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The real vector space of symmetric matrices has dimension $N=n(n+1)/2$. By the conic Carathéodory theorem, every member of $C_0$ is a conic combination of at most $N$ generators. Absorb each nonnegative coefficient into its vector through $\lambda x{x}^T=(\sqrt{\lambda }x)(\sqrt{\lambda }x{)}^T$.

If $B_{\ell }\in C_0$ converges to $B$, write, padding with zero vectors if necessary,The matrix trace satisfiesThe left side is bounded because $B_{\ell }$ converges. Thus the finite tuple $(x_{\ell 1},\dots ,x_{\ell N})$ is bounded. The Bolzano-Weierstrass theorem gives a subsequence on which every vector converges, say $x_{\ell j}\to x_j\ge 0$. Continuity of the outer product now gives $B={\sum }_{j=1}^N{x}_j{x}_j^T\in C_0$. HenceThis proves closedness of the completely positive cone. The uniform bound on the number of factors and the matrix trace bound are both essential: an arbitrary conic hull of a closed generating set need not be closed.

Let $U=\{x\ge 0:\Vert x{\Vert }_2=1\}$. This set is nonempty and compact for $n\ge 1$, and the quadratic form is continuous. Its minimum $\alpha$ is therefore attained.

For every nonzero $y$ in the nonnegative orthant, normalizing $y$ givesThe zero vector imposes no further condition. It follows that $Q-\lambda I$ is a copositive matrix exactly when $\lambda \le \alpha$. Thus the feasible scalar set is $(-\infty ,\alpha ]$, andBoth optima are attained. This copositive reformulation of an orthant Rayleigh minimum follows directly from homogeneity; it does not require invoking a duality theorem for the original nonconvex constraint.

The restriction to $x\ge 0$ matters. For the matrix in part (a), this minimum is $1$ whereas the unrestricted smallest eigenvalue is $-1$.

For $X$ in the dual cone $C$, copositivity of $Q-\lambda I$ gives $⟨X,Q-\lambda I{⟩}_F\ge 0$. An upper-bounding Lagrangian for the maximization is thereforeIts supremum over unrestricted $\lambda$ is finite precisely when $\mathrm{tr}X=1$. Minimizing that upper bound gives the conic programThis is completely positive optimization, not simply semidefinite programming: $C$ is the completely positive cone.

There is also a direct equality certificate. Every feasible $X={\sum }_j{u}_j{u}_j^T$, $u_j\ge 0$, has ${\sum }_j\Vert u_j{\Vert }_2^2=1$. Its objective is a weighted average of the nonnegative-sphere Rayleigh quotients, hence is at least $\alpha$. For a minimizing unit vector $x_{\ast }$ from part (g), $X_{\ast }=x_{\ast }{x}_{\ast }^T$ is feasible and has objective $\alpha$. Thus $\boxed{X_{\ast }=x_{\ast }{x}_{\ast }^T}$ attains the dual and both values agree, without relying on unverified regularity assumptions.

Split the sign vector as $x=(p^T,q^T{)}^T$. Multiplication of the two off-diagonal blocks givesbecause the two terms are the same real scalar. The map $(p,q)\mapsto x$ is a bijection between the feasible sign choices. ThereforeThis is the symmetric lifting of bipartite binary quadratic optimization to binary quadratic optimization. The factor $1/2$ is necessary: otherwise the two blocks would double the objective. No positive semidefiniteness of $A$ is assumed; generally its off-diagonal structure makes its quadratic form indefinite.

Every feasible sign vector gives the rank-one matrix $X=x{x}^T$. It is a positive semidefinite matrix, since $u^{T}Xu=(u^{T}x{)}^2\ge 0$, and $X_{ii}=x_i^2=1$. The matrix trace identity givesThus all original feasible objectives appear among the relaxed ones, provingThe semidefinite relaxation of binary quadratic optimization drops the rank-one requirement and keeps the elliptope constraints.

An optimal relaxed matrix exists. The feasible set is nonempty, since it contains $I$, and closed. Its two-by-two principal minors give $|X_{ij}{|}^2\le X_{ii}{X}_{jj}=1$, so it is bounded and therefore compact. The linear objective attains its maximum. Also $\mathrm{tr}(AI)=0$ and hence $p_{\mathrm{SDP}}^{\ast }\ge 0$, including the zero-objective case.

Write a test vector as $(u,v)$ and put $s={\sum }_i{u}_i$, $t={\sum }_j{v}_j$. Its quadratic form against the block-constant matrix isSince $a\ge |b|$, both $a$ and $a-|b|$ are nonnegative, andEquivalently it isTherefore the bipartite block-constant positive semidefinite matrix satisfies $\boxed{\left(\begin{array}{cc}a{J}_{n,n}& b{J}_{n,m}\\ b{J}_{m,n}& a{J}_{m,m}\end{array}\right)⪰0}$.

For positive block sizes, the condition is also necessary: choose $s=t$ and $s=-t$ to obtain $a+b\ge 0$ and $a-b\ge 0$. Equality is allowed, and vectors whose sums vanish in both blocks lie in the kernel. Positive semidefiniteness, rather than positive definiteness, is the conclusion.

Put $t=\log (1+\sqrt{2})=\mathrm{arsinh}(1)$, so $c_K=2t/\pi$. The power series coefficients of the hyperbolic sine preprocessing areThus $f_j=|g_j|$ for every $j\ge 0$, and the series converge on the full interval. Applying coefficient-dominated entrywise positivity gives $Y⪰0$.

Every diagonal entry belongs to one of the diagonal blocks and equals $f(X_{ii})=f(1)=\sinh t$. Because $e^t=1+\sqrt{2}$ and $e^{-t}=\sqrt{2}-1$, this is $(e^t-e^{-t})/2=1$. ThereforeThe matrix lies in the elliptope and admits a Gram matrix representation by unit vectors. This preprocessing is the Krivine rounding scheme; the equality of absolute coefficients is what preserves positive semidefiniteness even though the cross-block sine coefficients alternate in sign.

The Gram matrix representation and $Y_{ii}=1$ give $\Vert v_i{\Vert }_2=1$. For a standard Gaussian random vector $Z$, each $⟨v_i,Z⟩$ is a standard normal scalar, so the zero event has probability zero. Choose either sign convention at zero; the resulting vector is almost surely a feasible sign vector. Consequently $y^{T}Ay\le v^{\ast }$ almost surely.

The Gaussian sign-correlation identity givesThe last equality uses symmetry of the real matrices. HenceThe inverse sine is applied entrywise, not through spectral matrix functional calculus.

For completeness, the correlation coefficient identity has a geometric proof. If the angle between two unit vectors is $\theta$, the isotropic Gaussian random vector direction in their two-dimensional span gives opposite signs on angular sectors with probability $\theta /\pi$. Thus the sign product has expectation $1-2\theta /\pi =(2/\pi )\arcsin (\cos \theta )$. Parallel and antiparallel pairs give the endpoint values $1$ and $-1$ directly. This is Gaussian hyperplane rounding, and needs no independence between the rounded coordinates.

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.

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 $p_{-k}$.

The strict null space property of order $s$ isHere $v_S$ agrees with $v$ on $S$ and is zero elsewhere. Suppose a nonzero null vector had at most $2s$ nonzero entries. Split its support of a vector into disjoint sets $S,T$ of size at most $s$. Applying the null space property to both sets would give $\Vert v_S{\Vert }_1<\Vert v_T{\Vert }_1$ and $\Vert v_T{\Vert }_1<\Vert v_S{\Vert }_1$, a contradiction. Empty parts cause the same contradiction. Thus no such nonzero null vector exists.

For an $s$-sparse $x$, the feasible vector $x$ has $\Vert x{\Vert }_0\le s$, where the zero-subscript quantity counts nonzero entries and is not a norm. Any feasible competitor $z$ with $\Vert z{\Vert }_0\le \Vert x{\Vert }_0$ also has at most $s$ nonzero entries. The difference $z-x\in \ker A$ has at most $2s$ nonzero entries, so $z=x$. Competitors with larger support have strictly larger objective. Therefore every s-sparse vector is the unique sparsest feasible vector. This is sparse injectivity; for $s=0$ the only sparse vector is zero and the conclusion is immediate.