subspaces of singular matrices 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$(Q) > n$ but $PAQ=PBQ=0$?
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)?
 A: 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 
$$
\dim H^1(P^1,K(-1)) + \dim H^0(P^1,C) \ge r(C) + d(C) + n - d(C) = n + r(C) > n.
$$
A: 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 
$$
A = \left(\begin{smallmatrix}0 & 1 & 0\cr -1 & 0 & 0\cr 0 & 0 & 0\end{smallmatrix}\right),
\qquad
B = \left(\begin{smallmatrix}0 & 0 & 0\cr 0 & 0 & 1\cr 0 & -1 & 0\end{smallmatrix}\right).
$$
A: 
`But how can subspaces of singular matrices be described (if they can)?'

I doubt they can. For instance, there is the following counter-intuitive result:

for infinitely many $n$, there exists
an 8-dimensional space of $n\times n$ matrices
(over any field of characteristic zero) that is maximal singular (i.e. inclusion-maximal subspace of singular matrices).

A: Here is a non-answer to the more general question. All the examples noted in the question are generalized by the following construction. For each decomposition $n=p+q+r$ with $q$ odd, matrices of the following form are singular:
$$
  \begin{bmatrix}
    *&0&0\\
    *&A&0\\
    *&*&*\\
  \end{bmatrix}
$$
where the diagonal blocks are square of size $p,q,r$ and $A$ is anti-symmetric (for characteristic $2$ we require $v^tAv=0$). We can describe the construction in a basis-independent way. Suppose $U$ is a subspace of $V^*\oplus V$ with $\dim U=\dim V$ and $\dim\pi_1(U)+\dim\pi_2(U)+\dim V$ odd, where $\pi_1,\pi_2$ are the component projections from $V^*\oplus V$. Then
$$
  \{X\in\mathrm{End}(V)\mid\lambda Xu=0\text{ for }(\lambda,u)\in U\}
$$
is a space of singular matrices.
The second form seems promising because of the following result. For any space $L$ of singular matrices over an infinite field and $X\in L$ of rank $n-1$, we have $\lambda Lu=0$ where $\lambda,u$ span the kernels of $X^*$ and $X$ (to see this, note that $\mathrm{adj}(X)\propto u\lambda$ and $\mathrm{tr}(\mathrm{adj}(X)Y)$ is the coefficient of $yx^{n-1}$ in $\det(xX+yY)$). However, the construction still isn't exhaustive. Indeed any $L$ produced by the above construction further satisfies
$$
  \mathrm{tr}(\mathrm{adj}(X)Y\mathrm{adj}(Z)W)+
  \mathrm{tr}(\mathrm{adj}(X)W\mathrm{adj}(Z)Y)=0
$$
for $X,Y,Z,W\in L$. But the following four matrices fail this identity and span a space of singular matrices.
$$
X=\begin{bmatrix}
1&0&0&0&0\\
0&0&0&0&0\\
0&0&0&1&0\\
0&0&-1&0&0\\
0&0&0&0&1\end{bmatrix},\,
Y=\begin{bmatrix}
1&0&0&0&0\\
0&0&0&1&0\\
0&0&0&0&0\\
0&-1&0&0&0\\
0&0&0&0&1\end{bmatrix},\,
Z=\begin{bmatrix}
1&0&0&0&0\\
0&0&1&0&0\\
1&-1&0&0&0\\
0&0&0&0&0\\
0&1&0&1&0\end{bmatrix},\,
W=\begin{bmatrix}
0&1&0&1&-1\\
1&0&1&0&1\\
-1&-1&0&0&-2\\
0&1&-0&1&0\\
0&0&0&0&0\end{bmatrix}.
$$
