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..


First, I have to emphasise that $A=USU^t$ is, in general, not a SVD. It is a symmetric eigenvalue decomposition.

For all practical purposes, the assertion $$ \|U-U_T\| \rightarrow \;\;\; \mbox{as} \;\;\; \|\Delta\| \rightarrow 0 $$ is true. However, one has to be careful when the eigenvectors are not uniquely defined. As an example, any orthogonal matrix $U$, $U^tU = I$ where $I$ an identity matrix is an eigenvector matrix of the identity matrix $I$. In that case, as $\Delta$ goes to zero, solution might migrate to a different (but valid) eigenvector matrix. This cannot happen if the eigenvectors are unique. However, note that if $x$ is a normalised eigenvector then $-x$ is also an eigenvector.

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