On matrices in linear forms with vanishing determinant

Question

This is a cross-post from my original question at math.se. I decided to post here because it seems more difficult than I originally thought.

Let $R=\mathbb C[x_1,\ldots,x_r]$ be a polynomial ring. Assume that $f_{ij}\in R$ are homogeneous linear polynomials for $1\le i,j\le n$. If $\det(f_{ij})=0$, I can consider this equation over $K:=\mathrm{Frac}(R)$ and get $\lambda_1,\ldots,\lambda_n\in K$, not all zero, with the property that $$ \forall i:\quad \sum_{j=1}^n \lambda_j\cdot f_{ij} = 0 $$ Now, I can clear denominators and assume $\lambda_j\in R$. Since the $f_{ij}$ are homogeneous, I can also assume that the $\lambda_j$ are homogeneous and of minimal degree with the above property. Unfortunately, the $\lambda_j$ do not have to be constant (which is my desire), made visible by the simple counterexample $$\begin{pmatrix} x&y\\x&y\end{pmatrix} .$$ However, note that it is true for the transpose. So we similarly choose homogeneous $\mu_1,\ldots,\mu_n\in R$ which are not all zero and of minimal degree with $$ \forall j:\quad \sum_{i=1}^n \mu_i\cdot f_{ij} = 0$$

Question: Is it true that if the $\lambda_j$ are not all constant, then all the $\mu_i$ are constant?

I think the answer is affirmative, so my search for better counterexamples might have been half-hearted. However, so far, I cannot give a proof either. Searching the literature has yielded almost nothing, maybe I am looking in the wrong places. Your help is greatly appreciated!

Answer 1

The simplest counterexample is the following bivariate one: $$ \begin{bmatrix} 0 & x & y\\ x & 0 & 0\\ y & 0 & 0 \end{bmatrix}, $$ with left and right kernel spanned by $v=\begin{bmatrix}0 \\ -y \\ x\end{bmatrix}$.

See also my comment for a literature pointer on how to construct bivariate examples with arbitrary kernel structure.

Answer 2

The answer is negative. Your problem amounts to the following one: given a linear subspace $V$ of $M_n(\mathbb{C})$ consisting entirely of singular matrices, is it true that either all the matrices of $V$ vanish at some common non-zero vector, or the same holds for all the matrices of $V^T$?

For $2 \times 2$ matrices, the result is known to hold (this is a special case of Schur's theorem on vector spaces of matrices with rank at most $1$). For greater values of $n$, the following example shows that your conjecture fails: one takes the space $V$ of all $n \times n$ matrices of the form $\begin{bmatrix} ? & [?]_{1 \times (n-1)} \\ [?]_{(n-1) \times 1} & [0]_{(n-1) \times (n-1)} \end{bmatrix}$. Obviously, all these matrices have rank at most $2$ and hence they are singular if $n>2$. On the other hand, no non-zero vector is annilated by all the matrices in $V$, and the same holds for $V^T$ as $V^T=V$.

Answer 3

Typically there is no (nontrivial) "constant" solution of either $A\vec{\lambda}=\vec{0}$ or $A^\dagger\vec{\mu}=\vec{0}$. For instance, consider the following $6\times 6$ matrix, $$ A= \left[ \begin{array}{rrrrrr} y & -x & 0 & 0 & 0 & 0 \\ z & 0 & 0 & -x & 0 & 0 \\ 0 & y & -x & 0 & 0 & 0 \\ 0 & z & 0 & -y & 0 & 0 \\ 0 & 0 & 0 & 0 & z & -y \\ 0 & 0 & z & 0 & -y & 0 \end{array} \right]. $$ One nontrivial "nullvector" is $$ \vec{\lambda} = \left[ \begin{array}{r} x^2 \\ xy \\ y^2 \\ xz \\ yz \\ z^2 \end{array} \right]. $$ Thus, $\text{det}(A)$ is zero. However, there is no nontrivial constant solution of $A\vec{\lambda} = \vec{0}$, nor of $A^\dagger \vec{\mu} = \vec{0}$.