Given a symmetric matrix $A\in \mathbb{R}^{n\times n}$, with all the entries greater than zero $A_{i,j}>0$ with rank $k<n$, we can calculate its SVD decomposition:
$$ A = USU' $$
Assuming now that $A$ is equal to $A_T + \Delta$ where also $A_T$ is a matrix with the same properties of $A$ (real, symmetric, all positive terms, rank $k$) and $\Delta$ is a perturbation of $A_T$, we would have that
$$ A_T = U_TS_TU_T' $$
I am interested in understanding in which conditions the following implication holds
$$ || \Delta || \to 0 \Rightarrow ||U-U_T||\to 0 $$
I know that in the general case this implication might not be true, but I am interested in the case when this actually holds. Here Continuity of an "SVD" operator there is an example where the implication does not hold. However I found this paper
where actual perturbations bound are provided, so I am confused..