Is $U = V$ in the SVD of symmetric matrices? Consider the SVD of matrix $A$:
$$A = U \Sigma V^\top$$
If $A$ is a symmetric, real matrix, is there a guarantee that $U = V$?
There is a similar question here that also posits $A$ is positive semi-definite. But I'm wondering whether $U$ would be equal to $V$ if $A$ is symmetric?
 A: No: if $\Sigma$ is diagonal with non-negative entries, then $U \Sigma U^T$ will necessarily be positive semidefinite. Indeed, we note that for any column vector $x$, we have
$$
x^\top(U\Sigma U^\top)x = (U^\top x)^\top \Sigma (U^\top x) \geq 0.
$$
A: In this question, $A$ is symmetric and therefore must be square.
However, in the general case, if $A$ is not a square matrix, it can be converted into a square matrix $AA^T$ or $A^TA$ and an eigendecomposition can be done on that to get the SVD components. The $V$ are eigenvectors of $A^TA$ and the $U$ are eigenvectors of $AA^T$. The singular values in $\Sigma$ are the square roots of the non-zero eigenvalues of $A$. This (the square root of the eigenvalues) is where the positive semi-definite condition for the SVD to be equivalent to the eigendecomposition comes from. See The SVD and the EVD in http://www.math.kent.edu/~reichel/courses/intr.num.comp.1/fall11/lecture7/svd.pdf for a more detailed explanation.
[Edited to incorporate Joppy's comment.]
A: If $A=USU^T$ is a SVD, $A$ has to be positive semidefinite, as shown in another answer here.
The converse is not true. If $A$ is positive semidefinite and $A=USV^T$ is a SVD, $U$ is not necessarily equal to $V$. E.g. $A=U0V^T$ is a SVD for any two unitary matrices $U$ and $V$.
However, if $A$ is positive definite, we must have $U=V$ in its SVD, as shown in the answers to Is $U=V$ in the SVD of a symmetric positive semidefinite matrix?
