Rank-1 decomposition conjecture for matrix with linear function elements 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)
$
 A: 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.

