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-4+2y=0 and y=2x comes from Eigenvectors

  1. Apr 1, 2009 #1
    I don't understand where -4+2y=0 and y=2x comes from
    Is it obtained after finding the determinant or are the equations reconstructed that the matrix was created from?
    http://users.on.net/~rohanlal/eigen1.jpg [Broken]

    I also don't understand what's happening here. Where did the x(1,2) come from and how is e1=(1,2)
    derived from this?
    http://users.on.net/~rohanlal/eigen2.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 1, 2009 #2

    lanedance

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    Re: eigenvectors

    hi Ry22

    if A is you matrix first solve the characteristic equation given by
    [tex]det(\textbf{A} - \lambda.\textbf{I}) [/tex]
    this give you the eigenvalues

    Then to solve for the eignevectors corresponding to the ith eginevalue lambda, solve
    [tex]\textbf{A}.\textbf{x} = \lambda_i.\textbf{x} [/tex]

    it is not clear whether your original matrix is the one given, of which one of the eigenvalues is zero, or this is actually the matrix B related to an eigenvalue given by
    [tex]\textbf{B}.\textbf{x} = (\textbf{A}- \lambda_i.\textbf{I}).\textbf{x}= \textbf{0}[/tex]

    probably the 2nd case where the eigenvalue is known and included in the matrix - do you have any more info?
     
  4. Apr 2, 2009 #3
    Re: eigenvectors

    the matrix does include the eigenvalues and i understand how they're obtained. im just not sure about how the eigenvectors are obtained
     
  5. Apr 2, 2009 #4

    lanedance

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    Re: eigenvectors

    defintion of eignevectors for eignevalue lambda i, are that they mapped into a scalar multiple (= lamda i) of themselves.
    [tex]\textbf{A}.\textbf{x} = \lambda_i.\textbf{x} [/tex]

    re-writing the equation gives:
    [tex]\textbf{B}.\textbf{x} = (\textbf{A}- \lambda_i.\textbf{I}).\textbf{x}= \textbf{0}[/tex]

    In your 2d case x = (x,y) where x and y are unknown, so an equation from B.x to get x in terms of y. (note both equations give you the same relation)

    note the eigenvector is not unique, if u is an eigenvector, then so is k.u for any scalar k. So you only determine the eigenvector upto a constant.

    does this answer your question?
     
    Last edited: Apr 2, 2009
  6. Apr 2, 2009 #5

    HallsofIvy

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    Re: eigenvectors

    Do the matrix multiplication:
    [tex]\begin{bmatrix}-4 & 2 \\ -2 & 1\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}-4x+ 2y \\-2x+ y\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}[/tex]
    and, since two matrices are equal if and only if corresponding terms are equal, that gives -4x+ 2y= 0, -2x+ y= 0. Both of those equations reduce to the same thing, y= 2x. Now, by scalar multiplication of a matrix, it is true that
    [tex]x\begin{bmatrix}1 \\ 2\end{bmatrix}= \begin{bmatrix}x \\ 2x\end{bmatrix}[/tex]

    Actualy, we could have just chosen x= 1, or any other non-zero value, to get an eigen vector.
     
  7. Jun 21, 2009 #6
    Re: eigenvectors

    where did the x come from in the matrix on the RHS?
    [tex]
    x\begin{bmatrix}1 \\ 2\end{bmatrix}= \begin{bmatrix}x \\ 2x\end{bmatrix}
    [/tex]
     
  8. Jun 21, 2009 #7

    HallsofIvy

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    Re: eigenvectors

    From the "x" on the LEFT side, of course! I just multiplied the x into the vector.
     
  9. Jun 21, 2009 #8
    Re: eigenvectors

    well where did the 1 come from for the vector?
     
  10. Jun 21, 2009 #9

    lanedance

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    Re: eigenvectors

    which 1?

    the key to understanding this I think is to recognise each eignvalue has in fact a subspace associated with it & the eigenvectors for a given eigenvalue is a set of vectors that spans that subspace.

    In this case the subspace is a line, so you can pick any vector that spans the line given by -4x + 2y = 0, or equivalently y=2x. This is what I meant by determining the vector upto a constant.


    Hey Halls, on a seprate issue, any new latex seems to display as black for me & I can;t read it any more - any ideas on how I can fix it?
     
  11. Jun 22, 2009 #10
    Re: eigenvectors

    so i have y=2x
    and a matrix is derived from it.
    it consists of x and 2x. i can understand where 2x came from because y=2x.
    but why do we make the other element of this matrix x?
     
  12. Jun 22, 2009 #11

    lanedance

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    Re: eigenvectors

    I think Halls was using an easy way to show the line y = 2x is parallel to the vector [1,2] as follows

    i can't read tex at the moment so here goes, these should be column vectors to match above
    [x,y]
    =[x,2x] from the equation of the line through the origin y=2x
    =x[1,2]
     
  13. Jun 22, 2009 #12
    Re: eigenvectors

    It's simple matrix multiplication, a scalar times a matrix is just the product of each term in the matrix and the scalar (in this case x).
     
  14. Jun 22, 2009 #13

    Fredrik

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    Re: eigenvectors

    We don't. It's been x the whole time. You started by calling the first component x and the second y. Then you found out that they satisfy the relationship y=2x. That means that you can write your two components as x and 2x. If you prefer, you can write them as y/2 and y instead.
     
  15. Jun 22, 2009 #14

    HallsofIvy

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    Re: eigenvectors

    You are asking the right person because the same thing happened to me! I asked the administrators and they said they had put in an update of the LaTex and suggested that I was using an old version of "Internet explorer". It turned out I was (I was using version 6 and the latest is version 8). I downloaded the new version and now the LaTex works fine.
     
  16. Jun 22, 2009 #15

    lanedance

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    Re: eigenvectors

    thanks - will try the update
     
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