Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 339 / 3 / c

Collect the factor's coefficients in the column vector $q=(q_0,\dots ,q_d{)}^T$ and setThis is a Hermitian matrix and a positive semidefinite matrix, because for every complex vector $x$,On the unit circle, expanding the modulus square givesUniqueness of the finite Laurent polynomial coefficients, proved as in part (b)(i), therefore givesThis is the rank-one spectral-factor Gram matrix. The matrix $M$ has matrix rank one when $q\ne 0$ and zero when $q=0$. The orientation $i-j=k$ is the PDF's convention: using $\overline{q_i}{q}_j$ instead would generally interchange $p_k$ and $p_{-k}$.

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.