 Rademacher rounding for a semidefinite relaxation

Use an orthogonal eigendecomposition $X=V\Lambda V^T$ and a vector $\xi$ of independent Rademacher random variables. Because $\Lambda$ is diagonal and ${\xi }_j^2=1$, every sign vector gives $\Vert \hat{x}{\Vert }_2^2=\mathrm{tr}X$, without taking an expectation. If the scaling denominator $M$ is positive, then $x$ satisfies every slab constraint and $\Vert x{\Vert }_2^2=\mathrm{tr}X/M^2$. Under spanning constraints and positive trace, $M>0$ automatically. A nonzero $\hat{x}$ with $M=0$ instead certifies an unbounded direction.