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Eigenvalues of Inverse Transformations

  1. Feb 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Screen_shot_2012_02_24_at_3_34_33_PM.png

    3. The attempt at a solution
    So I observed:

    T(B) = λB
    T-1(B) = λ'B

    Also,

    T-1(T(B)) = λ'λB = B

    This implies,

    λ'λ = 1

    And so, there should be a relation

    [itex]λ = \frac{1}{λ'}[/itex].

    Is that right?
     
  2. jcsd
  3. Feb 24, 2012 #2
    Looks OK to me. Note that this is quite obvious if T is diagonal, which can be achieved by a coordinate transformation such that the new coordinates are along the Eigenvectors.
     
  4. Feb 24, 2012 #3
    So this is a similar sort of problem, so I'll ask about it here:

    Screen_shot_2012_02_24_at_4_28_11_PM.png

    T(1,0,0) = (1,1,0)
    T(0,1,0) = (2,2,0)
    T(0,0,1) = (0,0,1)

    Thus, the matrix relative to the standard basis is:

    [itex]\left| \begin{array}{ccc}
    1 &2&0 \\
    1&2&0 \\
    0&0&1 \end{array} \right|[/itex]

    [itex]Δ_T (t) = det(\left| \begin{array}{ccc}
    1 &2&0 \\
    1&2&0 \\
    0&0&1 \end{array} \right| - tI) [/itex]

    I have this as equaling,

    [itex](1-t)[(1-t)(2-t)-2] = (1-t)(-3t+2t^2)[/itex]. First of all, I don't know if this is right. If it is right, I don't know how to interpret it.

    Of course, I know I also have to do the other parts of the problem, but I'll get there eventually :)
     
    Last edited: Feb 24, 2012
  5. Feb 24, 2012 #4

    Dick

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    Science Advisor
    Homework Helper

    You are computing the characteristic polynomial. Once you find that the roots are the eigenvalues. [itex](1-t)[(1-t)(2-t)-2][/itex] is right, the other side isn't. You want to factor it.
     
  6. Feb 24, 2012 #5

    kai_sikorski

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    Gold Member

    Your conclusion is correct but I'm not sure I agree with your proof. In particular this
    T-1(B) = λ'B
    is only true if B is an eigenvector of T, and thats not really clear a priori. Just say that B is an eigenvector of T so

    T(B) = λB

    Apply T-1 to both sides
    T-1(T(B)) = T-1(λB)

    T is a linear operator (I assume) ... I'll let you finish the argument. It's pretty trivial.

    Also as to the diagonalization comment, while I agree that's a good way to think about it intuitively it's not a good technique for a proof. Obviously there are issues with some finite dimensional space maps being non-diagonalizible; but even more fundamentally for a Hilbert space I think you would need the operator to be compact to have any hope of such an approach working.
     
    Last edited: Feb 24, 2012
  7. Feb 26, 2012 #6
    Should I write: [itex](1-t)[(1-t)(2-t)-2 = -(t-3)(t-1)(t)[/itex]. This is the characteristic polynomial. Thus, the roots are 3,1,0. These are the eigenvalues. If I have equations,

    (1-t)x + 2y = 0
    1x + (2-t)y = 0
    (1-t)z = 0,

    and I plug in for t=0,1,3, I find for t=3 that eigenvectors are multiples of (1,1,0). For t=1, eigenvectors are multiples of (0,0,1). For t=0, eigenvectors are multiples of (-2,1,0). The matrix is diagonalizable because T has three linearly indep. eigenvectors.

    Because these vectors are linearly independent, and because the number of vectors = dim(R3), these vectors span R3. Thus, R3 is the eigenspace of T (???)

    How does that look?
     
  8. Feb 26, 2012 #7
    Hmm. What's the problem with writing:

    T-1(T(B)) = T-1(λB) = λ'λB = B

    And then arriving at an identical conclusion?
     
  9. Feb 26, 2012 #8

    kai_sikorski

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    Gold Member

    T-1(λB) = λ'λB

    This is only true if B is an eigenvector of T-1, (sorry guess I forgot the exponent in my first post) and I don't think that's clear ahead of time.
     
  10. Feb 26, 2012 #9
    Okay. How about this:

    T-1(T(B)) = T-1(λB) = B. This implies,

    T-1λ = 1. And thus,
    T-1 = 1/λ,

    which confirms the relation,

    T-1(T(B)) = (1/λ)λ B = 1B =B

    It occurs to me that this isn't very good notation. But is the idea here right? Or am I still off?
     
  11. Feb 26, 2012 #10

    kai_sikorski

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    :confused:

    you have a matrix on one side of the equation and a scalar on the other
     
  12. Feb 26, 2012 #11
    Okay. How about this:

    T-1(T(B)) = T-1(λB) = B. This implies,

    T-1λ(B) = 1(B). And thus,
    T-1(B) = 1/λ (B),

    which confirms the relation,

    T-1(T(B)) = (1/λ)λ B = 1B =B
     
  13. Feb 26, 2012 #12

    kai_sikorski

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    Gold Member

    You're done here. This line means exactly that B is in fact an eigenvector of T-1 and the eigenvalue is 1/λ.

    This is self evident from the definition of the inverse, and does not need to be confirmed.
     
  14. Feb 26, 2012 #13
    This is a problem nearly identical to the 2nd problem in this thread. Since I'm still kind of sure about the material, I'd like my work checked here as well.
    Screen_shot_2012_02_26_at_3_59_09_PM.png

    T(1,0,0) = (3,-1,0)
    T(0,1,0) = (0,1,0)
    T(0,0,1) = (-1,2,4)

    Thus, we have the matrix,

    [itex]\left| \begin{array}{ccc}
    3 &0&-1 \\
    -1&1&2 \\
    0&0&4 \end{array} \right|[/itex]

    [itex]Δ_T (t) = det( \left| \begin{array}{ccc}
    3 &0&-1 \\
    -1&1&2 \\
    0&0&4 \end{array} \right| - tI)[/itex]

    I have this equaling: -(t-4)(t-3)(t-1), which is the characteristic polynomial. The roots are the eigenvalues, which are 4,3,1.

    To compute the eigenvectors:

    When t=4, we have,
    -x-z=0
    -x-3y+2z=0
    0z=0

    Which implies that eigenvectors are multiples of (-1,1,1).

    When t=3, we have,
    -z=0
    -x-2y=0
    z=0

    Which implies that eigenvectors are multiples of (-1,2,0)

    When t=1, we have,
    2x-z=0
    -x+2z=0
    3z=0

    Which implies that eigenvectors are multiples of (0,1,0).

    T is diagonizable because (-1,1,1),(-1,2,0),(0,1,0) are lin. indep.

    I'm still unsure if the eigenspace is the span of the eigenvectors. If so, I guess R3, otherwise, I'm not sure.
     
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