Recover matrix Z from XZX', Z es symmetric and n-by-n, while X is k-by-n where n>>k
1 vue (au cours des 30 derniers jours)
Afficher commentaires plus anciens
I have the matrices (XZX') and X and I want to recover Z. Dimension: X is kxn, Z is nxn, and n >> k. I know that Z is simetric
5 commentaires
David Goodmanson
le 9 Août 2020
Hi Patricio, so now Z is 27x27 which is back to an underdetermined solution for Z.
Réponses (3)
Matt J
le 8 Août 2020
Modifié(e) : Matt J
le 8 Août 2020
In the case where k truly is <<n, you can use my KronProd class to get the minimum norm solution
k=10; n=100;
X=rand(k,n);
Ztrue=rand(n); Z=Z+Z.';
M=X*Ztrue*X.';
K=KronProd({X,X});
tic;
Z = pinv(K)*M ;
toc; %Elapsed time is 0.005358 seconds.
Naturally, you should not expect the result to equal the under-determined Ztrue.
2 commentaires
David Goodmanson
le 8 Août 2020
Hi Matt, I tried to reply to your last comment but that answer is gone. I indeed did not see your updated answer when I posted my answer. Sorry I assumed wrongly, it makes sense now.
KSSV
le 9 Août 2020
Does this match your criteria?
k = 8 ; n = 5 ;
% create dummy data
X = rand(n,k) ;
Z = rand(n) ;
Z = Z+Z' ; % make Z symmetric
D = X'*Z*X ; % known value
%% solve for Z knowing D and X
Z0 = inv(X*X')*X*D*X'*inv(X*X') ; % this is same as Z
0 commentaires
Bruno Luong
le 9 Août 2020
You can't getback to 27x27 covariance matrix Z after reducing it it on 5 dimensional space (by X). The information lost forevver.
0 commentaires
Voir également
Catégories
En savoir plus sur Matrix Indexing dans Help Center et File Exchange
Produits
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!