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Thread: Self adjoint

  1. #1
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    Self adjoint

    Hi all !!

    Someone can show me if that demonstration is going right ??

    Let A, B self adjoint operators such that AB = BA. Show that exist unique ortonormal base that diagonalize simultaneously A and B.

    Solution:
    How AB = BA then A and B are commutative operators. Let $\displaystyle \lambda$ an eigenvalue of A and $\displaystyle E_{\lambda}$ the self space associate. Let $\displaystyle v \in E_{\lambda}$ such that:

    $\displaystyle A v = \lambda v$

    Then

    $\displaystyle AB v = BA v \Rightarrow B \lambda v = \lambda Bv \Rightarrow Bv \in E_{\lambda} $

    Then $\displaystyle E_{\lambda}$ is invariant by B. Then v is an eigenvector common the A and B, then:

    $\displaystyle Av = \lambda v $

    $\displaystyle Bv = \gamma v $

    But in this point my demonstration becomes strange, because I assume that exist $\displaystyle v^{\perp}$, but I'm not sure that is correct...

    One more time, thanks a lot...
    Last edited by Borseti; Dec 1st 2011 at 09:14 AM.
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  2. #2
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    Re: Self adjoint

    I rewrote my demonstration:

    Let $\displaystyle A,B: E \rightarrow E$ two self adjoint operators such that AB = BA, let $\displaystyle \lambda$ an eigenvalue of A and $\displaystyle E_{\lambda}$ a self space associate. Now let $\displaystyle v \in E$ not null vector such that:

    $\displaystyle Av = \lambda v$

    How AB = BA, then:

    $\displaystyle AB = BA \Rightarrow ABv = BAv \Rightarrow AB v = B \lambda v \Rightarrow B \lambda v = \lambda B v \therefore B v \in E_{\lambda}$

    Logo $\displaystyle E_{\lambda}$ is invariant by B. Then v is an eigenvector common on A and B, then exist $\displaystyle \gamma$ such that:

    $\displaystyle B v = \gamma v$

    How $\displaystyle \lambda$ and $\displaystyle \gamma$ are real roots of characteristic polynomials of A and B, then $\displaystyle A - \lambda \textrm{I}$ and $\displaystyle B - \gamma \textrm{I}$ are both not invertible, then:

    $\displaystyle Av = \lambda v \Rightarrow (A - \lambda \textrm{I}) v = 0$

    $\displaystyle Bv = \gamma v \Rightarrow (B - \gamma \textrm{I})v = 0$

    Then v belong to orthonormal base $\displaystyle \mathbb{B} \subset E$ of eigenvectors of A and B, therefore the base $\displaystyle \mathbb{B}$ diagonalize A and B simultaneously.
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  3. #3
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    Re: Self adjoint

    It looks like you are saying this:

    Let u be the eigenvectors of A
    Au = su
    A(Bu) = s(Bu).......A and B commute
    Bu = pu......... Bu is an eigenvector of A
    u is an eigenvector of B
    The eigenvectors of A and B are the same.

    The matrix whose column vectors are u diagonalizes A and B.
    Last edited by Hartlw; Dec 1st 2011 at 10:28 PM. Reason: shorten
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