$ and $b$ (modelling the "field" as numbers from $0$ to $ab-1$ as usual) would be non-zero but multiply to $0$, and fields can't have zero divisors.2012-06-14