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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)?

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  • $\begingroup$ 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. $\endgroup$ Nov 25, 2010 at 19:18
  • $\begingroup$ Definitely not for arbitrary subspaces: try the subspace of $3\times 3$ matrices of the form $\left(\begin{array}{ccc} a&0&0 \\ b&0&0 \\ c&d&e \end{array}\right)$. $\endgroup$ Nov 25, 2010 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. $\endgroup$ Nov 25, 2010 at 19:24
  • $\begingroup$ oh, thanks, indeed! I need to think a bit to understand what I really wanted to ask instead:) $\endgroup$ Nov 25, 2010 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). $\endgroup$ Nov 26, 2010 at 0:25

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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? $\endgroup$ Nov 25, 2010 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, 2010 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, 2017 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, 2017 at 6:45
  • $\begingroup$ @Sasha thank you for this proof, eventhough I do not understand it completely since it uses algebraic geometry notation. Do you have a reference of an advanced book where I can find similar proofs? More precisely I'm looking for proofs of linear algebra facts using advanced techniques juste like the one you used. I'm wondering if this is very common to solve such questions in a non elementary fashion, and undergraduate courses in linear algebra obviously do not teach such techniques. Thank you. $\endgroup$
    – J.Mayol
    Apr 27, 2020 at 6:56
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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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  • $\begingroup$ 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, ${\rm rk}(P)+{\rm rk}(Q)=2<n=3$. $\endgroup$ Nov 25, 2010 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, 2010 at 20:21
  • $\begingroup$ Yes, thanks, Sasha, your (and Darij's) answer is correct, so I edited a question. $\endgroup$ Nov 25, 2010 at 20:23
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`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).

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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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