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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!

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Literature pointer: for the case with 2 homogeneous variables, check out the Kronecker canonical form and its relation to minimal indices. This will tell you how to construct counterexamples with arbitrary minimal degrees for the kernel (polynomial) bases. –  Federico Poloni Sep 26 '13 at 12:17
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3 Answers

up vote 4 down vote accepted

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.

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Pretty devastating news, but thanks for the minimal example. –  Jesko Hüttenhain Sep 26 '13 at 16:22
    
Don't get discouraged! What did you need this property for? Maybe that makes another interesting question for Math.se or Mathoverflow. –  Federico Poloni Sep 26 '13 at 16:35
    
Consider the polynomial ring $\newcommand{\C}{\mathbb C}R=\C[x_{ij}\mid 1\le i,j\le n]$ and the group $G=Gl(\C^{n\times n})$ acting on $R.$ We have $\det_n\in R_n$. Now, the action of $G$ extends to an action of $M:=\C^{n\times n}$ on $R_n$. For any $a\in M$ with $a.\det_n=0$, I would like to find a sequence of matrices $\beta_k\in\mathrm{Sl}_n$ such that $ab_k$ converges to the zero matrix, where $b_k$ is multiplication by $\beta_k$ from the right (left). –  Jesko Hüttenhain Sep 27 '13 at 6:45
    
If the above had been true, the construction would have been rather easy. Anyway, I am not so sure it is possible any more if the above does not hold. –  Jesko Hüttenhain Sep 27 '13 at 6:45
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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 $V^T=V$.

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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 $A^\dagger \vec{\mu} = \vec{0}$.

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