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If we know basis vectors for $K=XX^T$ (e.g. will be eigenvectors here since $K$ is symmetric), how can we find base vectors for $X$?

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Tell you what: $$ K \; = \; \left( \begin{array}{cc} 2 & 3 \\ 3 & 5 \end{array} \right). $$ Now you tell me what $X$ is and we can proceed. –  Will Jagy Nov 5 '12 at 23:52
    
ignore if meaningless, but can't you do Cholesky factorization on $K$ to get $X$? –  user2468 Nov 5 '12 at 23:59
    
Yes, or eigen value decomposition, for $K= U\Sigma U^T$ let $X=U sqrt(\Sigma)$. Clearly, $X$ is not unique. –  user25004 Nov 5 '12 at 23:59

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up vote 2 down vote accepted

Not generally, for example, if $X$ is orthogonal, then $X X^T = I$, and we can choose any basis we want for $I$, but that tells us nothing about $X$.

I misunderstood the question:

Since we have ${\cal R}(X) = {\cal R}(X X^T) = {\cal R}(K) $ (since ${\cal R}(X^T) = \ker X^\bot$), any basis for ${\cal R}(K)$ will do, for example, any maximal linearly independent set of columns of $K$.

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To copper.hat: Thanks. For $n\times n$ identity matrix, any set with cardinality $n$ of $n$ dimensional orthogonal vectors, is a basis set. On the other hand, if $XX^T=I$, rank of $X$ is also $n$ so the same set of basis is a basis for $X$ as well. So I think I do not see your point. –  user25004 Nov 5 '12 at 22:52
    
Suppose the original matrix $X$ is orthonormal, for example. Then $K$ is the identity, and hence any orthonormal basis is a set of eigenvectors for $K$. But these have no relation to the original $X$. The point is that the 'squaring' (well, $X X^T$) loses information that cannot be recovered. –  copper.hat Nov 5 '12 at 23:35
    
Maybe I misunderstand the question; are you asking if given the matrix $K$ can we find eigenvectors (or some information) for $X$? –  copper.hat Nov 5 '12 at 23:37
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Well, ${\cal R}(X) = {\cal R}(X X^T)$, so the answer is yes. I have updated my answer. –  copper.hat Nov 5 '12 at 23:45
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Another dead horse for my collection... –  copper.hat Nov 5 '12 at 23:57

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