Show $U = V$ in SVD of Hermitian, positive definite matrice $A_{n \times n}$
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Since eigenvalues and singular values coincide in Hermitian matrices,
$$
AV = U \Sigma = U \Lambda
$$
where $\Sigma$ is the singular value diagonal matrix with decreasing diagonal, and $\Lambda$ is the eigenvalue diagonal matrix, i.e. $\Sigma = \Lambda$.
This implies that
$$
Av_i = \lambda_i u_i \ \ (\ast)
$$
where $i=1,\cdots, n$, which does not necessarily imply that $v_i$ are eigenvectors, and $v_i = u_i$, $\forall i$.
Let
$$
v_i = c_{i1}\xi_1 + \cdots + c_{in}\xi_n \\
u_i = d_{i1}\xi_1 + \cdots + d_{in}\xi_n \\
$$
where $\xi_i$ is the eigenvectors corresponding to $\lambda_i$.
The $(\ast)$ implies that, $\forall i$,
$$
c_{i1}\lambda_1\xi_1 + \cdots + c_{in}\lambda_n\xi_n = d_{i1}\lambda_i\xi_1 + \cdots + d_{in}\lambda_i\xi_n
$$
thus, since $[\xi_1 | \cdots | \xi_n]$ : invertible by spectral theorem,
$$
\begin{bmatrix}
c_{11} & c_{12} & \cdots & c_{1n} \\
c_{21} & c_{22} & \cdots & c_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
c_{n1} & c_{n2} & \cdots & c_{nn} \\
\end{bmatrix}
\begin{bmatrix}
\lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_n
\end{bmatrix} =
\begin{bmatrix}
d_{11} & d_{12} & \cdots & d_{1n} \\
d_{21} & d_{22} & \cdots & d_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
d_{n1} & d_{n2} & \cdots & d_{nn} \\
\end{bmatrix}^T
\begin{bmatrix}
\lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_n
\end{bmatrix}
$$
I notice that the matrices of $c_{ij}$ and $d_{ij}$ are invertible by the invertibility of $U$ and $V$, but I'm stuck at showing the identity.
Any help will be appreciated.
Best Answer
Here is one way to prove it. Note that every positive semidefinite matrix $P$ has a unique positive semidefinite square root $P^{1/2}$. In fact, $P^{1/2}=f(P)$ where $f$ is the Lagrange interpolation polynomial that maps each eigenvalue of $P$ to its square root.
Now, in your case, since $A^2=AA^\ast=US^2U^\ast$, by the uniqueness of positive definite square root, we have $A=USU^\ast$. Hence $U=V$.