We fix a $\mathbb{C}$-vector space $E$ of dimension $n \geqslant 2$. Let $\mathcal{A}$ be an irreducible subalgebra of $\mathcal{L}(E)$ (i.e., the only vector subspaces stable by all elements of $\mathcal{A}$ are $\{0\}$ and $E$). Let $u_{0} \in \mathcal{A}$ be of rank 1. We can therefore choose a basis $\mathcal{B} = (\varepsilon_{1}, \ldots, \varepsilon_{n})$ of $E$ such that $(\varepsilon_{2}, \ldots, \varepsilon_{n})$ is a basis of $\ker u_{0}$. Conclude (that $\mathcal{A} = \mathcal{L}(E)$).
We fix a $\mathbb{C}$-vector space $E$ of dimension $n \geqslant 2$. Let $\mathcal{A}$ be an irreducible subalgebra of $\mathcal{L}(E)$ (i.e., the only vector subspaces stable by all elements of $\mathcal{A}$ are $\{0\}$ and $E$). Let $u_{0} \in \mathcal{A}$ be of rank 1. We can therefore choose a basis $\mathcal{B} = (\varepsilon_{1}, \ldots, \varepsilon_{n})$ of $E$ such that $(\varepsilon_{2}, \ldots, \varepsilon_{n})$ is a basis of $\ker u_{0}$.
Conclude (that $\mathcal{A} = \mathcal{L}(E)$).