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.

  • 2
    $\begingroup$ 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. $\endgroup$ – Peter Mueller Feb 28 '13 at 15:22
  • $\begingroup$ 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. $\endgroup$ – Will Sawin Feb 28 '13 at 16:15
  • 1
    $\begingroup$ An individual block is diagonalizbale if and only if its characteristic polynomial is the characteristic polynomial of a separable field extension. $\endgroup$ – Will Sawin Feb 28 '13 at 16:17

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!

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.