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I'm going through the proof for the Orthogonal Procrustes problem and I wanted to see how they got the following relation.

$$\langle \Omega A - B, \Omega A - B \rangle = ||A||^2_F+||B||^2_F-2\langle\Omega A, B\rangle$$

After using inner product rules, I arrived at the following quantity

$$\langle \Omega A, \Omega A\rangle + \langle B, B \rangle - 2 \langle \Omega A, B \rangle$$

which gives me

$$||\Omega A||_F + ||B||_F -2\langle\Omega A, B \rangle$$

which is wrong because of the extra $\Omega$. I'm not sure if there's a way to simplify beyond this, or if this is just wrong. It might help to know that $\Omega$ is an orthogonal matrix, since $||\Omega||_F = n$ if $\Omega$ is an $n \times n$ matrix.

I'm thinking that since $||\Omega A||_F = n||A||_F$, the $ \underset{\Omega}{\text{arg min}}$ doesn't change regardless of constant $n$.

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  • $\begingroup$ What do you define $<a,b>$ ? $\endgroup$ – Mostafa Ayaz Jun 16 '18 at 18:31
  • $\begingroup$ @MostafaAyaz according to the wikipedia article, it's the "matrix inner product" which for some reason links to the "matrix multiplication" article $\endgroup$ – Carpetfizz Jun 16 '18 at 18:33
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For unitary $\Omega$ $$\langle \Omega A - B, \Omega A - B \rangle=\langle \Omega A , \Omega A\rangle-2\langle \Omega A,B \rangle+\langle B, B \rangle$$ Also $$< \Omega A,\Omega A>=trA^H\Omega^H\Omega A=trA^HA=<A,A>$$which completes our proof

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