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Eigenspaces and geometric reasoning

  1. Nov 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Let T be the reflection about the line 6x + 1y = 0 in the euclidean plane. Find the standard matrix A of T. Then, write down one of the eigenvalues and its corresponding eigenspace (in the form span {[ ]}). Then, find the other eigenvalue of A and its corresponding eigenspace.

    2. Relevant equations
    -


    3. The attempt at a solution
    I found the matrix T of the transformation, which is:

    (-35/37 -12/37)
    (-12/37 35/35)

    I actually found the eigenvalues for the matrix T by finding the characteristic polynomial (they are -1, 1) and then equating to 0 and solve the homogeneous system. However, I was told that i can use "geometric reasoning" to find the answer quickly, and I have no idea where to start using it to find the eigenvalues and eigenspaces (firstly, what do they represent in this case)?

    Thank you.
     
  2. jcsd
  3. Nov 13, 2013 #2

    CompuChip

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    If T is a linear transformation, then the eigenvalue equation reads
    $$T \vec v = \lambda \vec v$$
    This means precisely that ##\vec v## is a direction that will be unaffected by the transformation - applying T to it will give back the same vector up to scalar prefactors.

    So can you think of a vector that will not change if you reflect it about 6x + y = 0?
     
  4. Nov 13, 2013 #3

    HallsofIvy

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    You want the matrix
    [tex]\begin{pmatrix}a & b \\ c & d\end{pmatrix}[/tex]
    satisfying two properties:
    [tex]\begin{pmatrix}a & b\\ c & d \end{pmatrix}\begin{pmatrix}x \\ 6x\end{pmatrix}= \begin{pmatrix}0 \\ 0\end{pmatrix}[/tex]
    for any x, and
    [tex]\begin{pmatrix}a & b\\ c & d \end{pmatrix}\begin{pmatrix}-6y \\ y\end{pmatrix}= \begin{pmatrix}6y \\ -y\end{pmatrix}[/tex]
    for any y.

    Do you see why?
     
  5. Nov 13, 2013 #4
    I think they would be vectors that go along the line y = -6x. However, how can we find the eigen values in this case?

    How does this relate to eigenvalues and eigenspaces though? And how can we establish those two propositions that you stated? (aren't eigenvalues the solution to the homogeneous system Av - cv = 0?
     
  6. Nov 13, 2013 #5

    Dick

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    Yes, and if a vector doesn't change under the reflection, wouldn't that mean it's eigenvalue is 1? And under a reflection some vectors will change into their negatives, yes? Which would those be and what would their eigenvalues be? I think this the 'geometric reasoning' part of the exercise.
     
    Last edited: Nov 14, 2013
  7. Nov 14, 2013 #6
    Okay i kind of get it (the other vector literally gets rotated by pi degrees so its negative is the same vector. However, can't we as well have have another eigenvalue of 1 with direction (1 -6)? The answers are -1 and eigenspace (1 1/6) and 1 and eigenspace (-1 6). Can't we have more combinations of these?

    Thank you.
     
  8. Nov 14, 2013 #7

    Dick

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    Eigenvectors aren't unique. If (-1 6) is an eigenvector with eigenvalue 1 then k*(-1 6) is also an eigenvector with eigenvalue 1 for any constant k. The set of all such vectors is what the eigenspace means. (1 -6) is (-1)*(-1 6). It doesn't define a different eigenspace.
     
  9. Nov 15, 2013 #8
    So we can simply state that any scalar of that vector is the eigenspace for that eigenvalue? (As well for the second vector with the other eigenvalue of -1)
     
  10. Nov 15, 2013 #9

    Dick

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    Better to say "any multiple of that vector" rather than "any scalar of that vector", but yes.
     
  11. Nov 15, 2013 #10

    HallsofIvy

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    or "scalar multiple of that vector".
     
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