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I'm trying to prove the following:

  1. Given a matrix, the singular values of that matrix are uniquely determined.
  2. If that matrix is square, and the singular values are distinct, the left and right singular vectors are uniquely determined up to a sign.

These are suppose to be corollaries of the relation between SVD's and eigenvalues decomposition, i.e. that if $A = U\Sigma V'$, then $A'A = V\Sigma^2 V'$.

Rodrigo de Azevedo
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Maverick Meerkat
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    You may find this useful. – Rodrigo de Azevedo Mar 23 '20 at 13:30
  • For 2), supposing we're working in $\mathbb C$, then if $A = U\Sigma V^* = UD\Sigma D^V^$ for some unitary diagonal matrix $D$. You also need to specify an enforced ordering $\sigma_1\gt \sigma_2 \gt....\gt \sigma_n$. A better, closely related exercise is to prove if $A$ is invertible then it's (left) polar decomposition is unique. – user8675309 Mar 23 '20 at 19:08
  • @RodrigodeAzevedo, Sorry but I don't find it useful. I found a proof in the Numerical Linear Algebra book, but it uses the construction method, not the eigenvalue decomposition. So maybe it's not a corollary. Also, the proof there is quite cryptic and has a lot of "jumps" in it. This solves some of the jumps: https://mathoverflow.net/questions/163200/about-partial-uniqueness-of-svd but it's still difficult as hell, and I'm afraid I still don't understand the proof. – Maverick Meerkat Mar 24 '20 at 10:58
  • SVD and the columns..... Post generalizes the ambiguity as a complex phase factor. – dantopa Mar 28 '20 at 00:38

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