Constructive factorization of an admissible QSP polynomial pair

If the pair comes from a phase list, Unitarity, determinant, and polynomial form of a QSP sequence gives exactly the stated matrix form, degree/parity bounds, and polynomial identity for the sequence defined in [F8]. This proves the forward implication.

Let $d=0$ and suppose the pair is admissible. Then $Q=0$ and $P$ is constant with $\mathrm{\mid }P\mathrm{\mid }=1$. Set ${\varphi }_0=\mathrm{Arg}(P)$ from the unique principal polar representation [F4]; [F5] gives $P=e^{i{\varphi }_0}$. By [F8], $U_{\mathrm{\Phi }}=Z({\varphi }_0)=\mathrm{diag}(P,P^{\ast })$, the required matrix.

Let $d=1$. Parity gives $P=ax$ and $Q=b$ for constants $a,b\in \mathbb{C}$. Evaluating the identity at $x=0$ and $x=1$ gives $\mathrm{\mid }b\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }=1$. Set $\alpha =\mathrm{Arg}(a)$ and $\beta =\mathrm{Arg}(b)$ by [F4], so [F5] gives $a=e^{i\alpha }$ and $b=e^{i\beta }$; then set ${\varphi }_0=(\alpha +\beta )\mathrm{/}2$, ${\varphi }_1=(\alpha -\beta )\mathrm{/}2$. By the order in [F8], direct multiplication of $Z({\varphi }_0)W_R(x)Z({\varphi }_1)$ gives entries $ax$, $s(x)b$, $s(x)\bar{b}$, and $-x\bar{a}$, exactly the four entries in the Statement for $d=1$.

Now let $d\ge 2$ and assume the converse has been constructed at every smaller nominal degree. Put $P_S=i^{-d}P$ and $Q_S=i^{-(d+1)}Q$. Multiplication by unit complex scalars preserves the bounds, parities and coefficient-conjugate identity in [F1]. The corresponding source-convention matrix is $M_S(x)=\left(\begin{array}{cc}{P}_S(x)& is(x)Q_S(x)\\ is(x)Q_S^{\ast }(x)& P_S^{\ast }(x)\end{array}\right)\mathrm{.}$ For real $x$, the diagonal entries of $M_S^{†}{M}_S$ are $\mathrm{\mid }{P}_S{\mathrm{\mid }}^2+s^2\mathrm{\mid }{Q}_S{\mathrm{\mid }}^2=1$, and the off-diagonal entries cancel as $is{P}_S^{\ast }{Q}_S-is{Q}_S{P}_S^{\ast }=0$ and its conjugate. Thus $M_S^{†}{M}_S=I$; together with $\det M_S=P_S{P}_S^{\ast }+(1-x^2)Q_S{Q}_S^{\ast }=1$ from [F1], this makes $M_S$ unitary by [F9]. Also $W_S^{†}{W}_S=W_S{W}_S^{†}=I$ and $\det W_S=1$ by direct multiplication using $x^2+s^2=1$, while every $Z(\psi )$ is unitary with determinant $1$ by [F2, F9]. Thus the peel multiplies by unitary matrices.

If $P_S$ is constant, then the identity at $x=1$ gives $\mathrm{\mid }{P}_S\mathrm{\mid }=1$. For every $x\in (-1,1)$ it then gives $(1-x^2)\mathrm{\mid }{Q}_S(x){\mathrm{\mid }}^2=0$, hence $Q_S(x)=0$ throughout that interval. If $Q_S$ were nonzero of degree $m$, choose $m+1$ distinct points in $(-1,1)$; repeated division by their linear factors using [F7] would force a divisor of degree $m+1$ of $Q_S$, contradicting [F6]. Thus $Q_S=0$. The parity of the nonzero constant $P_S$ forces $d$ even. Set ${\psi }_0=\mathrm{Arg}(P_S)$ by [F4], so [F5] gives $P_S=e^{i{\psi }_0}$. Direct multiplication using $x^2+s^2=1$ gives $W_{S}Z(\pi \mathrm{/}2)W_S=Z(\pi \mathrm{/}2)$ and therefore $W_{S}Z(\pi \mathrm{/}2)W_{S}Z(-\pi \mathrm{/}2)=I$. This identity is the length-two phase product $Z(0)W_{S}Z(\pi \mathrm{/}2)W_{S}Z(-\pi \mathrm{/}2)$. Insert it on the left of the current sequence; the adjacent $Z(-\pi \mathrm{/}2)$ and initial $Z({\psi }_0)$ combine to $Z({\psi }_0-\pi \mathrm{/}2)$ by [F3], so the result is again a standard phase list and has two additional signal factors. Repeating this $d\mathrm{/}2$ times pads the constant sequence to nominal degree $d$.

Otherwise let $\mathrm{\ell }=\mathrm{deg}{P}_S\ge 1$, with leading coefficients $p_{\mathrm{\ell }}$ for $P_S$ and $q_m$ for $Q_S$. The identity forces $Q_S\mathrm{\ne }0$ and $m=\mathrm{\ell }-1$: if $m\ge \mathrm{\ell }$, the highest term of $-(x^2)Q_S{Q}_S^{\ast }$ has degree $2m+2>2\mathrm{\ell }$ and cannot cancel; if $m<\mathrm{\ell }-1$, the degree-$2\mathrm{\ell }$ term $\mathrm{\mid }{p}_{\mathrm{\ell }}{\mathrm{\mid }}^2{x}^{2\mathrm{\ell }}$ cannot cancel. At degree $2\mathrm{\ell }$, the coefficients therefore satisfy $\mathrm{\mid }{p}_{\mathrm{\ell }}{\mathrm{\mid }}^2-\mathrm{\mid }{q}_{\mathrm{\ell }-1}{\mathrm{\mid }}^2=0$. Thus $r:=p_{\mathrm{\ell }}\mathrm{/}{q}_{\mathrm{\ell }-1}$ has modulus one. Set ${\psi }_d=\mathrm{Arg}(r)\mathrm{/}2$ using [F4]; then [F5] gives $e^{2i{\psi }_d}=r$.

Multiply $M_S$ on the right by $Z(-{\psi }_d)W_S^{†}$, where $W_S=\left(\begin{array}{cc}x& is\\ is& x\end{array}\right)$. The result has the same form with $P^{\mathrm{\prime }}=e^{-i{\psi }_d}x{P}_S+e^{i{\psi }_d}(1-x^2)Q_S,Q^{\mathrm{\prime }}=e^{i{\psi }_d}x{Q}_S-e^{-i{\psi }_d}{P}_S\mathrm{.}$ The possible degree-$(\mathrm{\ell }+1)$ coefficient of $P^{\mathrm{\prime }}$ and degree-$\mathrm{\ell }$ coefficient of $Q^{\mathrm{\prime }}$ both equal zero by $e^{2i{\psi }_d}=p_{\mathrm{\ell }}\mathrm{/}{q}_{\mathrm{\ell }-1}$. Their parities then give $\mathrm{deg}{P}^{\mathrm{\prime }}\le \mathrm{\ell }-1$ and $\mathrm{deg}{Q}^{\mathrm{\prime }}\le \mathrm{\ell }-2$, the bounds for nominal degree $d-1$. The right multiplier and $M_S$ are unitary with determinant $1$, so their product $N=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$ is also unitary with determinant $1$. The cofactor formula gives $N^{-1}=\left(\begin{array}{cc}D& -B\\ -C& A\end{array}\right)$, whereas [F9] gives $N^{-1}=N^{†}=\left(\begin{array}{cc}{A}^{\ast }& C^{\ast }\\ B^{\ast }& D^{\ast }\end{array}\right)$. Equating entries yields $D=A^{\ast }$ and $C=-B^{\ast }$. Since $A=P^{\mathrm{\prime }}$ and $B=is{Q}^{\mathrm{\prime }}$, this is the required lower row $[is{Q}^{\mathrm{\prime }\ast },P^{\mathrm{\prime }\ast }]$. If $a=e^{-i{\psi }_d}$, then $P^{\mathrm{\prime }}=ax{P}_S+\bar{a}(1-x^2)Q_S$ and $Q^{\mathrm{\prime }}=\bar{a}x{Q}_S-a{P}_S$; expanding their coefficient-conjugate products cancels the two mixed terms and leaves $P^{\mathrm{\prime }}{P}^{\mathrm{\prime }\ast }+(1-x^2)Q^{\mathrm{\prime }}{Q}^{\mathrm{\prime }\ast }=(x^2+1-x^2)(P_S{P}_S^{\ast }+(1-x^2)Q_S{Q}_S^{\ast })=1.$ So $(P^{\mathrm{\prime }},Q^{\mathrm{\prime }})$ is admissible at nominal degree $d-1$.

Put $k=d-1\ge 1$ and $M^{\mathrm{\prime }}=M_{S}Z(-{\psi }_d)W_S^{†}$, the matrix in step 2.1. The pair $(i^k{P}^{\mathrm{\prime }},i^{k+1}{Q}^{\mathrm{\prime }})$ is admissible at nominal degree $k$, since the unit scalars preserve degree, parity and the coefficient-conjugate identity. Apply the reflection-convention induction hypothesis to this pair. Its displayed matrix is $i^k{M}^{\mathrm{\prime }}$ entrywise: the lower-row factors are $(-1{)}^{k+1}{i}^{-(k+1)}=i^{k+1}$ and $(-1{)}^k{i}^{-k}=i^k$. By [F3], convert that reflection phase list to a source-convention list with $i^k{U}_S=i^k{M}^{\mathrm{\prime }}$, hence $U_S=M^{\mathrm{\prime }}$. Reversing the peel gives $M_S=M^{\mathrm{\prime }}{W}_{S}Z({\psi }_d)$, so appending ${\psi }_d$ constructs the required finite source-convention list. Each induction call has strictly smaller nominal degree; steps 1.2, 1.3 and 1.5 supply the bases and constant padding.

For $d\ge 2$, apply the inverse affine phase map in [F3] to obtain a reflection phase list $\mathrm{\Phi }$. Since $U_R(\mathrm{\Phi })=i^d{U}_S(\mathrm{\Psi })$, its upper-left entry is $i^d{P}_S=P$, its upper-right entry is $i^{d+1}s{Q}_S=sQ$, and its lower entries are $(-1{)}^{d+1}s{Q}^{\ast }$ and $(-1{)}^d{P}^{\ast }$. Thus it is the required reflection sequence and the converse follows. Together with the forward implication and both base cases this proves the equivalence for all $d\ge 0$.