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May 10, 2020 at 0:14 vote accept Santiago Armstrong
May 3, 2020 at 20:46 comment added Santiago Armstrong Thanks! You are absolutely right! I will add a proof here in case someone is looking for it in the future: Suppose $X^*=uv^*$ where $u$ and $v$ are the left and right singular vectors of $M$ associated with the largest singular value. To show that $X^*$ attains the optimum, it suffices to show that $\langle M, X^*\rangle = \|M\| = \sigma_{max}$. But this follows from the cyclicity of the trace: $trace(M^*uv^*) = trace(V^*\Sigma U u v^*) = trace(v^*V^*\Sigma U u) = \sigma_{max}$
May 3, 2020 at 14:27 history edited alesia CC BY-SA 4.0
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May 3, 2020 at 13:31 history answered alesia CC BY-SA 4.0