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)\ $?



When does a space of endomorphisms contain invertibles?
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.


