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January 4th, 2013, 12:08 PM  #1 
Newbie Joined: Feb 2012 Posts: 6 Thanks: 0  Orthogonal projection of eigenvector
Hi All, I'm studying for my Linear Algebra exam, but there is a part of the solution manual that I don't get. The problem is as follows: we have a symmetric matrix A which is Code: [ 5 4 2 ] [ 4 5 2 ] [ 2 2 2 ] The first thing I did was determining the eigenvalues, which are 10 (with multiplicity 1) and 1 (with multiplicity 2). v1 is corresponding to 10 and v2 is corresponding to v2. The next thing to do is to find a second eigenvector for the basis of the eigenspace corresponding to eigenvalue 1. Computations led to the vector v3 = (1,0,2), just like the solution manual said. However, to eventually get to the matrix P (to form A = PDP^(1) ), they convert v3 via an orthogonal projection to (1,1,4). My question is how they get there. Obviously the reason is that v2 and v3 are not orthogonal, but how do they come to the (1,1,4) vector? I hope everything is clear, any help would be appreciated! 
January 4th, 2013, 02:28 PM  #2  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: Orthogonal projection of eigenvector Quote:
so (2, 2 ,1) is an eigenvector corresponding to eigenvalue 10. However, but that is NOT a multiple of <1, 1, 1> so <1, 1, 1> is NOT an eigenvector. But it is easy to see that 1 is an eigenvalue. In order that (x, y, z) be a eignvector corresponding to eigenvalue 1, we must have That gives the equations 5x 4y 2z= x, 4x+ 5y+ 2z= y, 2x+ 2y+ 2z= z or 4x 4y 2z= 0, 4x+ 4y+ 2z= 0, 2x+ 2y+ z= 0, all of which are equivalent to z= 2x 2y. That is, we have (x, y, z)= (x, y, 2x 2y)= (x, 0, 2x)+ (0, y, 2y)= x(1, 0, 2)+ y(0, 1, 2). That tells us that (1, 0, 2) and (0, 1, 2) span the space of all eigenvectors corresponding to eigenvalue 1. That does NOT incude (1, 1, 1)! Quote:
 
January 5th, 2013, 04:24 AM  #3 
Newbie Joined: Feb 2012 Posts: 6 Thanks: 0  Re: Orthogonal projection of eigenvector
Oops, I have made a mistake; the eigenvector that the book gave should be (1,1,0), which is an eigenvector corresponding to the eigenvalue 1. So, for eigenvalue 1, the eigenspace would be spanned by (1,1,0) and (1,0,2) in my findings. However, then I'm still not convinced how they can convert (1,0,2) into (1,1,4). It seems that the eigenspace would change dramatically when doing this right? I do understand the part about normalizing the eigenvectors though! Thanks for all the help though! BTW, is there a guide on how to type in matrix notation, like you do? 
January 5th, 2013, 07:58 AM  #4 
Newbie Joined: Feb 2012 Posts: 6 Thanks: 0  Re: Orthogonal projection of eigenvector
Nevermind about the matrix code; I see you can use LaTex codes here! About the problem at hand, here is the full solution from the manual. The underlined part is the part I don't fully understand. 

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