
> 
> 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](https://mathoverflow.net/q/47362/24165). But for algebras, Burnside's theorem gives me some hope...



Subalgebras of singular matrices
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 $\begin{bmatrix}a&b\\0&0\end{bmatrix}$.


