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)?

  • $\begingroup$ 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. $\endgroup$ – darij grinberg Nov 25 '10 at 19:18
  • $\begingroup$ Definitely not for arbitrary subspaces: try the subspace of $3\times 3$ matrices of the form $\left(\begin{array}{ccc} a&0&0 \\ b&0&0 \\ c&d&e \end{array}\right)$. $\endgroup$ – darij grinberg Nov 25 '10 at 19:21
  • $\begingroup$ 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. $\endgroup$ – darij grinberg Nov 25 '10 at 19:24
  • $\begingroup$ oh, thanks, indeed! I need to think a bit to understand what I really wanted to ask instead:) $\endgroup$ – Fedor Petrov Nov 25 '10 at 19:39
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    $\begingroup$ Fedor: Since your problem is invariant under simultaneous multiplication of $A$ and $B$ by invertible matrices from the left and the right, it is a problem about representations of the tame $1$-Kronecker quiver $\tilde{A}_1$. See cs-linux.ubishops.ca/~bruestle/Publications/… for this (it's one of the so-called tame quivers). $\endgroup$ – darij grinberg Nov 26 '10 at 0:25

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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    $\begingroup$ 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? $\endgroup$ – Gerry Myerson Nov 25 '10 at 21:48
  • $\begingroup$ Certainly you can explain the answer in the other languages, but I believe this is the shortest way to explain. $\endgroup$ – Sasha Nov 26 '10 at 7:22
  • $\begingroup$ Do you have an answer for the more general question "how can subspaces of singular matrices be described (if they can)"? For subspaces of dimension $>2$ the given criterion is not necessary; eg consider the space $\left\{\begin{bmatrix}a&0&b\\0&a&c\\-c&b&0\end{bmatrix}\mid a,b,c\in k\right\}$ $\endgroup$ – stewbasic Sep 18 '17 at 23:29
  • $\begingroup$ @stewbasic: Geometrically, that is the question about linear spaces on the discriminant variety $\mathfrak{D} \subset \mathbb{P}^{n^2-1}$ of degenerate maps. I guess any such space lies inside a maximal one, but I am not sure that the classification of maximal subspaces is known for all $n$. Definitely, there are subspaces $L_{P,Q}$ as above, but for odd $n$ there also subspaces of skew-symmetric matrices up to a change of basis in one of the spaces (I guess your example is equivalent to this). Maybe there are other maximal subspaces as well. $\endgroup$ – Sasha Sep 19 '17 at 6:45

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

  • $\begingroup$ 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, ${\rm rk}(P)+{\rm rk}(Q)=2<n=3$. $\endgroup$ – Denis Serre Nov 25 '10 at 20:16
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    $\begingroup$ That was the answer to the previous version of the question which did not include $P$ and $Q$. $\endgroup$ – Sasha Nov 25 '10 at 20:21
  • $\begingroup$ Yes, thanks, Sasha, your (and Darij's) answer is correct, so I edited a question. $\endgroup$ – Fedor Petrov Nov 25 '10 at 20:23

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. $$ X=\begin{bmatrix} 1&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&1&0\\ 0&0&-1&0&0\\ 0&0&0&0&1\end{bmatrix},\, Y=\begin{bmatrix} 1&0&0&0&0\\ 0&0&0&1&0\\ 0&0&0&0&0\\ 0&-1&0&0&0\\ 0&0&0&0&1\end{bmatrix},\, Z=\begin{bmatrix} 1&0&0&0&0\\ 0&0&1&0&0\\ 1&-1&0&0&0\\ 0&0&0&0&0\\ 0&1&0&1&0\end{bmatrix},\, W=\begin{bmatrix} 0&1&0&1&-1\\ 1&0&1&0&1\\ -1&-1&0&0&-2\\ 0&1&-0&1&0\\ 0&0&0&0&0\end{bmatrix}. $$


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