# Sum of tensor-product of polynomials

Suppose you've got two degrees of freedom $x$ and $y$ and a function of the form \begin{align} F(x,y) = p_1(x) p_2(y) + p_3(x) p_4(y) + \ldots + p_n(x) p_{n+1}(y), \end{align} where the $p_i$ are polynomials (of arbitrary degree).

My question is: under which circumstances can $F(x,y)$ be written in the form \begin{align} F(x,y) = Q(x) R(y) \end{align} with $Q$ and $R$ being polynomials, too. Is there any general statement one can make ? Is it possible at all ? I have googled but I couldn't find anything but I am sure that people have thought about it. I don't know where to start looking though.

## 1 Answer

It's rather uncommon for this to be possible.

In your specific example, we can completely characterize the problem in terms of linear algebra; every polynomial in two variables can be written uniquely as a matrix product

$$\left[ \begin{matrix}1 & x & x^2 & \ldots \end{matrix} \right] P \left[ \begin{matrix} 1 \\ y \\y^2 \\ \vdots \end{matrix} \right]$$

where $P$ is a matrix with finitely many nonzero entries.

Then, $F(x,y)$ can be written in the form $Q(x) R(y)$ if and only if $P$ is a rank 1 matrix.

In fact, $P$ has rank $n$ if and only it is possible to write

$$F(x,y) = \sum_{i=1}^n Q_i(x) R_i(y)$$

but impossible to write

$$F(x,y) = \sum_{i=1}^{n-1} Q_i(x) R_i(y)$$

• that was very helpful, thank you for your answer – hickslebummbumm Feb 17 '17 at 14:13