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What I have:

$A \in \mathbb{R}^{n \times n}$ is a symmetric positive-definite matrix and has Singular Value Decomposition $A = U\Sigma V^T$.

What I want: prove that $U = V$.

Also... does $A$ has to be positive-definite for this to be true?

Edit: There is the exactly same question here, but none of the answers there satisfy my doubts. I'm talking in comments with people here and there. If I come up with a detailed enough answer I might write it there. But I suggest also reading the comments on my question here, they are helpful.

Figurinha
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    Does this answer your question? Is $U=V$ in the SVD of a symmetric positive semidefinite matrix? – FBruzzesi Sep 10 '20 at 14:16
  • @FBruzzesi Thanks! I hadn't found this question before... It's exactly my question. But the answers there aren't suficient for me. How should I answer the community member suggestion? – Figurinha Sep 10 '20 at 14:39
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    Try to show that if $U\ne V$ then the product is not symmetric. Remember that $(ABC)^\top = C^\top B^\top A^\top,$ i.e. the order of multiplication get reversed. – Michael Hardy Sep 10 '20 at 16:28
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    An alternative route that gives some useful results along the way: use SVD to construct the Polar decomposition for any square matrix (i.e. SVD implies existence of Polar decompostion). Then prove the Polar decomposition is unique for any invertible . Set := and if ≠ then you'd have (at least) two distinct polar decompositions for , which is impossible since is invertible. If we relax the PD criterion for : in general you lose uniqueness considerations when dealing with singular matrices; and if is indefinite then U=V implies A is PD so ≠ – user8675309 Sep 10 '20 at 18:11
  • @MichaelHardy That isn't exactly true. For instance, the product is still symmetric when $U = -V$. Things get even more complicated if $A$ is not necessarily invertible – Ben Grossmann Sep 11 '20 at 14:47

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