= Solution
Collect the factor's <coefficients> in the column <vector> $q=(q_0,\ldots,q_d)^T$ and set
$$
\boxed{M=qq^*,\qquad M_{ij}=q_i\overline{q_j}}.
$$
This is a <Hermitian matrix> and a <positive semidefinite matrix>, because for every complex <vector> $x$,
$$
x^*Mx=|q^*x|^2\geq0.
$$
On the <unit circle>, expanding the modulus square gives
$$
|q(z)|^2=\sum_{i,j=0}^d q_i\overline{q_j}z^{i-j}.
$$
Uniqueness of the finite <Laurent polynomial> <coefficients>, proved as in part (b)(i), therefore gives
$$
\boxed{p_k=\sum_{\substack{0\leq i,j\leq d\\i-j=k}}M_{ij}}.
$$
This is the <rank-one spectral-factor Gram matrix>. The <matrix> $M$ has <matrix rank> one when $q\ne0$ 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 $p_{-k}$.
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