/sci/ - Science & Math » Thread #11595284

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Anonymous Thu 23 Apr 16:55:28 2020 No.11595284 View Reply Original Report

Quoted By: >>11595362

Ok, so there is pic related. I can't fucking understand how you get the first two, and how does the third comes from the first two. So Anons, please help me out with this shit.
The problem is to construct the matrix of a linear projection, that creates the the component of r parallel to n.

Anonymous

Anonymous Thu 23 Apr 2020 17:13:16 No.11595362 Report

Quoted By: >>11595432

>>11595284
The first is just the definition, if r = (x_j) then the multiplication by the matrix P gives the first formula.
The second is just computing the projection with the inner products.
The third equates 1 and 2

Anonymous

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Anonymous Thu 23 Apr 2020 17:33:21 No.11595432 Report

Quoted By: >>11595670

>>11595362
Thanks Anon! Now I can finish my assigment. I was too dumb to think this through, and the teacher just thought that this is trivial for us.

Anonymous

Anonymous Thu 23 Apr 2020 18:50:44 No.11595670 Report

Quoted By:

>>11595432
it should be pretty obvious
I like TeX so here we go.
so we have
$ x=\left[ \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right] $
and $ P^{||} = \left[ \begin{array}{ccc} P^{||}_{1,1} & P^{||}_{1,2} & P^{||}_{1,3} \\ P^{||}_{2,1} & P^{||}_{2,2} & P^{||}_{2,3} \\ P^{||}_{3,1} & P^{||}_{3,2} & P^{||}_{3,3} \end{array} \right] $
so

$ r_{||} = P^{||}x = \left[ \begin{array}{ccc} P^{||}_{1,1} & P^{||}_{1,2} & P^{||}_{1,3} \\ P^{||}_{2,1} & P^{||}_{2,2} & P^{||}_{2,3} \\ P^{||}_{3,1} & P^{||}_{3,2} & P^{||}_{3,3} \end{array} \right] \left[ \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right] = \left[ \begin{array}{c} P^{||}_{1,1}x_1 + P^{||}_{1,2}x_2 + P^{||}_{1,3}x_3 \\ P^{||}_{2,1}x_1 + P^{||}_{2,2}x_2 + P^{||}_{2,3}x_3 \\ P^{||}_{3,1}x_1 + P^{||}_{3,2}x_2 + P^{||}_{3,3}x_3 \end{array} \right] $
and for each line $i$ of vector $r_{||}$ one has
$r_{|| i} = P^{||}_{i,1} x_1 + P^{||}_{i,2} x_2 + P^{||}_{i,3} x_3 = \sum\limits_{j=1}^{3} P^{||}_{ij}x_j $
which is the first expression.

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