I'm trying to compute the singular value decomposition of the Pauli matrix $$\sigma_x = \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}.$$ According to the SVD theorem, this matrix can be decomposed as $UDV^T$ where $U$ and $V$ are orthogonal matrices and $D$ is diagonal. The matrix $U$ is constructed from the orthonormal eigenvectors of $\sigma_x \sigma_x^*$ and the matrix $V$ is constructed from the orthonormal eigenvectors of $\sigma_x^* \sigma_x$. The problem I'm having is that $\sigma_x^* \sigma_x = \sigma_x \sigma_x^* = I$, the identity matrix, and any vector in $\mathbb C^2$ is an eigenvector of the identity matrix. This implies that I can take $U$ and $V$ to be any $2\times 2$ orthogonal matrix, and I should have a valid singular value decomposition. This of course does not work, and the problem seems to arise from the degeneracy of the singular values, which in this case are both $1$, leading to an SVD which is not unique.
My question is, how does one compute the SVD of a matrix of this type, where the right and left eigenvectors are not unique?