Recover matrix Z from XZX', Z es symmetric and n-by-n, while X is k-by-n where n>>k
1 view (last 30 days)
Show older comments
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 Comments
David Goodmanson
on 9 Aug 2020
Hi Patricio, so now Z is 27x27 which is back to an underdetermined solution for Z.
Answers (3)
Matt J
on 8 Aug 2020
Edited: Matt J
on 8 Aug 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 Comments
David Goodmanson
on 8 Aug 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
on 9 Aug 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 Comments
Bruno Luong
on 9 Aug 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 Comments
See Also
Categories
Find more on Matrix Indexing in Help Center and File Exchange
Products
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!