A semisimple matrix that is not diagonalizable over algebraic closure?

Question

In this quetion, it has been proved that an operator is semisimple iff it's diagonalizable over the algebraic closure of the base field.

However, when the minimal polynomial is not separable, it seems that there's something going wrong. Consider operator on $\mathbb F_2(t)$ to be

$$ T=\begin{bmatrix} &&1&\\ &&&1\\ t&&&\\ &t&& \end{bmatrix}. $$

Clearly it has minimal polynomial $m(x)=x^2+t$, which is irreducible but not separable on $\mathbb F_2(t)$.

Then by definition, $T$ is semisimple since $m(x)$ is a product of irreducible factors. While $T$ only has two linearly independent eigenvectors $(1,0,\sqrt t,0)$ and $(0,1,0,\sqrt t)$ in $\mathbb F_2(\sqrt t)$.

What's wrong here? Thanks in advance!

Answer 1

Reordering the bases, you just have the diagonal sum of two copies of $\begin{bmatrix}0&1\\t&0\end{bmatrix}$. This has characteristic polynomial $x^2+t$, as you noted, which factorises over the algebraic closure as $(x+\sqrt t)^2$. This has a repeated root, so the matrix is not necessarily diagonalisable.

The result you (mis-)quoted says semisimplicity is the same as the minimal polynomial splitting into distinct linear factors over the algebraic closure. In our example the minimal polynomial is the characteristic polynomial, so the matrix is not diagonalisable.

It must therefore yield the Jordan block $\begin{bmatrix}\sqrt t&1\\0&\sqrt t\end{bmatrix}$. Taking the basis $u=\begin{bmatrix}1\\\sqrt t\end{bmatrix}$ and $v=\begin{bmatrix}1\\1+\sqrt t\end{bmatrix}$ shows that this is indeed the case.