invers from covariance of a matrix*matrix'
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given a is a matrix, is the matrix of covariance of (a*a') is always singular?
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the cyclist
am 2 Jan. 2012
Can you please clarify? Are you interested in the singularity of cov(a) for arbitrary a, or of cov(b), for b = (a*a')?
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Teja Muppirala
am 4 Jan. 2012
cov(a) is ALWAYS singular for ANY square matrix a, because you subtract off the column means. This guarantees that you reduces the rank by one (unless it is already singular) before multiplying the matrix with its transpose.
a = rand(5,5); % a is an arbitrary square matrix
rank(a) %<-- is 5
a2 = bsxfun(@minus, a, mean(a));
rank(a2) %<-- is now 4
cova = a2'*a2/4 %<-- (rank 4) x (rank 4) = rank 4
cov(a) %<-- This is the same as "cova"
rank(cova) %<-- verify this is rank 4
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the cyclist
am 2 Jan. 2012
a = [1 0; 0 1]
is an example of a matrix for which (a*a') is not singular.
Did you mean non-singular?
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Walter Roberson
am 3 Jan. 2012
Just don't ask me _why_ it is singular. I didn't figure out Why, I just made sure square matrices could not get to those routines.
Walter Roberson
am 3 Jan. 2012
Experimentally, if you have a matrix A which is M by N, then rank(cov(A)) is min(M-1,N), and thus would be singular for a square matrix.
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