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Let $K$ be a field. For a matrix $A\in GL_n(K)$ we can find the Jordan normal form $A'$ in $GL_n(\overline{K})$, where $\overline{K}$ is the algebraic closure of $K$. We write $j_\alpha(A)$ for the number of $1\times 1$ Jordanblocks with eigenvalues $\alpha$ in $A'$. Now consider $\alpha\notin K$ and the corresponding number $j_\alpha(A)$. If $K$ is a perfect field, then $\overline{K}:K$ is a separable extension and for each occurence of $\alpha$ there must be at least one eigenvalue $\beta$ of $A$ which is a Galois conjugate of $\alpha$, and hence in particular $\alpha\neq \beta$. So we can conclude that $j_\alpha(A)\leq\frac{1}{2}n$. The typical example is $$\left(\begin{matrix} 0 & 1\\\ -1 &0\end{matrix}\right)$$ over $\mathbb{R}$ with complex eigenvalues $\alpha=i$, $\beta=-i$.

The question is: If $K$ is an imperfect field, is it still true that $j_\alpha(A)\leq \frac{1}{2}n$ if $\alpha\notin K$? If not, how can one find a counterexample?

I tried more or less randomly computing Jordan normal forms of matrices over $\mathbb{F}_2(x^2)$ and similar fields, starting from the (non-counter)example $$\left(\begin{matrix} 0 & 1\\\ x^2 &0\end{matrix}\right)$$ with eigenvalues $x$ and $x$ (and one $2\times 2$ Jordan block). Unfortunately I am lacking ideas how to constructively build a working counterexample.

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The Frobenius normal form shows that there cannot be a counterexample which is diagonalizable over $\bar K$. On the other hand, I believe that if there is a counterexample to your question, then there should be a diagonalizable one too. –  Peter Mueller Feb 28 '13 at 15:22
    
The Frobenius normal form also shows that we can reduce to the diagonalizable case. We can treat each block separately, and each either is diagonalizable or has a Jordan normal form with no $1 \times 1$ blocks. –  Will Sawin Feb 28 '13 at 16:15
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An individual block is diagonalizbale if and only if its characteristic polynomial is the characteristic polynomial of a separable field extension. –  Will Sawin Feb 28 '13 at 16:17
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up vote 2 down vote accepted

To draw a conclusion from the comments, assume $\alpha$ is an eigenvalue of $A$, where $K$ has characteristic $p$. If $K(\alpha):K$ is inseparable, the minimal polynomial of $\alpha$ over $K$ is inseparable, hence $f'=0$. On the other hand, if $\alpha$ is the only root of $f$, $f$ divides the characteristic polynomial $(x-\alpha)^s$, which is the characteristic polynomial of the Jordan block (of size $s$) corresponding to $\alpha$. Hence $f=(x-\alpha)^k$, $0=f'=k(x-\alpha)^{k-1}$, and therefore $p|k$. This implies $s\geq k\geq p\geq 2$, which is all I wanted to know.

Thank you for giving the above hints!

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