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Let $K$ be a field such that $ \mathrm{char}(K) \neq 2 $. Let $ p(x) $ be an arbitrary irreducible polynomial over $K$ of degree $n$. Using the rational canonical form, we can always construct an $ n \times n $ matrix $A$ such that $p(A)=0$.

Can we always construct a symmetric matrix $A$ such that $p(A) = 0$ ?

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    $\begingroup$ If $p(x)=x^2-s$ then $s$ has to be a sum of two squares. This is not the case, for instance when $p$ is $-1$ mod $4$, in $\mathbf{F}_p((s))$ (I believe that trying the case $n=2$ would have been the right thing to do before asking). $\endgroup$
    – YCor
    Commented Nov 4, 2023 at 11:52

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Let $\mathbb{F}$ be a field of characteristic two, and let $K=\mathbb{F}(t)$, the field of rational functions over $\mathbb{F}$. Let $p(x)=x^2-t$. If $A$ is a symmetric matrix with entries in $K$ then $A^2$ has its diagonal entries in $\mathbb{F}(t^2)$, so it can't be $t$ times the identity.

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  • $\begingroup$ Is there a counterexample for characteristic greater than 2? $\endgroup$
    – Max Alekseyev
    Commented Nov 4, 2023 at 12:22
  • $\begingroup$ My example shows that there are no symmetric matrices of any size, which may just be a characteristic two phenomenon. If you want the size to be the degree, as in the question, you can use YCor's example: just take $x^2-t$ where $t$ is not a sum of two squares. $\endgroup$
    – Dave Benson
    Commented Nov 4, 2023 at 13:17
  • $\begingroup$ Sorry for writing $\text{char}(K) \neq 0$, it should be $\text{char}(K) \neq 2$. I have edited the question. $\endgroup$
    – Sky
    Commented Nov 4, 2023 at 13:35

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