Can Anyone prove the following conjecture?


Consider $k$ rational function vectors $V\_1(x\_1,\cdots,x\_n),\cdots,V\_k(x\_1,\cdots,x\_n)$. They are called \textbf{linearly dependent} if there exists rational functions $\alpha\_1(x\_1,\cdots,x\_n),\cdots,\alpha\_k(x\_1,\cdots,x\_n)$ which are not identically zero such that
\begin{align}
\alpha\_1(x\_1,\cdots,x\_n)V\_1(x\_1,\cdots,x\_n)+\cdots+\alpha\_k(x\_1,\cdots,x\_n)V\_1(x\_1,\cdots,x\_n)=0
\end{align}
This defines the rank of a matrix.


Conjecture:
(Rank-1 Decomposition Conjecture)
Let a $l \times m$ polynomial matrix
\begin{align}
A(x\_1,\cdots,x\_n)=
\left(
\begin{array}{ccc}
a\_{11}(x\_1,\cdots,x\_n) & \cdots & a\_{1m}(x\_1,\cdots,x\_n) \newline
\vdots & \ddots & \vdots \newline
a\_{l1}(x\_1,\cdots,x\_n) & \cdots & a\_{lm}(x\_1,\cdots,x\_n) 
\end{array}
\right)
\end{align}
is rank $k$ where $a\_{ij}$ are linear functions whose coefficients are taken from $\mathbb{C}$. Then, there exists $l \times m$ polynomial matrices $A^{(1)}(x\_1,\cdots,x\_n),\cdots,A^{(k)}(x\_1,\cdots,x\_n)$ that satisfies the following three properties:


(i) $A(x\_1,\cdots,x\_n)=A^{(1)}(x\_1,\cdots,x\_n)+ \cdots + A^{(k)}(x\_1,\cdots,x\_n)$. 


(ii) The rank of $A^{(i)}(x\_1,\cdots,x\_n)$ is $1$.


(iii) The elements of $A^{(i)}(x\_1,\cdots,x\_n)$ are linear functions on $x\_1,\cdots,x\_n$ with coefficients taken from $\mathbb{C}$.


Examples :
(1)
$\left(
\begin{array}{ccc}
x\_1 & x\_2 & x\_3 \newline
x\_4 & x\_5 & x\_6 
\end{array}
\right)$


=
$\left(\begin{array}{ccc}
x\_1 & x\_2 & x\_3 \newline
0 & 0 & 0 
\end{array}\right)$
+
$\left(\begin{array}{ccc}
0 & 0 & 0 \newline
x\_4 & x\_5 & x\_6 
\end{array}\right)
$


(2)
$\left(
\begin{array}{cccc}
x\_1 & x\_2 & x\_5 & x\_5 \newline
x\_1 & x\_2 & x\_6 & x\_6 \newline
x\_1 & x\_2 & x\_7 & x\_7 \newline
x\_8 & x\_9 & x\_{10} & x\_{11}
\end{array}\right)
$


$=
\left(
\begin{array}{cccc}
x\_1 & x\_2 & 0 & 0 \newline
x\_1 & x\_2 & 0 & 0 \newline
x\_1 & x\_2 & 0 & 0 \newline
0 & 0 & 0 & 0 \\
\end{array}
\right)
+
\left(
\begin{array}{cccc}
0 & 0 & x\_5 & x\_5 \newline
0 & 0 & x\_6 & x\_6 \newline
0 & 0 & x\_7 & x\_7 \newline
0 & 0 & 0 & 0
\end{array}
\right)
+
\left(
\begin{array}{cccc}
0 & 0 & 0 & 0 \newline
0 & 0 & 0 & 0 \newline
0 & 0 & 0 & 0 \newline
x\_8 & x\_9 & x\_{10} & x\_{11}
\end{array}
\right)
$


(3)
$\left(
\begin{array}{cc}
x\_1 & x\_4 \newline
x\_2 & x\_5 \newline
x\_3 & x\_6 
\end{array}\right)$


$=
\left(
\begin{array}{cc}
x\_1 & 0 \newline
x\_2 & 0 \newline
x\_3 & 0 
\end{array}
\right)
+
\left(
\begin{array}{cc}
0 & x\_4 \newline
0 & x\_5 \newline
0 & x\_6 
\end{array}\right)
$


(4)
$
\left(
\begin{array}{ccc}
x\_1 & 2 x\_1 & 3 x\_1 \newline
2x\_2 & 4 x\_2 & 6 x\_2 \newline
x\_3 & 2 x\_3 & 4 x\_3 
\end{array}
\right)$


$=
\left(
\begin{array}{ccc}
x\_1 & 2 x\_1 & 3 x\_1 \newline
2x\_2 & 4 x\_2 & 6 x\_2 \newline
x\_3 & 2 x\_3 & 3 x\_3 
\end{array}
\right)
+
\left(
\begin{array}{ccc}
0 & 0 & 0 \newline
0 & 0 & 0 \newline
0 & 0 & x\_3
\end{array}
\right)
$



Rank-1 decomposition conjecture for matrix with linear function elements
The answer is **NO**. Take $l=m=3$ and the generic skew-symmetric matrix
$$A:=\begin{pmatrix} 0 & x & y \\\\ -x & 0 & z \\\\ -y & -z & 0 \end{pmatrix}.$$
The rank is $k=2$, yet it cannot be written $A=A^1+A^2$ where both $A^j$ would be linear in the indeterminates $x,y,z$ and rank-one.



> 
> Sketch of the proof. Such matrices $A^j$ must be of the form $CR$ where one of the column $C$ or the row $R$ is constant and the other one is linear (Hint: use Gauss' Lemma). If the decomposition holds, we have $A=C^1R^1+C^2R^2$ with three possible cases. Either both rows $R^j$ are constant, or both column $C^j$, or one row and one column. In the first case, the range of $A(x,y,z)$ would be contained in the fixed plane spanned by $R^1$ and $R^2$, which is false. The second case is treated as well. There remains the case where $C^1$ and $R^2$ are constant, whereas $C^2$ and $R^1$ are linear. But then the fixed vector $C^1$ belongs to the range of $A(x,y,z)$ for every $(x,y,z)$ such that $R^1$ is not parallel to $R^2$, therefore for almost every $(x,y,z)$. Again, this is absurd.
> 
> 
> 


