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I Eigenvectors for degenerate eigenvalues

  1. Oct 24, 2016 #1

    dyn

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    I am looking at some notes on Linear algebra written for maths students mainly to improve my Quantum Mechanics. I came across the following example - $$ \begin{pmatrix} 2 & -3 & 1 \\ 1 & -2 & 1 \\ 1 & -3 & 2 \end{pmatrix} $$
    The example then gives the eigenvalues as 0 and 1(doubly degenerate). It then calculates the eigenvectors using Gaussian elimination. This is where my problem arises - coming from a physics background I tried to find the eigenvectors for the repeated eigenvalue 1 using back substitution but it doesn't seem to produce a solution this way. Am I doing something wrong or is it possible for back substitution not to work while Gaussian elimination works ?
    The answer given for the eigenvector is a linear combination of the 2 vectors ( 3 1 0 )T and (-1 0 1)T. In the Quantum Mechanics textbook I am using it says for degenerate eigenvalues to choose 2 mutually orthogonal vectors. The 2 vectors I have listed are not orthogonal. Is the orthogonal part just a preference for QM and not a requirement ?
    Thanks
     
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  3. Oct 24, 2016 #2

    andrewkirk

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    You can turn them into an orthogonal pair by subtracting from one the projection of the other onto it.

    Given two linearly independent vectors ##\vec u,\vec v##, the pair ##\vec u-\frac{\vec u\cdot \vec v}{\vec v\cdot\vec v}\vec v, \vec v## is orthogonal. You can check that by calculating ##(\vec u-\frac{\vec u\cdot \vec v}{\vec v\cdot\vec v}\vec v)\cdot \vec v##
     
  4. Oct 24, 2016 #3

    dyn

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    So choosing the eigenvectors as orthogonal is just a matter of preference. Thanks. Any thoughts on why I can't calculate the eigenvectors by back substitution but it can be done by Gaussian elimination ?
     
  5. Oct 24, 2016 #4

    dyn

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    If I apply a general vector ( a b c )T to the eigenvalue equation with eigenvalue 1 , I end up with 3 equations exactly the same a-3b+c=0. How do I then proceed to end up with the answer given which is equivalent to ( 3x-y , x , y )T
     
  6. Oct 24, 2016 #5

    andrewkirk

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    The equation 'a-3b+c=0' can be written as 'a=3b-c' which just says that for any an eigenvector with Eigenvalue 1, whose2nd and 3rd components are b,c, the first component is 3b-c.

    Relabel a,b,c as x,y,z and you have the given answer.
     
  7. Oct 25, 2016 #6

    Mark44

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    Elaborating on what andrewkirk said, relabel the equation above as x - 3y + z = 0.

    Then
    x = 3y - z
    y = y
    z = .... z
    If you look at the right sides as a sum of two vectors, you get
    ##\begin{bmatrix} x \\ y \\ z \end{bmatrix} = y\begin{bmatrix} 3 \\ 1 \\ 0 \end{bmatrix} + z\begin{bmatrix} -1 \\ 0 \\ 1\end{bmatrix}##

    Here y and z on the right side can be considered arbitrary constants.
     
  8. Oct 25, 2016 #7

    dyn

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    Thanks for your replies. So essentially because I end up with 3 equations that are the same I really have just one equation with 3 unknowns. So I take 2 of those unknowns to have arbitrary values and the express the remaining unknown in terms of the 2 arbitrary values
     
  9. Oct 25, 2016 #8

    Mark44

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    Yes. In the work I showed, you can take y = 1 and z = 0, and get one solution, and you can take y = 0, z = 1, to get another solution. Since y and z are completely arbitrary, you get a double infinity of solutions.

    Geometrically, the two vectors I showed determine a plane in R3. Every point in this plane is some linear combination of those two vectors.
     
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