Singular value decomposition of positive definite matrix 
Let $A$ be a positive definite matrix, and let $A = U \Sigma V^*$ be its singular value decomposition (SVD). Show that $U=V$.

What I have done: $A$ is Hermitian, so $A$ is unitarily diagonalizable, say, $A=WDW^*$ where $D$ consists of the eigenvalues (decreasing order). Also $D=\Sigma$ since $A$ is positive definite. From $A^2=AA^*=UD^2U^*$, and similarly I have $A^2=UD^2U^*=VD^2V^*=WD^2W^*$ so the column vectors of $U,V,W$ corresponds to same eigenvalues of $A^2$. And I'm now stuck. How could I proceed?
 A: Not sure how you could proceed. However, without loss of generality, suppose $\Sigma=(\sigma_1 I_{r_1})\oplus\cdots\oplus(\sigma_k I_{r_k})$ where $\sigma_1,\ldots,\sigma_k$ are distinct. If $A=U\Sigma V^\ast$ is positive definite, then so is $U^\ast AU=\Sigma W$, where $W=V^\ast U$. In particular, $\Sigma W$ is Hermitian and
$$
W^\ast\Sigma^2W = (\Sigma W)^\ast(\Sigma W) = (\Sigma W)(\Sigma W)^\ast = \Sigma^2.
$$
Therefore $W$ must be a block diagonal matrix $W_1\oplus\cdots\oplus W_k$ such that $W_j$ and $I_{r_j}$ have the same size and
$$
A=U\Sigma V^\ast = V(\sigma_1 W_1)\oplus\cdots\oplus(\sigma_k W_k)V^\ast.\tag{1}
$$
Since $A$ is positive definite, its eigenvalues coincide with its singular values (and hence they are positive). Now you may argue from $(1)$ that $W=I$ and hence $U=V$.
A: As you already figured out, $A = U D V^*$ gives $A^2 = UD^2 U^* = V D^2 V^*$. Now, since positive Hermitian matrices only have one positive Hermitian square root, $A = UDU^* = UDV^*$, and since $U$ and $D$ are invertible, $U^* = V^*$.
Now, we need to show that positive Hermitian matrices have only one positive Hermitian square root. Suppose that $A$ is Hermitian. Consider the eigenspaces of $A$. Suppose that $Av = \lambda v$. Then $A^2 v = \lambda^2 v$. Since the map $\lambda \rightarrow \lambda^2$ is one-to-one on positive reals, this shows that $A$ and $A^2$ have exactly the same eigenspaces, with an eigenvalue of $\lambda$ in $A$ corresponding to an eigenvalue of $\lambda^2$ in $A^2$. This is enough to uniquely characterize $A$, given $A^2$. (We are implicitly using the fact that the vector space is a direct sum of eigenspaces of $A$. This is true since $A$ is Hermitian.) 
