Recover matrix Z from XZX', Z es symmetric and n-by-n, while X is k-by-n where n>>k
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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 件のコメント
David Goodmanson
2020 年 8 月 9 日
Hi Patricio, so now Z is 27x27 which is back to an underdetermined solution for Z.
回答 (3 件)
Matt J
2020 年 8 月 8 日
編集済み: Matt J
2020 年 8 月 8 日
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 件のコメント
David Goodmanson
2020 年 8 月 8 日
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
2020 年 8 月 9 日
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 件のコメント
Bruno Luong
2020 年 8 月 9 日
You can't getback to 27x27 covariance matrix Z after reducing it it on 5 dimensional space (by X). The information lost forevver.
0 件のコメント
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