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If $A, B \in \mathbb{R}^{n \times r}$ are two matrices, the solution to the so-called Orthogonal Procrustes Problem

$$\min_{O^TO=I_r} \|AO-B\|$$

is given by the polar factor of $A^TB$ whenever the norm is the Frobenius norm. The minimization here is taken over $r \times r$ orthogonal matrices.

I've read that the solution is the same if we use the operator norm instead, but for some reason I can't produce a proof myself. Is this fact true? If so, is it easy to see?

Rodrigo de Azevedo
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squattyroo
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1 Answers1

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This is not true. Counterexample: $$ A=\pmatrix{2&1\\ 1&1},\ B=\pmatrix{1\\ &2},\ A^TB=\pmatrix{2&2\\ 1&2}. $$ The polar decomposition of $A^TB$ is $PO$, where $$ P=\frac1{\sqrt{17}}\pmatrix{10&6\\ 6&7},\ O=\frac1{\sqrt{17}}\pmatrix{4&1\\ -1&4}. $$ One can verify that $$ \|AO-B\|=\left\|\frac1{\sqrt{17}}\pmatrix{7&6\\ 3&5}-B\right\|\approx1.6866>\sqrt{2}=\|AI-B\|. $$

user1551
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