Singular values of difference of two matrices in specific setting

Question

Suppose we have the SVD $X = U\Sigma V^T$, where we can assume $\Sigma$ is nonzero. Suppose we have a matrix $G$ satisfying $$Tr(X^TG) = Tr(\Sigma)$$ $$\sigma_1(G) \leq 1$$ where $\sigma_1(G)$ is its largest singular value. I am trying to prove that then $$\sigma_1(G - UV^T) \leq 1$$ Both conditions must be necessary, because otherwise we could take $G = -UV^T$ and get a counterexample. The first condition says that $$Tr(V\Sigma U^TG) = Tr(\Sigma) \implies Tr((U^TGV - I)\Sigma) = 0$$ This doesn't quite say $G = UV^T$ (if it did, the problem would be trivial), but I'm thinking it might say something close enough so that we can solve the problem.

Answer 1

$\Sigma$ needs to be full rank, otherwise, Wolog let $\Sigma_{00}=0$, entries of $G'=U^TGV$ be all $0$ except $G'_{00}=-0.99$, all conditions are satisfied and $\sigma_1(G-UV^T)=1.99$.

If $\Sigma$ is full rank, because $x^T(I-G')x\geq 0$ for any vector $x$, the diagonal terms of $I-G'$ must be non-negative, hence all zero. Hence, the diagonal terms of $G'$ must be all $1$. Because $\sigma_1(G')\leq 1$, $G'$ must be diagonal. Hence, $I-G'=0$.