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Eigenvectors and eigenspaces

  1. Sep 8, 2011 #1
    1. The problem statement, all variables and given/known data

    Let T: M22 - M22 be defined by

    T(a b = 2c a + c
    c d) b-2c d

    (a) Find the eigenvalues T.
    (b) Find bases for the eigenspaces of T.

    2. Relevant equations

    det(lambdaI - A) = 0

    3. The attempt at a solution

    I just wanted to know how to start here. I am used to using vector transformations...

    Would I use identity matrices 1 0 0 1 0 0 0 0
    0 0 0 0 1 0 0 1

    And then form a matrix somehow? I just don't know how big or how this matrix would be formed...?
     
  2. jcsd
  3. Sep 8, 2011 #2

    micromass

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    One small thing: you shouldn't call the matrices

    [tex]\left(\begin{array}{cc} 1 & 0\\ 0 & 0\end{array}\right), \left(\begin{array}{cc} 0 & 1\\ 0 & 0\end{array}\right), \left(\begin{array}{cc} 0 & 0\\ 1 & 0\end{array}\right), \left(\begin{array}{cc} 0 & 0\\ 0 & 1\end{array}\right)[/tex]

    identity matrices. We reserve the term identity matrix for something else.
    You can call them elementary matrices however.

    Anyway, your approach is good: try to find the images of the elementary matrices to find the matrix form of T. T will have the form of a 4x4-matrix.
     
  4. Sep 8, 2011 #3
    Ok. So I find the matrix for T to be:

    0 1 0 0
    0 0 1 0
    2 1 0 0
    -2 0 0 1

    The characteristic equation of which I find to be:

    (lambda)4 - (lambda)3 - (lambda)2 - (lambda) + 2

    Just wondered if there was an easy way to solve this?
     
  5. Sep 8, 2011 #4

    micromass

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    Hmm, that's not what I get:

    [tex]T\left(\begin{array}{cc} 1 & 0\\ 0 & 0\end{array}\right)=\left(\begin{array}{cc} 0 & 1\\ 0 & 0\end{array}\right)[/tex]

    So the first column needs to be (0,1,0,0), no?
     
  6. Sep 8, 2011 #5
    O ok. So I must turn the resulting matrix into a column vector? I just put them together as matrices...my bad...ok I shall try this agen...shall get back to you to confirm my answer...thanks alot...

    Ciao...I feel mentally ill that I am such a dumb ***!...but then again...have to learn somehow...
     
  7. Sep 8, 2011 #6
    Ok I think I have solved it:

    The standard matrix for T is:

    0 0 2 0
    1 0 1 0
    0 1 -2 0
    0 0 0 1

    Whose characteristic equation is:

    (lambda - 1)2(lambda + 1)(lambda + 2) = 0

    This right?

    Thanks

    Derryck
     
  8. Sep 8, 2011 #7

    micromass

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    Yes, that's also what I've got!!
     
  9. Sep 9, 2011 #8
    Thanks alot micromass

    Tell you what if it wasn't for physicsforums I don't know how I would have passed my correspondence linear algebra course...

    Cheers

    Derryck
     
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