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When the matrix is symmetric positive definite, I know it has positive eigenvalues. With this condition, can we say the SVD of the matrix is unique?

To say SVD of a matrix is unique, as I know, it's needed to have distinct eigenvalues (up to signs). But I'm not sure the condition (symmetric positive definite) guarantee that

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  • $\begingroup$ You may want to take a look at this. $\endgroup$ Commented Aug 19, 2020 at 11:16

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Observe that the identity matrix is positive definite and symmetric but it hasn't distinct eigenvalues

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SVD is unique up to the diagonal singular value matrix when values are written in descending order. To see an example for square non-singular matrices, consider matrix $A=U\Sigma V^H$. Let $D$ be a diagonal matrix whose diagonal entries are points on the unit circle on the complex plane. Thus $DD^H = I$. Then, note that $$A=UDV^H=UD\Sigma D^HV^H = (UD) \Sigma (VD)^H$$ where $UD$ and $VD$ are unitary as well.

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  • $\begingroup$ @RodrigodeAzevedo now? $\endgroup$ Commented Aug 20, 2020 at 7:48
  • $\begingroup$ Suppose that I flip the signs of both $u_1$ and $v_1$. Still unique? The idea is developed here. $\endgroup$ Commented Aug 20, 2020 at 8:28
  • $\begingroup$ I didn't mean to say SVD is unique. The only part of SVD that remains unique is the diagonal singular matrix (up to ordering). Singular vectors are not as mentioned in the answer. $\endgroup$ Commented Aug 20, 2020 at 8:31

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