Re:Singular Value Decomposition Part 2: Theorem, Proof, Algo...
by amelius on Tuesday May 17, 2016 @07:33AM
(#11712692)
In 2 and 3 dimensions the answer is exactly the amount of positions below the main diagonal. So in a SVD decomposition, the degrees-of-freedom actually add up nicely, i.o.w. there is no loss/creation of information. I'm guessing it generalizes to higher dimensions.
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