Subalgebras of singular matrices 
Is it true that any subalgebra of singular matrices have a common null-vector?

In other words, is it true that, for any subalgebra $\cal S$
of the algebra of linear operators in a finite-dimensional vector space over a field,
$$
\bigcap_{A\in\cal S}\ker A=\{0\}\quad\hbox{implies that}
\quad\ker A=\{0\}\hbox{ for some $A\in\cal S\;$? } 
$$
(I am interested in finite prime fields mostly.)
Note that for subspaces (instead of subalgebras) nothing similar is true. But for algebras, Burnside's theorem gives me some hope...
 A: It's false.  Take the subalgebra of $M_3(K)$ generated by the matrices $\begin{bmatrix} 0 & 0&0\\ 1 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$ and $\begin{bmatrix} 0 & 0 & 0\\ 0 &1&0\\ 1& 0&1\end{bmatrix}$.  These two elements form a two element right zero semigroup and so the algebra they generate is just their span which is $2$-dimensional and obviously singular since all elements have first row zero.  These two matrices have no common zero except zero.
Conceptually, let $S$ be the two-element right zero semigroup $\{a,b\}$.  Then $KS$ acts faithfully on the left of $KS^1$ where $S^1$ is obtained by adjoining an identity. But these two right zeroes have no common vector they annihilate.  For $a(c_11+c_2a+c_3b) = c_1a+c_2a+c_3b$ and $b(c_11+c_2a+c_3b) =c_1b+c_2a+c_3b$ and for these both to be zero you need $c_3=0=c_2$ and you need $c_1=-c_3$ and $c_1=-c_2$ and so $c_1=c_2=c_3=0$.   The representation is by singular matrices since $1$ is not in $KS\cdot KS^1$.
Update I guess you can go one dimension lower and consider the algebra of matrices of the form $\begin{bmatrix}a&b\\0&0\end{bmatrix}$.
