Matrix with no eigenvalues 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 $F$. 

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.
