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Spectral Decomposition

  1. Apr 6, 2012 #1
    [itex]
    \begin{bmatrix}
    1 & 2 \\
    2 & 1 \\
    \end{bmatrix}[/itex] = A

    The eigenvalues of A are 3 and -1. The eigenvectors are (1,1) and (-1,1), respectively. I'm not sure how to proceed.
     
  2. jcsd
  3. Apr 7, 2012 #2
    Anyone?
     
  4. Apr 7, 2012 #3
    What you wrote is the same as [tex]x^{2}-2x-3[/tex] try to use this probably might make it easier for you.
     
  5. Apr 7, 2012 #4
    I don't know how to show it possesses spectral decomposition.

    What do I orthogonally project onto the eigenspaces?
     
  6. Apr 7, 2012 #5
    Well I'd say you should draw out that parabola that i gave you and it should look easier.
     
  7. Apr 7, 2012 #6
    I don't see how the characteristic polynomial comes into play. Am I projecting the parabola onto the lines associated with the eigenspaces?
     
  8. Apr 7, 2012 #7
    1)What does it mean for something to have a spectral decomposition?
    Since you have your eigenvectors, how can you show that it has one?
    2)You should be able to find the projections easily enough
    3)I'm not quite sure what 'use the spectral theorem' means since there is no 'spectral theorem' on it's own. I assume it means just work out the decomposition and show that it agrees with A.
     
  9. Apr 7, 2012 #8
    The set = {λ1, λ2,..., λk} of eigenvalues of T is the spectrum of T.

    T = λ1T1 + λ2T2 + ... + λkTk

    Ti is the orthogonal projection of V on Wi.
     
  10. Apr 7, 2012 #9
    Yes, and what does that matrix T look like?
    (I should say, what does T look like in a certain basis ;))
     
  11. Apr 7, 2012 #10
    Is T a diagonal matrix with its eigenvalues along the diagonal?
     
  12. Apr 7, 2012 #11
    Yes, you just need to show that this is so
     
  13. Apr 7, 2012 #12
    So, just find an invertible matrix Q such that D = Q-1AQ?
     
  14. Apr 7, 2012 #13
    Yes
    Although I wouldn't say 'find', you've already got everything written down that you need to know in order to write Q out.
     
  15. Apr 7, 2012 #14
    Q is composed of the eigenvectors of A, right?
     
  16. Apr 7, 2012 #15
    Yes, do you understand WHY this is so?
    Do you know what Q is actually doing other than sitting there being an invertable matrix?
     
  17. Apr 7, 2012 #16
    Well, A is normal and so it is diagonalizable. What is Q doing?
     
  18. Apr 7, 2012 #17
    The Q is changing the basis you are working in.
    If [itex]a[/itex], [itex]b[/itex] are the eigenvectors of [itex]T[/itex] and [itex]\lambda_1[/itex], [itex]\lambda_2[/itex] the eigenvalues we have

    [itex]Ta = \lambda_1 a[/itex]
    [itex]Tb = \lambda_2 b[/itex]

    If we use the basis where the vector (1,0) ~ a and (0,1) ~ b then T is obviously diagonal. What the Q does is takes the vectors a and b and turns them into (1,0) and (0,1) respectively.

    If we have a as the first row and b as the second row in Q then [itex]Q.(1,0) = a[/itex] and [itex]Q.(0,1) = b[/itex]. So then [itex]Q^{-1}.a = (1,0)[/itex] and [itex]Q^{-1}.a = (0,1)[/itex]

    What your [itex]T=Q\ D\ Q^{-1}[/itex] decomposition is doing is when it acts on vector x is this;
    1. Change the basis of x from (1,0) and (0,1) to a and b
    2. Apply the diagonal matrix to a and b, a simple multiplication [itex]a \rightarrow \lambda_1 a[/itex], [itex]b \rightarrow \lambda_2 b[/itex]
    3. Change the basis back to (1,0) and (0,1) from a and b
     
  19. Apr 7, 2012 #18
    Are you calling my original matrix A matrix T?
     
  20. Apr 7, 2012 #19
    Yes, sorry for mixing notations :shy:

    But what I wrote is still true for any matrix which has a full set of eigenvalues and vectors
     
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