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Eigenvalue of Sum of Eigenvectors

  1. Mar 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Skjermbilde_2012_03_10_kl_10_33_27_AM.png

    3. The attempt at a solution
    So, first I wrote,

    [itex]T(X) = λ_1 X, T(Y) = λ_2 Y[/itex]

    If [itex]λ_1 = λ_2[/itex]:

    [itex]T(X+Y) = T(X) + T(Y) = λ_1 X + λ_2 Y = λ_1 (X+Y)[/itex],

    so this does indeed seem to be an eigenvector. But I'm less convinced for the case [itex]λ_1 ≠ λ_2[/itex]. Again, I get the transformation down to the form:

    [itex]λ_1 X + λ_2 Y[/itex]

    But if eigenvalues are necessarily one constant, then I don't see how I pull constants out as above. How do I go about showing this?
     
  2. jcsd
  3. Mar 10, 2012 #2

    HallsofIvy

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    Did you read the problem carefully? It doesn't ask you to show that this is true, it asks you to determine whether or not it is true and the show that.

    Okay, if you have doubts about this being true, how about trying to show its not true? If it is not, you should be able to show a counterexample.
    Obviously
    [tex]\begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}[/tex]
    has 1 and 2 as eigenvalues. What are corresponding eigenvalues? What do you get if you multiply that matrix by their sum?
     
  4. Mar 10, 2012 #3

    Mark44

    Staff: Mentor

    I'm pretty sure HallsOfIvy meant "corresponding eigenvectors".
     
  5. Mar 10, 2012 #4

    HallsofIvy

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    Yes, thanks.
     
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