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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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    Yes, or eigen value decomposition, for $K= U\Sigma U^T$ let $X=U sqrt(\Sigma)$. Clearly, $X$ is not unique.2012-11-05

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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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    Another dead horse for my collection...2012-11-05