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First suppose $A=P+N$, with $P$ a real positive semidefinite matrix and $N$ a symmetric nonnegative matrix. Factor $P=B^{T}B$, and write $v(z)=(z_1^2,\dots ,z_n^2{)}^T$. ThenEach term is a square multiplied by a nonnegative scalar, so $p$ is a sum of squares polynomial.

Conversely, suppose $p={\sum }_{\ell }{q}_{\ell }^2$. Because $p$ is a degree-four homogeneous polynomial, the homogeneous sum of squares representation allows every $q_{\ell }$ to be quadratic and homogeneous. Explicitly, higher-degree parts cannot cancel in a sum of squares; constant parts vanish because $p(0)=0$, and the degree-two part ${\sum }_{\ell }(q_{\ell }^{(1)}{)}^2$ forces all linear parts to vanish. WriteSince $p$ is invariant under every coordinate sign change, sign averaging of a sum of squares over independent Rademacher random variables givesThe cross terms vanish because their sign products contain an odd power of at least one independent sign. SetThen $P$ is a positive semidefinite matrix and $N$ is a symmetric nonnegative matrix. Comparing the coefficients of $z_i^4$ and $z_i^2{z}_j^2$ gives $A=P+N$. This proves the sum of squares criterion for a biquadratic form.

For a real symmetric matrix $A$, set $v(z)=(z_1^2,\dots ,z_n^2{)}^T$. ThenFor sufficiency, factor $P=B^{T}B$. Its contribution is a sum of squares of linear combinations of $z_i^2$, while the contribution of $N$ is ${\sum }_i{N}_{ii}(z_i^2{)}^2+{\sum }_{i<j}2N_{ij}(z_i{z}_j{)}^2$. For necessity, use a homogeneous sum of squares representation and sign averaging of a sum of squares. Writing each quadratic summand with coefficients $a_{\ell i},b_{\ell ij}$ gives $P={\sum }_{\ell }{a}_{\ell }{a}_{\ell }^T$, $N_{ii}=0$ and $N_{ij}=\frac{1}{2}{\sum }_{\ell }{b}_{\ell ij}^2$ for $i<j$. Comparing coefficients gives $A=P+N$.

This is the basic semidefinite programming certificate of copositivity discussed in Parrilo's paper on matrix copositivity.