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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 trcos(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  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: Finding the matrix in the standard basis for the project Quote:
I can't help but wonder if you are not using "projection" where you should be using "matrix".  

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