Here is a polynomial $p(x,y) = (ax + by)^2$, it can be written like this $$p(x,y) = \left(\left[ \begin{array}{cc} a & b \\ \end{array} \right] \left[ \begin{array}{c} x\\ y\\ \end{array} \right]\right)^2$$ and I know that it can also be written as something like $v^TMv$, here $v = [x,y]^T$, and $$M = \left[ \begin{array}{cc} a^2 & ab \\ ab & b^2\\ \end{array} \right]$$. But how could I find out $M$, any technique?
Furthermore, here $p(x,y)$ has degree $2$, and it can be represented with the multiplication of matrix and vector, what if the degree is $3,4...$?
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$\begingroup$Note that if $M=(m_{ij})$, then $$v^TMv=\sum_{i,j}m_{ij}v_iv_j$$ Then compare the coefficients of $v_iv_j\ \forall i,j$ of both sides of the equation to get $m_{ij}$.
$\endgroup$ 1 $\begingroup$$$\left[ \begin{array}{cc} a & b \\ \end{array} \right] \left[ \begin{array}{c} x\\ y \end{array} \right] = \left[ \begin{array}{cc} x & y \\ \end{array} \right] \left[ \begin{array}{c} a\\ b \end{array} \right]$$ Then $M= \left[ \begin{array}{c} a & b \end{array}\right] \left[ \begin{array}{c} a\\ b \end{array}\right]$
Letting $v=\left[ \begin{array}{c} x\\ y \end{array} \right]$ and $A=\left[ \begin{array}{cc} a & b \\ \end{array} \right]$ we use the first line to note that $Av=v^TA^T$., and hence that $$(Av)^2 = (Av)^T(Av) = v^TA^TAv = v^T(A^TA)v$$
So $M=A^TA$.
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