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Rank of Matrices and Eigen Vectors

  1. Nov 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the rank off matrices?

    i)A=[2 0 9 2; 1 4 6 0; 3 5 7 1 ] 3X4

    ii)A=[3 1 4; 0 5 8; -3 4 4; 1 2 4;] 4X3

    Find Eigen Vectors and Values of A;

    A = [3 2 4; 2 0 2; 4 2 3 ]


    2. Relevant equations

    -when det(A) is not equal to zero it will the rank of matrices.
    - (A-λ)X' = 0

    3. The attempt at a solution
    I have calculate all minors ;

    det(a) = [ 2 0 9; 1 4 6; 3 5 7] , [2 0 2; 1 4 0; 3 7 1] , [0 9 2; 4 6 0; 5 7 1]
    And I found different numbers from zero.Does it mean that rank is 3?

    My original question about Eigen Vectors and Values is to understand what is it? And where can we use it?
     
  2. jcsd
  3. Nov 30, 2012 #2

    haruspex

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    I would think that if any of the minors is nonzero the rank must be 3.
    An Eigenvector, as you can see from the equation, is one which is not rotated at all by the matrix. Instead, the vector is merely expanded/shrunk linearly (but not to zero). These turn out to have many uses in getting to grips with the core features of the transformation.
     
  4. Nov 30, 2012 #3
    Ok,thank you.
    a is true.Tested with matlab.
    For b) all det. is nonzero? so how can I write on 2X2 det form?
     
  5. Nov 30, 2012 #4

    vela

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    An eigenvalue can equal 0 so that Ax=0x. The eigenvector, however, can not be 0.
     
  6. Dec 3, 2012 #5
    A = [-1 1 0; 1 -1 0; 0 0 0] what is the eigen vectors of this matrices?

    I found eigenvalues as 0 and -2.

    Eigen vector of 0?
     
  7. Dec 3, 2012 #6

    haruspex

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    As vela rightly said, an eigenvalue can be 0 but an eigenvector cannot be.
    For an eigenvalue of 0, the eigenvectors will be in the null space of the transformation.
     
  8. Dec 3, 2012 #7
    Ok but how we can find eigenvector when eigenvalue is zero?

    Also there's another problem..

    A=[7 2; 0 1 ] this matrice eigenvalues is not zero.Eigenvalues are 7 and 1.

    (A-λI)*X=0

    [7 2 ; 0 1] - 7 * [ 1 0; 01 ] = [ 0 2; 0 -6} --> [0;0;0]

    If X1 = [ x1;y1]

    [0 2; 0 -6] [x1;y1] = [0;0]

    2y1 = 0
    -6y1 = 0

    x1 = 0? or not? If = 0 where to go from here?
     
  9. Dec 3, 2012 #8

    haruspex

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    The same way as when it is nonzero: you know A and λ, so solve (A-λI)x = 0.
    This is telling you x1 can be anything nonzero. Remember that although we speak of eigenvectors they're really eigenspaces. The dimension of the eigenspace is the number of times the eigenvalue repeats in the roots of the polynomial. Here, neither root repeats so each eigenspace has only one dimension. You can pick any nonzero vector in that space as a representative eigenvector, but any nonzero scalar multiple of it would do as well.
     
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