Re:Singular Value Decomposition Part 2: Theorem, Proof, Algo...
by amelius on Tuesday May 17, 2016 @06:10AM
(#11712374)
This got me thinking. How many numbers do you need to describe any unitary matrix of size N?
Re:Singular Value Decomposition Part 2: Theorem, Proof, Algo...
by mathpepe on Tuesday May 17, 2016 @01:45PM
(#11715664)
In the case of matrices with real coefficients, these matrices constitute the famous ortogonal group. The ortogonal group is a Lie group of dimension n*(n-1)/2. A table of Lie groups with their dimensions is here https://en.wikipedia.org/wiki/Table_of_Lie_groups
Every Lie group has an associated vector space with the same dimension, so many question about Lie groups can be addressed by studying the associate vector space.
To suggest an application of Lie groups and their associated vector spaces, in machine learning one can reduce the number of parameters when a group act over a space (rotations and the like), so you obtain new features to train your machine in a more efficient way.
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.
Although a useful metric, measuring only the amount of input and output misses other useful characteristics.
The SVD produces more output than input, but the output is very heavily constrained.