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B Elementary eigenvector question

  1. Mar 27, 2017 #1
    Hi
    If I have this matrix:
    \begin{array}{cc}0&1\\1&0\end{array}
    and I want to find its eigenvectors and eigenvalues, I can try it using the definition of an eigenvector which is:

    A x = λ x , where x are the eigenvectors

    But if I try this directly I fail to get the right answer, for example using a column eigenvector (a b) , instead I get:

    (b a) = λ (a b) , (These are column vectors.)

    THere is no lambda able to make this correct, unless it is zero which is not the right answer. Why is that this approach didn't work?

    I have to use the identity matrix, and the determinant of A - λ I, to get the right result.

    Thanks!
     
  2. jcsd
  3. Mar 27, 2017 #2

    DrClaude

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    The part I bolded is false. The set of two equations of two unknowns is not hard to solve. Try it!
     
  4. Mar 27, 2017 #3

    BvU

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    You can check that this must be wrong by comparing it with the outcome of
    where you find ##\lambda^2-1 = 0 \Rightarrow \lambda = \pm 1## and can then determine the corresponding eigenvectors from (b a) = λ (a b) !
     
  5. Mar 27, 2017 #4
    Ups, ok, you are right. I get the same results, and then for the eigenvectors I get:

    Lambda=1 ---> (a a) , I simply get a=b, so the eigenvector is (1 1)

    Lambda=-1---> (a -a), I get a=-b, so the eigenvector is (1 -1)
     
  6. Mar 27, 2017 #5

    BvU

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    Well done! (apart from normalization) you have found the eigenvectors of the first of the Pauli matrices; they play an important role in quantum mechanics for particles with spin.
     
  7. Mar 27, 2017 #6
    Yes, that is what I was reading about.

    Thanks.
     
  8. Mar 30, 2017 #7

    mathwonk

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    you could just picture the map, since from the matrix columns, it is obvious it interchanges the standard unit vectors on the x and y axes. Hence it is a reflection of the plane in the line x=y, hence it leaves invariant both that line and its orthocomplement, acting as the identity on the line x=y and as - the identity on the line x=-y. but the algebraic method is more sure, if less geometric.
     
  9. Mar 30, 2017 #8

    WWGD

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    This is a good idea for transformations but it seems harder for other situations. Can you (or anyone else, of course) see some geometric interpretations for e.g. a correlation or covariance matrix?
     
  10. Apr 4, 2017 #9

    WWGD

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    Sorry, I don't mean to hijack the thread, it is just that I am curious about a related issue: the interpretation of eigenvalues in correlation/covariance matrices. These supposedly describe directions of maximal variability of the data, but I just cannot see it at this point. I thought since the OP seems satisfied with the answers given, it may make sense to extend the thread beyond the original scope.
     
  11. Apr 5, 2017 #10

    StoneTemplePython

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    I'm not totally sure I understand your question as they are a lot of interpretations here. In all cases I assume we're working with centered (read: zero mean, by column) data. I also assume we're operating in reals.

    If you have your data in a matrix ##\mathbf A##, and you have some arbitrary vector ##\mathbf x##, where ##\big \vert \big \vert \mathbf x \big \vert \big \vert_2^{2} = 1##, then to maximize ##\big \vert \big \vert \mathbf{Ax} \big \vert \big \vert_2^{2}##, you'd allocate entirely to ##\lambda_1##, ##\mathbf v_1## the largest eigenpair in ##\mathbf A^T \mathbf A## (aka largest singular value (squared) of ##\mathbf A## and the associated right singular vector). This is a quadratic form interpretation of your answer. Typically people prove this with a diagonalization argument or a Lagrange multiplier argument. I assume the eigenvalues are well ordered for this symmetric positive (semi)definite covariance matrix, so ##\lambda_1 \geq \lambda_2 \geq ... \lambda_n \geq 0##. Where
    ##\mathbf A =
    \bigg[\begin{array}{c|c|c|c}
    \mathbf a_1 & \mathbf a_2 &\cdots & \mathbf a_{n}
    \end{array}\bigg]
    ##
    That is, ##\mathbf a_j## refers to the jth feature column in ##\mathbf A##

    using the interpretation of matrix vector multiplication as the scaled sum across the column space of a matrix, we see that ##\mathbf {Ax} = x_1 *\mathbf a_1 + x_2 *\mathbf a_2 + ... + x_n *\mathbf a_n## .

    Thus when someone asks to do a constrained maximization of ##\big \vert \big \vert \mathbf{Ax} \big \vert \big \vert_2^{2}##, what they are saying is come up with the vector that is a linear combination of features from data matrix ##\mathbf A## that has the maximal length, subject to the constraint that ##x_1^2 + x_2^2 + ... + x_n^2 = 1## (or some other constant > 0, but we use one for simplicity here). Since all features are zero mean (i.e. you centered your data), what you have done is extract a vector with the highest second moment / variance from your features -- again subject to the constraint ##x_1^2 + x_2^2 + ... + x_n^2 = 1##.

    here is another interpretation

    If you wanted to low rank approximate -- say rank one -- your matrix ##\mathbf A##, and you were using ##\big \vert \big \vert \mathbf A \big \vert \big \vert_F^{2}## as your ruler (i.e. sum up the squared value of everything in ##\mathbf A## which is a generalization of a L2 norm on vectors), you'd also allocate entirely to ##\lambda_1##, where ##\big \vert \big \vert \mathbf A \big \vert \big \vert_F^{2} = trace\big(\mathbf A^T \mathbf A\big) = \lambda_1 + \lambda_2 + ... + \lambda_n##, where each associated eigenvector is mutually orthonormal, so we have a clean partition, and for each eigenvalue your allocate to, you increase the rank of your approximation, thus for rank 2 approximation you'd allocate to ##\lambda_1## and ##\lambda_2## and so forth.
     
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