invers from covariance of a matrix*matrix'

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eri 2012-1-2

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采纳的回答： Teja Muppirala

given a is a matrix, is the matrix of covariance of (a*a') is always singular?

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the cyclist 2012-1-2

Can you please clarify? Are you interested in the singularity of cov(a) for arbitrary a, or of cov(b), for b = (a*a')?

eri 2012-1-3

cov(b) for b=a*a'

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Teja Muppirala 2012-1-4

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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