Subalgebra with semisimple centralizer itself semisimple?

Question

Let $V$ be a finite-dimensional $K$-vector space and suppose that $\mathcal{B} \subseteq \operatorname{End}_K(V)$ is a $K$-subalgebra such that the centralizer $$\mathcal{Z}_{\operatorname{End}_K(V)}(\mathcal{B}) = \{\varphi \in \operatorname{End}_K(V) \colon \varphi \circ \beta = \beta \circ \varphi \text{ for all } \beta \in \mathcal{B}\}$$ is semimsimple (equipped with the $K$-subalgebra structure from $\operatorname{End}_K(V)$). Is $\mathcal{B}$ then necessarily a semisimple $K$-algebra?

Answer 1

Take $V$ to be two-dimensional and $L<V$ a line in $V$, and let $$ \mathcal B = \{x \in \mathrm{End}(V): x(L) \subseteq L\}. $$ Then if $y$ centralizes $\mathcal B$, $y= \lambda.I_V$, since if we take any decomposition $V = L\oplus N$, then the projection $\pi_N\colon V \to L$ with kernel $N$ lies in $\mathcal B$, and hence $y$ commutes with $\pi_N$, and thus must preserve both $L$ and $N$. But then $y$ preserves every line in $V$, hence it must be a scalar. Thus $\mathcal Z_{\mathrm{End}_K(V)}(\mathcal B) = K.I_V$, which is semisimple, but $\mathcal B$ is clearly not semisimple.