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For a bipartite elliptope matrix $X$, put $t=\log (1+\sqrt{2})$ and apply $\sinh (tx)$ within the two diagonal blocks and $\sin (tx)$ across them. Matching absolute power series coefficients give a positive semidefinite matrix by coefficient-dominated entrywise positivity, while $\sinh t=1$ gives unit diagonal. Gaussian hyperplane rounding then turns each cross-block correlation coefficient into $(2/\pi )\arcsin (\sin (t{X}_{ij}))=(2t/\pi )X_{ij}$ because $t<\pi /2$. The zero diagonal blocks of the bipartite objective eliminate every other contribution.

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.

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.

For real symmetric $A$, replace sign-vector lifts $x{x}^T$ by all matrices in the elliptope:Every sign vector gives a feasible rank-one matrix with the same objective, so the relaxed maximum is an upper bound. The omitted rank-one condition is substantive: an arbitrary feasible Gram matrix need not come from signs.