# 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$(Q) > n$ 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)?

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This definitely holds if $A$ and $B$ commute. I don't think it holds generally, but I don't know a counterexample out of my hat. – darij grinberg Nov 25 '10 at 19:18
Definitely not for arbitrary subspaces: try the subspace of $3\times 3$ matrices of the form $\left(\begin{array}{ccc} a&0&0 \\ b&0&0 \\ c&d&e \end{array}\right)$. – darij grinberg Nov 25 '10 at 19:21
Ok, for two matrices it also doesn't work: $\left(\begin{array}{ccc}1&0&0 \\ 0&0&0 \\ 1&1&0 \end{array}\right)$ and $\left(\begin{array}{ccc}0&0&0 \\ 1&0&0 \\ 0&0&1 \end{array}\right)$. No warranty. – darij grinberg Nov 25 '10 at 19:24
oh, thanks, indeed! I need to think a bit to understand what I really wanted to ask instead:) – Fedor Petrov Nov 25 '10 at 19:39
@Darij: I edited a question a bit. – Fedor Petrov Nov 25 '10 at 19:42

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 $$\dim H^1(P^1,K(-1)) + \dim H^0(P^1,C) \ge r(C) + d(C) + n - d(C) = n + r(C) > n.$$

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Wow. The question can be explained to a student with a semester of linear algebra; is there no solution at a similar level of sophistication? – Gerry Myerson Nov 25 '10 at 21:48
Certainly you can explain the answer in the other languages, but I believe this is the shortest way to explain. – Sasha Nov 26 '10 at 7:22

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 $$A = \left(\begin{smallmatrix}0 & 1 & 0\cr -1 & 0 & 0\cr 0 & 0 & 0\end{smallmatrix}\right), \qquad B = \left(\begin{smallmatrix}0 & 0 & 0\cr 0 & 0 & 1\cr 0 & -1 & 0\end{smallmatrix}\right).$$

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Sasha, it is a little more subttle: you can take $P$ to kill $A$ (at left) and $Q$ to kill $B$ at right. Then you have $PAQ=PBQ=0_3$. However, ${\rm rk}(P)+{\rm rk}(Q)=2<n=3$. – Denis Serre Nov 25 '10 at 20:16
That was the answer to the previous version of the question which did not include $P$ and $Q$. – Sasha Nov 25 '10 at 20:21
Yes, thanks, Sasha, your (and Darij's) answer is correct, so I edited a question. – Fedor Petrov Nov 25 '10 at 20:23