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Another reminder on finding eigenvectors

  1. May 3, 2012 #1
    Another question with respect to finding eigenvectors.

    If I remember correctly, I should be able to look at certain 2 by 2 matrices and practically write down the eigenvalues and eigenvectors.

    For example, I have a diagonal matrix, I know immediately what the eigenvalues and eigenvectors are.

    E.g. M = \begin{bmatrix}
    1 &0 \\[0.3em]
    0 & x \\[0.3em]

    \end{bmatrix}

    Well, I know immediately λ_1 =1, λ_2 = x and that the eigenvectors are e_1 = (1 0) and e_2 = (0 1).

    Now, what about an upper diagonal matrix?

    Take M = \begin{bmatrix}
    -1 & -1 \\[0.3em]
    0 & x-(1/4) \\[0.3em]

    \end{bmatrix}

    I can see λ_1=1 and that e_1 = (1 0)

    How do you find the second eigenvalue and eigenvector?
     
  2. jcsd
  3. May 3, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    In a "triangular matrix", just as in a diagonal matrix, the numbers on the diagonal are the eigenvalues. In your example they are -1, NOT 1, and x- (1/4).

    More generally, if your matrix is
    [tex]\begin{bmatrix} a & 0 \\ -1 & b\end{bmatrix}[/tex]
    then a and b are the eigenvalues.

    The eigenvector corresponding to eigenvalue a must satisfy
    [tex]\begin{bmatrix} a & -1 \\ 0 & b\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}ax & ay\end{bmatrix}[/tex]
    so that we have equations ax- y= ax and by= ay. The first equation, on subtracting ax from both sides, gives -y= 0 so y= 0. The second equation, which is equivalent to (b- a)y= 0 also gives y= 0. An eigenvector corresponding to eigenvalue a is <1, 0> as you say.

    The eigenvector corresponding to eigenvalue b must satisfy
    [tex]\begin{bmatrix} a & 1 \\ 0 & b\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}bx \\ by\end{bmatrix}[/tex]
    which gives the equations ax- y= bx and by= by. The first equation is the same as y= (b- a)x while the second equation is satisfied by any y. Any eigenvector is of the form <x, y>= <x, (b-a)x>= x<1, b-a>.

    In your case, with a= 1 and b= x- 1/4, 1 is an eigenvalue with corresponding eigenvector <1, 0> (or any multiple) and x- 1/4 is an eigenvalue with corresponding eigenvector <1, x- 5/4>.
     
  4. May 3, 2012 #3
    This makes sense, but according to what I have here, the eigenvector should be
    λ_2 = <-1 x+ (3/4)>

    This is assuming x-(1/4) > 0. Would that make a difference or is what I have a typo?
     
  5. May 3, 2012 #4
    i.e. b-a is 1-(-1/4)=3/4, right?
     
  6. May 4, 2012 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, 1- (-1/4)= 1+ 1/4 = 5/4, not 3/4.
     
  7. May 4, 2012 #6
    Typo on my part.

    we get for the "b value" an equation -v_1-v_2 = (x-1/4)*v_1

    This is -v_2 =(x+3/4)*v_1
     
  8. May 4, 2012 #7
    That is, using your equation ax- y= bx

    a=-1, b=(x-1/4)
     
  9. Jul 20, 2012 #8
    perhaps you should credit the author for the quote at the bottom of your post -- Edna St. Vincent Millay
     
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