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 October 16th, 2013, 07:47 PM #1 Newbie   Joined: Sep 2013 Posts: 23 Thanks: 0 Finding the matrix in the standard basis for the projection The problem is to find the matrix in the standard basis for the projection R^2 of the projection onto the line 2y=x. So this means the solution should be a linearly independent set of 2 vectors, but how do I figure this out, is it just a matter of picking arbitrary points for x and y? October 17th, 2013, 05:43 AM #2 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Finding the matrix in the standard basis for the project Are you sure the question isn't to find the matrix for reflection in the line 2y = x? In any case, are you not looking for a matrix rather than a basis? October 17th, 2013, 11:10 AM #3 Senior Member   Joined: Feb 2013 Posts: 281 Thanks: 0 Re: Finding the matrix in the standard basis for the project A line is a one dimensional subspace. Choose a nonzero element of the line, for example (2,1). If the length of this vector is not 1 then divide it by its length so you get a unit vector t=(2/sqrt(3),1/sqrt(3)). For the projection you should decomposite every r vector into a vector parallel to the line and a vector orthogonal to the line. In fact the projection means that you maps the r to the parallel component of it. You can express it with the introduced t unit tangent vector as: r' = (tr)t. Indeed, the scalar product in the parenthesis is |t||r|cos(theta)=r cos(theta) that is the length of parallel component. Multiplying t by this length is the projected vector r'. Now you can conclude that the projection is a linear transformation, because scalar product is linear. You can also read the element of the matrix if you reexpress it with (x,y). The formula says: x' = (t1*x+t2*y)*t1 = (t1*t1)x + (t1*t2)y, so the first row of the matrix is [t1*t1 t1*t2]. Note: The method works in higher dimension, as well. Sometimes the transformation of one dimension projection is written as that is (Sadly I forget the name of this special production.) October 22nd, 2013, 08:01 AM #4 Newbie   Joined: Sep 2013 Posts: 23 Thanks: 0 Re: Finding the matrix in the standard basis for the project Thank you very much for the help. I wa wondering, if the matrix firme by the projection in part a by a 90 degree counter clockwise rotation would it just negate the values in the x column? October 22nd, 2013, 05:21 PM   #5
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Re: Finding the matrix in the standard basis for the project

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 Originally Posted by 84grandmarquis Thank you very much for the help. I wa wondering, if the matrix firme by the projection in part a by a 90 degree counter clockwise rotation would it just negate the values in the x column?
You are using words in very strange ways that make me wonder if you are clear on their definitions. You cannot talk about a "projection by a 90 degree counter clockwise rotation". A projection is not a "rotation". And a "projection onto a line" does NOT give "a linearly independent set of 2 vectors" because a line is one dimensional.

I can't help but wonder if you are not using "projection" where you should be using "matrix". Tags basis, finding, matrix, projection, standard Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post 84grandmarquis Linear Algebra 1 December 3rd, 2013 09:18 AM 84grandmarquis Linear Algebra 2 December 2nd, 2013 09:11 AM jsdieorksw Linear Algebra 1 November 2nd, 2010 12:07 PM needd Linear Algebra 0 November 18th, 2009 12:30 PM md288 Linear Algebra 1 October 23rd, 2009 01:55 AM

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