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Given matrix $V',W',Y'$ is of $d\times m (d\le m)$ ; column vector $c$ is of size $m$; $r_i, i=1,...,d$ are distinct; and each row of the matrix A is $A_i=(r_i^0 ... r_i^{d-1})$. So, A is of $d\times d$, and clearly A is invertible. We define matrix V, W, Y by $AV=V', AW=W', AY=Y'$ and $x^T=(x^0 x^1 ... x^{d-1})$. And all the elements are real number.

How to show the condition $V'c\circ W'c=Y'c$ ($\circ$ is for pointwise multiplication) is equivalent to the statement that the polynomial $p(x)=(x^TVc)(x^TWc)-x^TYc$ is divisible by $t(x)=\prod_{i=1,...,d}(x-r_i)$.

Some basic facts, and notation, so we do not have to use different notation: Clearly, $p(x)$ is of degree $2d-2$, $t(x)$ is of degree d; suppose $p(x)=t(x)*h(x)$, then $h(x)$ is of degree $d-2$.

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I read a little bit more, and I found the so called bifactor theorem as shown in Divisibility of polynomial and If I remove the premise $a\neq b$ in this question, will the statement still be true?, I know how to solve the problem.

If $p(r_i)=0, i=1,...,d$, so it is divisible by $t(x)$. And $p(r_i)=0$ means $A_iVc\circ A_iWc = A_iYc$, so. $AVc\circ AWc = AYc$ i.e. $V'c\circ W'c=Y'c$.

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