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Let $Q$ be an arbitrary non-zero matrix and let $x$ be a column vector.

It should be true that $xx^TQ$ and $Qxx^T$ are both rank 1 matrices.

It is a fact that all rank-one matrices can be factorized as $uv^T$, where $u$ and $v$ are vectors. This can be accomplished by means of the singular value decomposition rather slowly. Is there a quick way to factorize the above rank-one matrices with small time complexity? I would think there should be a quick way to factorize all one-rank matrices...

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  • $\begingroup$ If $Q$ is not square, you need different size vectors $x,y$ for $xx^TQ$ and for $Qyy^T$. $\endgroup$ Commented Sep 27, 2013 at 7:07
  • $\begingroup$ It is not true that $xx^TQ$ and $Qyy^T$ are always rank$~1$. Whenever $x\in\ker Q^T$ respectively $y\in\ker Q$ they are of rank$~0$. $\endgroup$ Commented Sep 27, 2013 at 7:08

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But they are already factorized as wished. We have $$ xx^tQ = x(Q^tx)^t, \quad Qxx^t = (Qx)x^t. $$

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  • $\begingroup$ D'oh! Why, of course. Thank you! $\endgroup$
    – user21725
    Commented Sep 27, 2013 at 7:14

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