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?
Best Answer
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.)