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Finding a matrix P that orthogonally diagonalizes

  1. Oct 16, 2009 #1
    1. The problem statement, all variables and given/known data
    Find a matrix P that orthogonally diagonalizes I - vvT if

    v = (1, 0, 1)


    2. Relevant equations
    Well, solving I - vvT will give me my A if I am correct, and the characteristic equation for A is

    det([tex]\lambda[/tex]I - A) = [tex]\lambda[/tex]3 - 2[tex]\lambda[/tex]2 = 0

    Solving this gives me [tex]\lambda[/tex]1 = 0 and [tex]\lambda[/tex]2 = 2

    Now when I substitute these values back into det([tex]\lambda[/tex]I - A) I only get 2 basis vectors for my P. (1, 0, 1) and (-1, 0, 1). Since I dont have 3 vectors I'm not going to yet normalize them, which has to be done before they become the column vectors of P.

    Any help for finding this P would be great

    EDIT : I guess technically I also have a 3rd basis vector (0, 0, 0) with [tex]\lambda[/tex]=0 am I able to use the zero vector?
     
    Last edited: Oct 16, 2009
  2. jcsd
  3. Oct 16, 2009 #2

    lanedance

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    are you sure about you determinant eqaution? maybe show how you got there... and your matrix

    hint: if i've done it right & i multiply your first eignevector by the matrix A, I get 1x the eigenvector...
     
    Last edited: Oct 16, 2009
  4. Oct 16, 2009 #3

    lanedance

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    also i notice if
    A = I - vvT

    and u is an eignevector of A, with value m, then you should be able to show that u is also an eignevector of vvT with related eigenvalues, and it should be reasonably easy to guess the eigenvectors of vvT ro maybe even A
     
    Last edited: Oct 16, 2009
  5. Oct 16, 2009 #4

    lanedance

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    and one more...

    to clarify in this case, 0 is an eignvalue of vvT but not A

    When 0 is an eigenvector, it means the matrix is singular, with the eigenvectors corresponding to the 0 eigenvalue, belonging to the null space of the matrix

    note that the zero vector, 0 is never included in the solutions to eigenvalue equation
    Au = mu
    as it it always a solution reagrdless of eigenvalue (considered the trvial solution).

    So the zero eigenvalue will have corresponding non-zero eigenvector/s.
     
    Last edited: Oct 16, 2009
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