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Given the SVD $A = U\Sigma V^T$, is $\Sigma$ uniquely determined up to permuting the rows and columns?

My take is that singular value matrix is uniquely determined. Its diagonal elements are square roots of eigenvalues of $A^TA$ and $AA^T$ (both of which have the same unique set of eigenvalues). Thus, $A$ has a unique set of singular values.

Permuting rows and columns of $\Sigma$ we just rearrange the singular values along the diagonal of $\Sigma$. SVD will continue to hold if we rearrange corresponding vectors of $U$ and $V$.

However, such reasoning seems to lack rigour.

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If an eigenvalue of $A^T A$ or $AA^T$ has multiplicity greater than 1, the corresponding left and right singular vectors are not unique. In this case, the columns of $U$ and $V$ can be permuted, but one can also choose entirely different left and right singular vectors.

A simple example. $$ \begin{align*} \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} &= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}^T \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix}^T \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}} \end{bmatrix}. \\ \end{align*} $$

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