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Eigenvalues homework

  1. Apr 28, 2006 #1
    matrix [tex]A = \left(\begin{array}{ccc}3&0&0 \\ 0&3&0 \\3&0&0 \end{array}\right)[/tex]

    has two real eigenvalues lambda_1=3 of multiplicity 2, and lambda_2=0 of multiplicity 1. find the eigenspace.

    [tex]A = \left(\begin{array}{ccc}3-3 &0&0 \\ 0&3-3&0 \\3&0&0-3 \end{array}\right)[/tex]

    [tex]A = \left(\begin{array}{ccc}0 &0&0 \\ 0&0&0 \\3&0&-3 \end{array}\right)[/tex]

    [tex]A = \left(\begin{array}{ccc}1&0&-1 \\ 0&0&0 \\0&0&0 \end{array}\right)[/tex]

    that means the eigenspace for lambda_1 is the column vector [0 1 0] and [1 0 1]

    for lambda_2, there will be two different eigenspaces such that...

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

    I know the answer has two vectors, how do i get two vectors when I have 4 that defines the eigenspace?
     
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  3. Apr 28, 2006 #2

    Gokul43201

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    Looks correct up to here ...
    How do you get that ?
     
  4. Apr 28, 2006 #3

    J77

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    Is your eigenvector for [tex]\lambda_3[/tex] not [tex]v_3=(0, 0, k)^T,\;k\in\mathbb{R}[/tex], and for [tex]\lambda_{1,2}[/tex] not [tex]v_{1,2}=(k_1,k_2,k_1)^T,\;k_{1,2}\in\matbb{R}[/tex]?
     
  5. Apr 28, 2006 #4

    HallsofIvy

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    If
    [tex]A = \left(\begin{array}{ccc}1&0&0 \\ 0&1&0 \\0&0&0 \end{array}\right)\left(\begin{array}{c}x \\y\\ z\end{array}\right)= \left(\begin{array}{c}0 \\ 0 \\0 \end{array}\right)[/tex]
    Which says x= 0, y= 0. Looks to me like you have just a one dimensional eigen space, spanned by (0, 0, 1)T.
     
  6. Apr 28, 2006 #5
    oh right, so three vectors then... [0, 0, 1], [0 1 0] and [1 0 1]

    the solution gives 2 vectors, how would I find the eigenspace from 3 vectors?
     
  7. Apr 28, 2006 #6

    HallsofIvy

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    WHAT solution gives 2 vectors and what eigenspace are you talking about?

    Not all texts use that terminology but I think by "eigenspace" you mean the sub-space consisting of eigenvectors haveing the same eigenvalue. In this case, you have two distinct "eigenspaces", one having dimension 2 and the other dimension 1.
     
    Last edited: Apr 28, 2006
  8. Apr 28, 2006 #7

    J77

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    Depending on the definition, I think you could also use the general eigenvectors that I gave and form your basis by sticking in ones and zeroes.
     
  9. Apr 28, 2006 #8

    yeah right.

    If you start with a three dimensional square hermitian matrix, you'll get a 3D eigenspace. period. If you have degenerate eigenvalues - like you do, then you get one unambiguous eigenvector (for the eigenvalue 0) and a 2D eigenspace (corresponding to the degenerate eigenvalue 1) from which you have the freedom of choosing any two linearly independent eigenvectors.
     
  10. Apr 28, 2006 #9

    HallsofIvy

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    That's not the question I was responding to.
    (And, the matrix we are talking about is not Hermitian.)
     
    Last edited: Apr 28, 2006
  11. Apr 28, 2006 #10
    so... there are three vectors the span the space... this is because:

    [tex]\lambda_1=3[/tex]

    [tex]A = \left(\begin{array}{ccc}1&0&-1 \\ 0&0&0 \\0&0&0 \end{array}\right)[/tex]

    [tex] \left(\begin{array}{c}a \\b \\c \end{array}\right) = s \left(\begin{array}{c}0 \\1 \\0 \end{array}\right) + t \left(\begin{array}{c}1 \\0 \\1 \end{array}\right) [/tex]

    [tex]\lambda_1=0[/tex]

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

    [tex] \left(\begin{array}{c}a \\b \\c \end{array}\right) = u \left(\begin{array}{c}0 \\0 \\1 \end{array}\right) [/tex]

    so, there are three linearly independent vetors that span the eigenspace

    they are: [tex] \left(\begin{array}{c}0 \\0 \\1 \end{array}\right) , \left(\begin{array}{c}0 \\1 \\0 \end{array}\right) , \left(\begin{array}{c}1 \\0 \\1 \end{array}\right)[/tex]

    however, my solution is telling me that I need two vectors instead of three such that:
    [tex] \left(\begin{array}{c}a \\b \\c \end{array}\right) , \left(\begin{array}{c}d \\e \\f \end{array}\right) [/tex]
     
    Last edited: Apr 28, 2006
  12. Apr 28, 2006 #11

    HallsofIvy

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    Then what was the question? Yes, the three vectors span the space- it is, after all, 3 dimensional.

    "my solution is telling me that I need two vectors". Your solution? What solution are you talking about? Your matrix has two distinct eigenvalues, 3 and 0. The eigenspace corresponding to 3 is two dimensional and is spanned by
    [tex] \left(\begin{array}{c}1 \\0 \\1 \end{array}\right)[/tex]
    and
    [tex]\left(\begin{array}{c}0 \\1 \\0 \end{array}\right) [/tex]
    The eigenspace corresponding to 0 is one dimensional and is spanned by
    [tex]\left(\begin{array}{c}0 \\0 \\1 \end{array}\right)[/tex]

    Those three vectors, of course, span all of R3.
     
  13. Apr 28, 2006 #12
    oh whoops, I was to find the eigenspace for only [tex]\lambda_1=3[/tex]

    so I didnt have to take into account [tex]\lambda_2[/tex]
     
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