Question

Let $A$, $B$ be square matrices over infinite field (we identify them with linear operators on the vector space of columns). It is given that for all scalars $a,b$ the matrix $aA+bB$ is singular. Does it follow that there exist matrices $P$, $Q$ such that rank$(P)$+rank$(Q) > n$ such that $PAQ=PBQ=0$?

If yes, is the same true for arbitrary subspaces of singular matrices? Well, the answer is no for antisymmetric matrices $3\times 3$... But how can subspaces of singular matrices be described (if they can)?

Answer 1

Since the question in the new formulation is quite different, I am adding a new answer. Now the answer is positive, but the proof is not so simple, I will sketch the basic steps.

First of all, assume $A$ and $B$ are matrices of size $n$. Let $V$ and $W$ be $n$-dimensional vector spaces, so $A,B \in Hom(V,W)$. Then consider $P^1$ with coordinates $(x:y)$ and consider the morphism $V\otimes O(-1) \to W\otimes O$ given by $xA + yB$. Let $K$ be its kernel and $C$ its cokernel. Thus we have an exact sequence $$ 0 \to K \to V\otimes O(-1) \to W\otimes O \to C \to 0. $$ The condition of singularity implies $r(K) = r(C) > 0$. Also from the exact sequence it follows that $d(K) = d(C) - n$. Now let us take $Q$ to be the induced map $$ H^1(P^1,K(-1)) \to H^1(P^1,V\otimes O(-2)) = V $$ and $P$ to be the induced map $$ W = H^0(P^1,W\otimes O) \to H^0(P^1,C). $$ Then one can check $Q$ is an embedding, $P$ is a surjection and that $PAQ = PBQ = 0$, so it remains to check that $\dim H^1(P^1,K(-1)) + \dim H^0(P^1,C) > n$. But this can be done like this. First, $$ \dim H^0(P^1,C) \ge \chi(C) = r(C) + d(C). $$ Further, $$ H^1(P^1,K(-1)) \ge - \chi(K(-1)) = - (r(K) + d(K) - r(K)) = -d(K) = n - d(C). $$ Summing up we see that $$ \dim H^1(P^1,K(-1)) + \dim H^0(P^1,C) \ge r(C) + d(C) + n - d(C) = n + r(C) > n. $$

Answer 2

No. For example you can take $A$ and $B$ to be skew-symmetric and $n$ odd. Then all linear combinations of $A$ and $B$ are skew-symmetric, hence degenerate. But for generic choice of $A$ and $B$ they would not have common kernel or cokernel vector. An explicit example is $$ A = \left(\begin{smallmatrix}0 & 1 & 0\cr -1 & 0 & 0\cr 0 & 0 & 0\end{smallmatrix}\right), \qquad B = \left(\begin{smallmatrix}0 & 0 & 0\cr 0 & 0 & 1\cr 0 & -1 & 0\end{smallmatrix}\right). $$