Is it true that, for any subalgebra $\cal S$
of the algebra of linear operators in a finite-dimensional vector space $V$ over a field,
$$
\bigcap\_{A\in\cal S}\ker A=\{0\}\hbox{ and }
\bigcup\_{A\in\cal S}A(V)=V
\quad\hbox{implies that}
\quad\hbox{some $A\in\cal S$ is non-singular? } 
$$


(A more naive version of this question [was answered by Benjamin Steinberg](https://mathoverflow.net/a/397766/24165).)



Subalgebras of singular matrices (less naive version)
Let $S$ b be the set of 3 x 3 matrices whose lower left 2 x 2 block equals zero. Then $S$ is an algebra satisfying the conditions, but containing no invertible matrix.


