Left singular vectors of an orthogonal matrix $X$ are the same as the right singular vectors of $X^T$

Question

I figured out the singular value decomposition of a square orthogonal matrix $X \in \mathbb{R}^{n\times n}$ is $X = U \cdot \Sigma \cdot V^*$, where $U, V^* \in \mathbb{R}^{n\times n}$. Taking transpose on both sides, we now have $X^T = V^{*T} \cdot \Sigma^T \cdot U^T$.

I mean it's definitely clear that the left singular vectors of X is the same as the transpose of the right singular vectors. But how the left singular vectors the same as the right singular vectors??

Answer 1

Write the SVD of $X$ as $X=\sum_k s_k u_k v_k^\dagger$. The left singular vectors are $(u_k)_k$, while the right singular vectors are the $(v_k)_k$.

Then the Hermitian conjugate reads $X^\dagger = \sum_k s_k v_k u_k^\dagger$. For real matrices you can replace the Hermitian conjugate with the transpose.

The left singular vectors of $X$ are the $u_k$, which are clearly also the right singular vectors of $X^\dagger$.