Why $||U\Sigma V^\intercal - X_k||_F = ||\Sigma - U^\intercal X_k V ||_F$

Question

Here $U\Sigma V^\intercal $ is the svd decomposition of M.I know that U and V are othogonal but i dont know which property it is using. This in reference to the post Proof of Eckart-Young-Mirsky theorem $$||M-X_k||_F = ||U\Sigma V^\intercal - X_k||_F = ||\Sigma - U^\intercal X_k V ||_F$$

Answer 1

The first step is to see that

$$ U\Sigma V^\intercal - X_k \iff (U\Sigma - X_kV)V^{\intercal} \iff U(\Sigma - U^{\intercal}X_kV)V^{\intercal}. $$

Answer 2

As @science said:

$$||U\Sigma V^\intercal - X_k||_F=||U(\Sigma- U^\intercal X_k V) V^\intercal ||_F$$

Now use the property of F-norm

$$||UAV^T||=\sqrt{\text{trace}((UAV^T)^T(UAT^T))}=\sqrt{\text{trace}(VA^TU^TUAV^T)}=\sqrt{\text{trace}(VA^TAV^T)}=\sqrt{\text{trace}(A^TAV^TV)}=\sqrt{\text{trace}(A^TA)}=||A||_F$$