invers from covariance of a matrix*matrix'
4 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
given a is a matrix, is the matrix of covariance of (a*a') is always singular?
2 commentaires
the cyclist
le 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')?
Réponse acceptée
Teja Muppirala
le 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
0 commentaires
Plus de réponses (1)
the cyclist
le 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?
8 commentaires
Walter Roberson
le 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
le 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.
Voir également
Catégories
En savoir plus sur Descriptive Statistics dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!