This is a [cross-post from my original question at math.se](https://math.stackexchange.com/questions/503724/determinant-vanishing-over-polynomial-ring). 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!



On matrices in linear forms with vanishing determinant
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.


