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Math Help - Eigenvalues

  1. #1
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    Eigenvalues

    I need to find the characteristic polynomial, eigenvalues, and eigenvectors or the following matrix.

    [1 1]
    [1 1]
    So far, this is what I have done:
    det(Lamda-A) And got

    [lamda - 1 -1]
    [-1 lamda - 1]

    The det of that gives (lamda - 1)(lamda-1) + 1 which is lamda squared - 2 lamda +2

    I used the quad formula on that and got 1 plus/minus i

    I dont know where to go from here. I think Now i do this:
    A-(1+i)I =

    [i 1]
    [1 i]

    i have no clue where to go from there

    Please help. THanks
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  2. #2
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    Israel
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    Hi!

    There really is no need to calculate the characteristic polynomial that way.

    First, you can easily observe that 0 is an eigenvalue of geometric multiplicty 1 (since rank(A) = 1. Now, note that trace(A) = 1+1 = 2 and A only has two eigenvalues, so the second one must be 2. Another way to know that 2 is an eigenvalue is to note that the sum of all rows is 2, therefore, according to a theorem which you should have learned, 2 is an eigenvalue with corresponding eigenvector (1 \  1)^T.

    So we know that 2 is an eigenvalue of geometric multiplicity \geq 1 and 0 is an eigenvector with geometric multiplicity 1. Since algebraic multiplicity \geq geometric multiplicity for each eigenvalue, we get that the algebraic and geometric multiplicities for each eigenvector are 1 and 1.

    Therefore: \Delta_A(x) = x(x-2)
    With 0,2 as eigenvalues
    and (1\ -1)^T, (1 \ 1)^T the corresponding eigenvectors.

    However, if you still want to know what you did wrong --

    det(\lambda I-A) = (\lambda-1)(\lambda-1) - (-1)(-1) = (\lambda-1)^2 -1 \neq (\lambda-1)^2+1

    =\Rightarrow det(\lambda I-A) = \lambda^2 -2\lambda +1 -1 = \lambda^2 -2\lambda = \lambda(\lambda-2)
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  3. #3
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    Thanks

    Thanks a bunch. It would be nice if i could add :P
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  4. #4
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    So I was trying to do it the way i was. I got the (1,1), but instead of (1,-1) I got (-1,1).

    0lamda - A =
    -1 -1
    -1 -1

    then rref that and got
    1 1
    0 0

    x1= -r
    x2 = r

    -1, 1
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  5. #5
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    Israel
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    That is fine. The eigenvector can be any non-zero scalar multiple of (1 \ -1)^T. In this case, (-1)(1 \ -1)^T = (-1 \ 1)^T.

    From this, (2\ -2)^T, \ (-8\ 8)^T,... could all be eigenvectors as well.
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