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Proof of Isometry

  1. Apr 25, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove or give a counterexample: if [tex]\mathcal{S} \in \mathcal{L} \left( V \right) [/tex] and there exists and orthonormal basis [tex]\left( e_{1} , \ldots , e_{n} \right)[/tex] of [tex]V[/tex] such that [tex]\left\| \mathcal{S} e_{j} \right\| = 1 [/tex] for each [tex]e_{j}[/tex], then [tex]\mathcal{S}[/tex] is an isometry.


    2. Relevant equations
    [tex] \mathcal{S} [/tex] is an isometry if
    [tex]\left\| \mathcal{S} v \right\| = \left\| v \right\| [/tex]
    for all [tex]v \in V[/tex].


    3. The attempt at a solution
    I figure that if I can prove that [tex]\mathcal{S}[/tex] is normal, then I can use the Spectral Theorem to prove that all every basis vector is an eigenvector.

    Then I can prove that the square of the absolute value of the eigenvalue for each eigenvector equals 1, and the proof comes easy after that.

    But I'm getting hung up on proving if it is normal.

    Thanks in advance for any help.
     
  2. jcsd
  3. Apr 25, 2008 #2

    HallsofIvy

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    You don't need eigenvalues. Write v in terms of the basis: [itex]\vec{v}= a_1\vec{e_1}+ a_2\vec{e_2}+ \cdot\cdot\cdot+ a_n\vec{e_n}[/itex]. What is its length? What is Lv? What is its length?

    You say [tex]\left\| \mathcal{S} e_{j} \right\| = 1 [/tex]. I would have interpreted [tex]\mathcal{S} \in \mathcal{L} \left( V \right) [/tex] to mean "linear transformations from V to itself". Do you mean the dual space of V: linear transformations from V to R?
     
  4. Apr 25, 2008 #3
    Sorry for my lack of clarity. [tex]\left\| \mathcal{S} e_{j} \right\| = 1 [/tex] is a given property of the linear transformation [tex] \mathcal{S} [/tex]. So the only thing I know about [tex] \mathcal{S} [/tex] is that it's a linear transformation and it doesn't change the length of the orthonormal basis vectors.


    I think I have a solution, but I'm not sure if it's the same thing you were saying:

    [tex] \left\| S e_{i} \right\| = \left\| e_{i} \right\| [/tex]
    [tex]\left\langle S e_{i} , S e_{i} \right\rangle = \left\langle e_{i} , e_{i}
    \right\rangle[/tex]
    [tex] \left\langle e_{i} , S^{*} S e_{i} \right\rangle = \left\langle e_{i} , e_{i} \right\rangle [/tex]
    [tex] \left\langle e_{i} , S^{*} S e_{i} - e_{i} \right\rangle = 0 [/tex]
    So since [tex] e_{i} [/tex] is not the zero vector,
    [tex]S^{*} S e_{i} - e_{i} = 0 \ \Leftrightarrow S^{*} S e_{i} = e_{i} [/tex]
    And from there, it's easy.

    Does that look right?
     
  5. Apr 25, 2008 #4

    Dick

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    That doesn't prove (S*)S(e_i)-e_i=0. It just shows (S*)S(e_i)-e_i is orthogonal to e_i. I would suggest you take some time out from trying to prove it and put a little more energy into looking for a counterexample.
     
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