Relation between SVD and eigen decomposition for symetric matrix.

Question

Suppose I have a symetric positive definite matrix $Q$;

I seems that the eigen values of $Q$ are equal to the singular values of $Q$, I did not found any counterexample but I may be wrong ^^

Does anyone can tell me if the proposal is true or not ?

That's what I tried :

$Q = U \Sigma V^*$

$QQ^* = (U \Sigma V^*)(U \Sigma V^*)^*$

$= U \Sigma V^* V \Sigma^* U^*$

$= U \Sigma \Sigma^* U^*$

$= (U \Sigma V^*)(U \Sigma V^*)$

$=...$

$=U \Sigma U*$

Answer 1

$QQ^*=(U\Sigma V^*)(U\Sigma V^*)^*=U\Sigma^2V^*$

$Q^*Q=(U\Sigma V^*)^*(U\Sigma V^*)=V\Sigma^2U^*$

If $Q$ is symetric then :

$U\Sigma^2V^*= V\Sigma^2U^*$

$=> U = V$

Thus :

$Q = U \Sigma V^* = U \Sigma U^*$