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Find the symmetric matrix from eigen vectors

  1. Apr 24, 2013 #1
    Hello to all of you,

    Is there a way to get the matrix A=[a b c d] from the eigenvectors (orthogonal) matrix
    H= sin(x) cos(x)
    cos(x) -sin(x)
    or to pose it differently to find a matrix that has these 2 eigenvectors ?

    Thank you in advance .
    Michael
     
  2. jcsd
  3. Apr 24, 2013 #2

    mfb

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    Just write the eigenvectors as columns of your matrix and watch what A*(1,0)^T does.
     
  4. Apr 25, 2013 #3
    ^T is the transpose? and why [1,0]? thanks!
     
  5. Apr 25, 2013 #4
    I have made the assumption that matrix A should be symmetric because of the orthogonality of eigen vectors matrix! is this true? so matrix A = (a,b ;b, d) rows seperated by ;
     
  6. Apr 25, 2013 #5

    mfb

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    Oh sorry, ignore my first post.
    Do you have fixed eigenvalues for those eigenvectors? Otherwise, there is a lot of freedom: all multiplies of the identity matrix (including the identity matrix) have all vectors as eigenvectors, for example.
    With fixed eigenvalues, I get 4 equations for 4 unknown parameters, so I would expect that there is a unique solution.
     
  7. Apr 25, 2013 #6
    no there are not fixed eigen values! how do I proceed?
     
  8. Apr 25, 2013 #7

    HallsofIvy

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    If you don't know the eigenvalues, it is impossible to find the matrix. There are an infinite number of matrices having the same eigenvectors but different eigenvalues.
    If D is the diagonal matrix, having the eigenvalues on the main diagonal and A is the matrix having the corresponding eigenvectors of the matrix as columns then [tex]ADA^{-1}[/tex] is the matrix having those eigenvalues and eigenvectors.
     
  9. Apr 25, 2013 #8
    I started from the eq A*H.1= lamda1*H.1 where H.1 is the first column of H (the A matrix of user HallsofIvy) .
    Then I did the same for lamda2 (the second eigen value that is unknown) and I got lamda1=]a*sin(x)+b*cos(x)]/sin(x) and another value lamda1=[b*sin(x)+d*cos(x)]/cos(x).
    Also two values for lamda2.
    Should I try to get 4 equations for the four pairs of lamda1 and lamda2 doing A=H*L*H^-1????
    my main objective is to set a,b,d from A as functios of sin and cos and not set numbers I suppose. . .
     
  10. Apr 25, 2013 #9
    Solved

    Problem solved !!!
    Matrix A=[a ,b]
    [b, d]
    From A*H.1=lamda1*H.1 I took
    lamda1=(a*sin(x)+b*cos(x))/sin(x)
    and lamda1=(b*sin(x)+d*cos)/cos(x)

    From A*H.2=lamda2*H.2 I took
    lamda2=(a*cos(x)-b*sin(x))/cos(x)
    and lamda2=(d*sin(x)-b*cos(x))/sin(x)

    For each set of lamda(i) i=1,2
    i do next [A-lamda(i)]*H.i=0
    All four sets of equations give me with free a ,d

    b=((a-d)*cos(x)*sin(x))/(sin(x)^2-cos(x)^2)

    It must be cos(x),sin(x) and sin(x)^2-cos(x)^2 not equal to zero

    Thus matrix A takes the form A= a b=((a-d)*cos(x)*sin(x))/(sin(x)^2-cos(x)^2)

    b=((a-d)*cos(x)*sin(x))/(sin(x)^2-cos(x)^2) d

    (This is a square symmetric matrix )

    In statistical program R i wrote a small script so as to validate the results

    Code (Text):
    #test1

    x=(pi)
    a=7
    d=1
    b=((a-d)*cos(x)*sin(x))/(sin(x)^2-cos(x)^2)

    l1=(a*sin(x)+b*cos(x))/sin(x)
    l2=(a*cos(x)-b*sin(x))/cos(x)
     
    A=matrix(c(a,b,b,d),2,2)
    H=matrix(c(sin(x),cos(x),cos(x),-sin(x)),2,2)
    L=matrix(c(l1,0,0,l2),2,2)
    #it must be A*H=H*L where L is the diagonial matrix with lamda1 and lamda2 as l1 and l2 #respectively

    A%*%H
    H%*%L

    #here I get A from H*L*H^-1
    H%*%L%*%ginv(H)
    #here I get A from H*L*H^T
    H%*%L%*%t(H)
    A
       
     
    Last edited: Apr 25, 2013
  11. Apr 25, 2013 #10
    I have one more but i ll post it in new thread !
    P.S I m new here so posting the solution is good , isn't it ???
     
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