SVD with symmetric positive definite

Question

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

You may want to take a look at this. – Rodrigo de Azevedo Aug 19 '20 at 11:16 — Rodrigo de Azevedo, Aug 19 '20 at 11:16

score 1 · Answer 1 · answered Aug 18 '20 at 04:09

1

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

answered Aug 18 '20 at 04:09

Alessandro Cigna

2,496

dineshdileep · Answer 2 · 2020-08-20T07:48:43.810

0

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.

edited Aug 20 '20 at 07:48

answered Aug 18 '20 at 04:24

dineshdileep

8,887

@RodrigodeAzevedo now? – dineshdileep Aug 20 '20 at 07:48
Suppose that I flip the signs of both $u_1$ and $v_1$. Still unique? The idea is developed here. – Rodrigo de Azevedo Aug 20 '20 at 08:28
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. – dineshdileep Aug 20 '20 at 08:31

SVD with symmetric positive definite

2 Answers2