Singular Value Decomposition of Symmetric Matrix

Question

I'm trying to understand the SVD of a real symmetric matrix.

Let $A$ be our $n\times n$ real symmetric matrix. And let an SVD be $A=U\Sigma V^T$.

Let $u_i$'s and $v_i$'s be the columns of $U$ and $T$

In the comment https://math.stackexchange.com/a/22832/764199, it says $u_i=\pm v_i$.

I could not get it why it's the case?

For example here https://math.stackexchange.com/a/3683742/764199 it says columns of $U$ and $V$ are eigenvectors of $A^2$, so they are eigenvectors of $A$. But this is not correct in general.

Addition:

For instance at https://math.berkeley.edu/~hutching/teach/54-2017/svd-notes.pdf, Example 2.2 there is the following explanation:

If $A$ is real symmetric matrix, then we can obtain an SVD from the eigenvalue decomposition $A=PDP^{-1}$. In that case, we can obtain $v_i=\pm u_i$. But how do we know that we can obtain every SVD from an eigenvalue decomposition?

Addition 2: Because we can say it for the SVD obtained from the eigenvalue decomposition, it suffices to show that, as it's written in Wikipedia page, "the SVD is unique up to arbitrary unitary transformations applied uniformly to the column vectors of both $U$ and $V$ spanning the subspaces of each singular value" to complete our proof. https://en.wikipedia.org/wiki/Singular_value_decomposition

But I couldn't find the proof of that "uniquenes up to" statement.

Answer 1

Without additional conditions, the statement is false. Here are two simple counterexamples:

1. $U0V^T$ is a SVD of the zero matrix for any pair of orthogonal matrices $U$ and $V$.
2. $IIA^T$ is a SVD of $A=\frac{1}{\sqrt{2}}\pmatrix{1&1\\ 1&-1}$.

However, the statement is true when all singular values of $A$ are distinct. Since $A$ is symmetric, we have $U\Sigma^2U^T=AA^T=A^2=A^TA=V\Sigma^2V^T$. Therefore, for each $j$, $u_j$ and $v_j$ are unit eigenvectors of $A^2$ corresponding the the $j$-th largest eigenvalue. Yet, $A^2$ have distinct eigenvalues (because its eigenvalues are the singular values of $A$). Therefore its eigenspaces are one-dimensional. Hence $u_j=\pm v_j$.

Alternatively, as positive semidefinite matrices have unique positive semidefinite square roots, the equality $U\Sigma^2U^T=V\Sigma^2V^T$ implies that $B:=U\Sigma U^T=V\Sigma V^T$. Hence $u_j$ and $v_j$ are unit singular vectors of $B$ corresponding to the singular value $\sigma_j$. As all singular values of $B$ are different (because the singular values of $B$ are just the singular values of $A$), the singular spaces of $B$ are one-dimensional. Hence $u_j=\pm v_j$.

A third argument is to note that $U\Sigma^2U^T=V\Sigma^2V^T$ implies $(V^TU)\Sigma^2=\Sigma^2(V^TU)$. Since all diagonal entries of $\Sigma^2$ are distinct, $D:=V^TU$ must be a diagonal matrix. As $V^TU$ is orthogonal, the diagonal entries of $D$ must be $\pm1$. Hence $u_j=\pm v_j$.