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This is an example of a quadratic form

$$x_1 x_2 + x_1x_3=(x_1,x_2,x_3)\begin{pmatrix}0&\frac{1}{2}&\frac{1}{2}\\\frac{1}{2}&0&0\\\frac{1}{2}&0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}.$$

How could I represent polynomials of degree $4$ by a matrix using linear $x_i$? I do not want to use $x_i x_j$ as left and right vectors. Do I have to use a 4-dimensional $4\times4\times4\times4$-matrix, if such a definition exists? Of course, I would need the inverse and transpose of such a construction.

$$x_1 x_2 x_3 x_4+ x_1 x_3 x_4^2=(x_1,x_2,x_3,x_4)\begin{pmatrix}&\\4\times4\times4\times4\\\text{matrix?}\\&\end{pmatrix}\begin{pmatrix}x_1\\ x_2 \\x_3 \\x_4\end{pmatrix}?$$

Or can I write a quartic polynom as a product of 2 quadratics that can then be joined to a single matrix?

$$x_1 x_2 x_3 x_4+ x_1 x_3 x_4^2=\left[(x_1,x_2,x_3,x_4)\begin{pmatrix}&\\4\times4\\\text{matrix1?}\\&\end{pmatrix}\begin{pmatrix}x_1\\ x_2 \\x_3 \\x_4\end{pmatrix}\right]\left[(x_1,x_2,x_3,x_4)\begin{pmatrix}&\\4\times4\\\text{matrix2?}\\&\end{pmatrix}\begin{pmatrix}x_1\\ x_2 \\x_3 \\x_4\end{pmatrix}\right]$$

granular_bastard
  • 563

1 Answers1

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The quadratic form is

$ f_2(x_1,x_2,x_3)=\sum_{i,j=1,3} A_{ij} x_i x_j$.

It involves 6 independent components of $A_{ij}$ due to symmetries: $(A_{12}+A_{21}) x_1 x_2$. We can therefore impose $A_{ij}=A_{ji}$.

The quartic form is

$f_4(x_1,x_2,x_3)=\sum_{i,j,k,l=1,3} B_{ijkl} x_i x_j x_k x_l$.

$B_{ijkl}$ has 81 components, but due to the symmetries $B_{ijkl}=B_{jikl}=B_{kjil}=B_{ljki}$ it has only 15 independent components.

By counting the number of independent components it is clear that not any quartic form (with 15 independent components) can be written as the product of two quadratic forms (with 6 independent components each).

Rainer Glüge
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