Example of $f$ semi-simple but not diagonalizable

Question

In our notes, we have written down: $f$ diagonalizable $\Rightarrow$ $f$ semi-simple (*)

However, we also mentioned that

$f$ semi-simple $\iff$ the minimal polynomial $\mu_{f}=\prod^{n}_{i=1}(X-\lambda_{i})$ and $\lambda_{i}\neq \lambda_{j}$, when $i \neq j$.

But it is also know that

$f$ is diagonalizable $\iff$ the minimal polynomial $\mu_{f}=\prod^{n}_{i=1}(X-\lambda_{i})$ and $\lambda_{i}\neq \lambda_{j}$, when $i \neq j$.

Using this logic, surely (*) should be "$\iff$", so equivalent???

As a counterexample, I have been given $M= \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} $. I realize that $\chi_{M}=X^{2}+1$, and so irreducible in $\mathbb R[X]$. And so also not diagonalizable in $\mathbb R$. But then somehow this $M$ must still be semi-simple in order to still be a counterexample. And if it is semi-simple, then by the definition above, it must also have $\mu_{M}$ with distinct values, but $\chi_{M}$ is not reducible so how do we find factors for $\mu_{M}$.

I am confused

Answer 1

I'm not sure where did you get that from:

$f$ semi-simple $\iff$ the minimal polynomial $\mu_{f}=\prod^{n}_{i=1}(X-\lambda_{i})$ and $\lambda_{i}\neq \lambda_{j}$, when $i \neq j$.

because that's not true. This is only true over algebraically closed fields and indeed, over algebraically closed fields the implication "semi-simple $\Rightarrow$ diagonalizable" holds.

In general though $f$ is semi-simple iff $\mu_f=\prod g_n$ where each $g_n$ is an irreducible polynomial and $g_n\neq g_m$ (up to scalar) for $n\neq m$.

That's why your $M$ is semi-simple, because $\mu_M=X^2+1$ is irreducible over $\mathbb{R}$.

Answer 2

A space can be semi-simple without being the sum of simples of dimension one. Diagonalizable operators are semi-simple and sums of simples of dimension one, while your example gives an operator $M$ on $V=\mathbb R^2$ so that $V$ has no $M$-invariant subspaces of dimension $1$, so it is a $2$-dimensional simple space, and in particular semi-simple.