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I was looking at this http://www.math.wsu.edu/faculty/watkins/pdfiles/francis.pdf

And this: http://www.math.wsu.edu/faculty/watkins/slides/ilas10.pdf

And it says that once you have the value of x you remove the zeroes and somehow create this enter image description here

I have x and I can compute alpha, but I'm not sure how to generate this (3x3) reflector. I am trying to do this with code, so I'm ultimately looking for an algorithm/formula and not a "figure it out" type thing - the paper seems to suggest the existence of one with Theorem 1 but I can't figure out how to calculate the reflector.

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    $\begingroup$ I'm trying to find the $\tilde Q_0$ you're talking about. I see a $Q_0$ in the slides, is that the same thing? $\endgroup$ Aug 26, 2016 at 21:57

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Apparently, you mean equation (4) from the paper on page 393. You're supposed to build $\tilde Q_0$ directly using the fact that $\tilde Q_0 x = \alpha e_1$ and Theorem 1 on page 390.

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  • $\begingroup$ Yea, I understand that, but I wanted to know HOW you build it? Like, I just am not getting how Q*x can equal two separate vectors. It says alpha is pos/neg ||x||*e1. How do you solve for Q? $\endgroup$ Aug 29, 2016 at 12:43
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    $\begingroup$ It's not that $Qx$ is equal to both, it's that you can choose a $Q$ for either one, and it doesn't matter which you choose. To find your $Q$, plug in your choice of $\|x\|e_1$ or $-\|x\|e_1$ to the formula. $\endgroup$ Aug 29, 2016 at 12:56

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