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Eigen values and Eigenvectors for a special case of a symmetric matrix

  1. Apr 25, 2013 #1
    Hey guys if i have a vector x=[x1,x2, ... xn]
    what are the eigenvectors and eigenvalues of X^T*X ?
    I know that i get a n by n symmetric matrix with it's diagonal entries in
    the form of Ʃ xii^2 for i=1,2,3,. . . ,n

    Thank you in advance once again!
     
  2. jcsd
  3. Apr 25, 2013 #2

    Office_Shredder

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    You can calculate them fairly directly. Let [itex] A = x^t x [/itex] Then for a vector [itex] v = (v_1,...,v_n) [/itex]
    [tex] Av^t = (x^t x) v^t = x^t (x v^t) = x^t \left( x\cdot v \right) [/tex]
    where [itex] x\cdot v [/itex] is the dot product of x and v. Based on this you should be able to spot that there can only be one eigenvector with a nonzero eigenvalue (and what the eigenvectors and eigenvalues are)
     
  4. Apr 25, 2013 #3
    I ll try it . thank you very much!
     
  5. Apr 26, 2013 #4
    I tried it but i get a nx1 in size matrix. Aren't eigenvalues and eigenvectors for nxn matrices . . . ?
    How can I get the values ?
    Thank you .
     
  6. Apr 26, 2013 #5

    HallsofIvy

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    Yes, Office Shredder said that would give the eigenvectors, not a matrix.
     
  7. Apr 30, 2013 #6
    Sorry but i still don't get it. What is the eigenvector in the second part of the equation? Could you provide the solution beyond that first step! Thank you !
     
  8. Apr 30, 2013 #7
  9. May 1, 2013 #8
    Solved

    Solved !!! lamda=x*x' and the corresponding eigenvector is x
     
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