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

Community
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biswpo
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2 Answers2

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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

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

KittyL
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The first step is to see that

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

science
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  • i am not getting why the first term is equivalent to the second term why $U\Sigma V^T - X_k \iff U\Sigma V^TV$ – biswpo Apr 11 '15 at 10:29
  • Just apply $V$ from the right as $ (U\Sigma V^\intercal - X_k)V $. – science Apr 11 '15 at 10:31
  • that is what i am not getting how can i apply V then the two term is not equal.Is it because multiplying with V does not change the optimization problem ? – biswpo Apr 11 '15 at 10:32
  • they have also stated that the frobenius norm is equal.I am not getting it is coming from which property of probenius norm.Is there any property of frobenius norm that it is orthogonal matrix invariant? – biswpo Apr 11 '15 at 10:35
  • I'll ben with you.. – science Apr 11 '15 at 10:37
  • sorry but i am not getting your comment. – biswpo Apr 11 '15 at 10:42