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

Answer 1

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

Answer 2

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.