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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 ] and they have given two eigenvectors, v1 = (-2,2,1) and v2 = (1,1,1). Now the question is to orthogonally diagonalize A. 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
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Re: Orthogonal projection of eigenvector

Quote:
 Originally Posted by Robertoo 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 ] and they have given two eigenvectors, v1 = (-2,2,1) and v2 = (1,1,1). Now the question is to orthogonally diagonalize A.
How nice of them to give you those eigenvectors- just don't believe everything you are told! Yes,
$\begin{bmatrix}5=&-4=&-2 \\ -4=&5=&2 \\ -2=&2=&2\end{bmatrix}\begin{bmatrix}-2 \\ 2 \\ 1\end{bmatrix}= \begin{bmatrix}5(-2)- 4(2)- 2(1) \\ -4(-2)+ 5(2)+ 2(1) \\ -2(-2)+ 2(2)+ 2(1)\end{bmatrix}= \begin{bmatrix}-10- 8- 2 \\ 8+ 10+ 2 \\ 4+ 4+ 2\end{bmatrix}= \begin{bmatrix}-20 \\ 20 \\ 10\end{bmatrix}= 10\begin{bmatrix}-2 \\ 2 \\ 1\end{bmatrix}$
so (-2, 2 ,1) is an eigenvector corresponding to eigenvalue 10.

However,
$\begin{bmatrix}5=&-4=&-2 \\ -4=&5=&2 \\ -2=&2=&2\end{bmatrix}\begin{bmatrix} 1\\ 1 \\ 1\end{bmatrix}= \begin{bmatrix}5(1)- 4(1)- 2(1) \\ -4(1)+ 5(1)+ 2(1) \\ -2(1)+ 2(1)+ 2(1)\end{bmatrix}= \begin{bmatrix}5- 4- 2 \\ -4+ 5+ 2 \\ -2+ 2+ 2\end{bmatrix}= \begin{bmatrix} -1 \\ 3 \\ 2\end{bmatrix}$
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
$\begin{bmatrix}5=&-4=&-2 \\ -4=&5=&2 \\ -2=&2=&2 \end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}= \begin{bmatrix}5x- 4y- 2z \\ -4x+ 5y+ 2z \\ -2x+ 2y+ 2z\end{bmatrix}= \begin{bmatrix}x \\ y \\ z\end{bmatrix}$
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:
 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!
Once you have three eigenvectors, to "orthogonally diagonalize" A, you need three orthonormal eigenvectors. It is easy to see that (-2, 2, 1) is orthogonal to both (1, 0, 2) and (0, 1, -2) so you can just divide (-2, 2, 1) by its length, 3. (1, 0, 2) has length $\sqrt{5}$ so you can divide by that. To find a vector in the same space as (1, 0, 2) and (0, 1, 2) that is perpendicular to (1, 0, 2), find the projection of (0, 1, 2) on (1, 0, 2) and then subtract that from (0, 1, 2). And, of course, divide by its length.

 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
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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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