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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)\ $?

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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.

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