## Given two linear operators A and B over a finite field, is there a third operator C whose kernel is the intersection of kernels of A and B?

Let $V$ be a finite dimensional linear space over a finite field $k$. Let $A$ and $B$ be two endomorphisms of $V$.

Question 1. Is there an endomorphism $C$ of $V$, which is expressed in terms of some "natural operations" on $A$ and $B$, and whose kernel is the intersection of kernels of $A$ and $B$?

By "natural operations" I mean some analogue of the recipe which works over complex numbers: choose a scalar product on $V$ and take $C:= A^{\ast}A + B^{\ast}B$. Indeed, $A^{\ast}A$ is a positive operator whose kernel is the same as the one of $A$, same for $B^{\ast}B$, and kernel of a sum of two positive operators is the intersection of kernels of the summands.

Actually, slightly less would serve my purpuses. Suppose that there is a chosen basis of $V$, and we fix one of the vectors from this basis. Call this vector $v$. Suppose $A$ is an endomorphism of $V$.

Question 2. Is there an endomorphism $C$ of $V$, which is expressed in terms of some "natural operations" on $A$, and whose kernel consists of those vectors of kernel of $A$ whose coefficient of $v$ in the chosen basis of $V$ is $0$?

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Is the following natural enough for you: take the left ideal $I$ generated by $A$ and $B$ and take a $C \in I$ with largest possible rank? – Johannes Ebert Oct 30 2010 at 14:15
@Johannes: Unfortunately not. I need a "formula", which is polynomial in $A$, $B$, and "functorial operations on A and B" (as in the complex case, the functors could be defined using some additional structure on the vector spaces). – Łukasz Grabowski Oct 30 2010 at 17:31
If $A$ is the all-$1$-matrix over $GF(2)$ and $B$ the matrix with first column zero and second column $1$ over $GF(2)$, then $A^2=0, B^2=B, AB=0, BA=A$ so that every polynomial (with trivial constant coefficient) in $A$ and $B$ is a linear combination of $A$ and $B$. However, it is easy to see that every such linear combination is singular. At the same time the kernels of $A$ and $B$ intersect only in the zero element in $GF(2)^{\oplus 2}$. – Andreas Thom Nov 2 2010 at 23:22