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Nov 18, 2019 at 22:56 comment added Gautam Hi guys, thanks for all the help! @YoavKallus was right, I already had the answer - just forgot a term when doing the optimization!
Nov 18, 2019 at 14:00 comment added Nathaniel Johnston @Guatam -- The polar decomposition and SVD for square matrices are "essentially" the same. If A = USV^* is an SVD then A = (UV^*)(VSV^*) is a polar decomposition, and if A = UP is a polar decomposition then A = (UV)(V^*PV)V^* is a singular value decomposition, where V is chosen so that V^*PV = S is a spectral decomposition of P.
Nov 18, 2019 at 9:50 comment added J. M. isn't a mathematician Have you seen Kahan's note?
Nov 18, 2019 at 6:53 comment added Yoav Kallus You already have the answer in your own post: for the case of a real nonnegative diagonal matrix, think of the projection to the less strict constraint of normalized rows. If this already lends you in the stricter constraint of unitarity, then the result is also the projection to the stricter constraint. Also, you can get the polar decomposition from SVD and check that the answers agree.
Nov 18, 2019 at 6:16 comment added Gautam Hi @YoavKallus, thanks for your response. I looked at the text you suggested - their theorem is in terms of polar decomposition, not SVDs. How did you pull out the SVD result? I had already reduced the problem to the case when $X$ is diagonal, using unitary invariance of the Frobenius norm. Intuitively, the identity matrix is the optimal choice in this case, which leads to your result. How do you prove that the identity matrix is the best one when $X$ is diagonal?
Nov 18, 2019 at 4:55 review Close votes
Nov 18, 2019 at 15:49
Nov 18, 2019 at 3:40 comment added Yoav Kallus Same answer applies: if X = WSV is the svd, U = WV is the projection.
Nov 18, 2019 at 3:39 comment added Yoav Kallus Related: mathoverflow.net/questions/86539/…
Nov 18, 2019 at 2:22 history asked Gautam CC BY-SA 4.0