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Suppose a symmetric matrix $L\in\mathbb{R}^{n\times n}$ is given, and a rectangular matrix $A\in\mathbb{R}^{n\times m}$, $m. A solution to the system $LAx=b, \tag 1$ is sought, for known $b\in\mathbb{R}^n$ and unknown $x\in\mathbb{R}^m$. Since $LA\in\mathbb{R}^{n\times m}$, a solution $x$ should be calculated in a least squares sense. Now, suppose that one premultiplies the above system by $A^T$, ie, $A^TLAx=A^Tb \tag 2$

Under which conditions on $A$ (or $A^T$) can the solution to (2) be obtained by solving (1) (by eliminating $A^T$ from the second)? In other words, is it safe to replace solving (1) by solving (2), or vice versa?

I suppose that in case $A^T$ is non-invertible the solution to the first and the second system differ. What is the condition in a general case of rectangular $A$? What about square symmetric $A=A^T$?

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The question reduces to: could one replace solving (1) by solving (2), or vice versa? Why couldn't one replace solving (2) by solving (1)? If a solution to (1) exists, then it is also a solution to (2). Could it be that the solution to (1) does not exits (ie, exists only in LS sense), but the exact solution to (2) exists?

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    I believe the answer is yes; I have to think a while, but I will post my thoughts shortly.2012-09-19

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