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Under the displayed diagonal-sum convention, set $w(z)=(1,z^{-1},\dots ,z^{-d}{)}^T$ on the unit circle. A Hermitian matrix $M$ then givesConsequently $M⪰0$ proves nonnegativity. Conversely the Fejér–Riesz theorem produces the rank-one spectral-factor Gram matrix. This gives a semidefinite programming representation of nonnegative trigonometric polynomials. The reversed convention $j-i=k$ instead uses the unconjugated monomial vector.

On the unit circle, $z^{-1}=\overline{z}$, soThus the trigonometric polynomial is nonnegative, with a zero at $z=-1$, andEquivalently, writing $z=e^{i\theta }$ gives $p(z)=2+2\cos \theta =4{\cos }^2(\theta /2)$. The factor has polynomial degree one as required. Multiplication of $q$ by any constant of modulus one leaves the factorization unchanged; uniqueness of $q$ is not claimed.

This is the simplest instance of the Fejér–Riesz theorem, where nonnegativity of a trigonometric polynomial on the unit circle admits a polynomial modulus-square factorization.

Pair the off-circle roots of a polynomial using reciprocal-conjugate root pairing, and split each unit-circle root of a polynomial's even multiplicity equally between the two members of a pair. The fundamental theorem of algebra and the leading coefficient $p_d$ give $P(z)=p_d{\prod }_{i=1}^d(z-{\zeta }_i)(z-1/\overline{{\zeta }_i})$. None of the selected ${\zeta }_i$ is zero.

On the unit circle, the identityturns $P(z)/z^d$ intoAt a point of the unit circle outside the finite set of roots of a polynomial, the product is positive and $p(z)$ is nonzero and nonnegative. Its ratio to the product is therefore real and strictly positive. This proves $c>0$, even though the algebraic expression initially permits a complex constant. ConsequentlyThis proves the Fejér–Riesz theorem for a nonzero trigonometric polynomial of actual order $d$. A positive constant has a constant square-root factor, and the identically zero trigonometric polynomial has $q=0$; if $p_d=0$ for a specified upper order $d$, reduce to the actual order first.

Although the PDF permits assuming even multiplicity, there is a short proof of even multiplicity of unit-circle roots of a nonnegative trigonometric polynomial. The real analytic function $p(e^{i\theta })\ge 0$ cannot have a zero of odd order. Near $\zeta =e^{i{\theta }_0}$, $e^{i\theta }-\zeta$ has a simple zero, and the nonzero factor $e^{-id\theta }$ leaves the zero order of $P(e^{i\theta })$ unchanged. Hence the multiplicity of a root must be even.

For $z$ on the unit circle, use the conjugated monomial vectorThen the prescribed diagonal sums givePositive semidefiniteness provesThis is the Gram matrix representation of a trigonometric polynomial. The Hermitian condition also implies $p_{-k}=\overline{p_k}$, so its values on the unit circle are real. The conjugated monomial vector is required by the source's $i-j=k$ convention; the unconjugated vector would represent $p(z^{-1})$ instead.

In particular $p_0=\mathrm{tr}M\ge 0$. If this matrix trace is zero, all nonnegative eigenvalues vanish and $M=0$, so $p=0$. A general feasible Gram matrix need not have matrix rank one; the Fejér–Riesz theorem ensures a rank-one representative exists whenever the nonnegative trigonometric polynomial is nonzero.