# When does a space of endomorphisms contain invertibles?

Let $V$ be a finite-dimensional vector space over a field $K$. Let $U$ be a linear subspace of $\mathrm{End}(V)$. Write $UV$ for the span of all $Av$ where $A\in U$ and $v\in V$. Suppose that $$\ker(U)=\bigcap_{A\in U}\ker(A)$$ is zero and that $$\mathrm{coker}(U)=V/UV$$ is zero. Is it true that $U$ must contain an invertible element of $\mathrm{End}(V)\$?

Nope: $$\left\{ \begin{bmatrix} 0 & x & y\\ x & 0 & 0\\ y & 0 & 0 \end{bmatrix}: x,y \in\mathbb{R} \right\}.$$ This is a classical counterexample in numerical linear algebra -- the simplest singular matrix pencil with a nontrivial Kronecker canonical form.