# Collection of matrices with every vector fixed

Suppose that $S$ is a collection of $n\times n$ matrices closed under addition, multiplication, and scalar multiplication, such that every vector is fixed by some element of $S$.

Must the identity matrix be in $S$?

This is clearly true for $n=1$, and I have been able to verify it is true for $n=2$.

I am particularly interested in the answer over a field, but I would also like to know if this is true over a commutative ring.

Working over a field, $S$ must contain the identity matrix.
Let $A$ be an element of $S$ such that $\mathrm{Id}_n-A$ has the smallest possible rank. I claim that $\mathrm{Id}_n-A=0$. If not, there is some vector $\vec{v}$ such that $(\mathrm{Id}_n-A)\vec{v}\neq\vec{0}$. By assumption, we can find $B\in S$ with $$B\big[(\mathrm{Id}_n-A)\vec{v}\big]=(\mathrm{Id}_n-A)\vec{v}.$$ Let $A'=A+B-BA \in S$. Then $\mathrm{Id}_n-A'=(\mathrm{Id}_n-B)(\mathrm{Id}_n-A)$, so $\ker(\mathrm{Id}_n-A')\supseteq \ker(\mathrm{Id}_n-A)$. Additionally, $(\mathrm{Id}_n-A)\vec{v}\neq\vec{0}$, but $$(\mathrm{Id}_n-A')\vec{v}=(\mathrm{Id}_n-B)\big[(\mathrm{Id}_n-A')\vec{v}\big]=\vec{0},$$ so $\ker(\mathrm{Id}_n-A')\supsetneq \ker(\mathrm{Id}_n-A)$. This implies $\mathrm{Id}_n-A'$ has smaller rank than $\mathrm{Id}_n-A$, which contradicts the choice of $A$. This proves the claim.