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Math Help - Eigenvalues of Self-Adjoint Operators

  1. #1
    Guy
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    Eigenvalues of Self-Adjoint Operators

    Hey guys,

    Let \langle , \rangle be an arbitrary inner product on \mathbb{R}^n. Suppose A is self adjointing with respect to this inner product. Prove that A has real eigenvalues.

    My instinct would be to do something like this: given an eigenvalue \lambda, let v be a corresponding eigenvector. Then
    \displaystyle<br />
\lambda \langle v, v\rangle = \langle \lambda v, v\rangle = \langle A v, v\rangle<br />
= \langle v, A v\rangle = \langle v, \lambda v\rangle = \bar{\lambda} \langle v, v\rangle<br />

    The thing about this is that the inner product is defined on \mathbb{R}^n, not \mathbb{C}^n , and I'm having a hard time changing my thought process away from what I did above. I've been thinking maybe it should be \mathbb{C}^n since maybe the question would be trivial otherwise. Any thoughts?
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  2. #2
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    Quote Originally Posted by Guy View Post
    Hey guys,

    Let \langle , \rangle be an arbitrary inner product on \mathbb{R}^n. Suppose A is self adjointing with respect to this inner product. Prove that A has real eigenvalues.

    My instinct would be to do something like this: given an eigenvalue \lambda, let v be a corresponding eigenvector. Then
    \displaystyle<br />
\lambda \langle v, v\rangle = \langle \lambda v, v\rangle = \langle A v, v\rangle<br />
= \langle v, A v\rangle = \langle v, \lambda v\rangle = \bar{\lambda} \langle v, v\rangle<br />

    The thing about this is that the inner product is defined on \mathbb{R}^n, not \mathbb{C}^n , and I'm having a hard time changing my thought process away from what I did above. I've been thinking maybe it should be \mathbb{C}^n since maybe the question would be trivial otherwise. Any thoughts?


    If the inner product is defined on \mathbb{R}^n then any existing eigenvalue is real, otherwise they meant \mathbb{C}^n and what you did is correct.

    Tonio
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  3. #3
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    Since the set of real numbers is a subset of the complex numbers, even though <u, v> is defined to be a real number, you can think of it as a function from pairs of vectors to a complex number.

    For example, R^2 with inner product <(a, b), (c, d)>= ac+ bd has inner product always real but the matrix \begin{bmatrix}0 & 1\\ -1 & 0\end{bmatrix} has characteristic equation \lambda^2+ 1= 0 which has i and -i as roots. Since we can't have Av= iv in R^2, the corresponding linear transformation does NOT have any eigevalues. But we can, temporarily switch to "over C^n so that any linear transformation has eigenvalues, show, as you did, that, for a self-adjoint linear transformation, they must be real and so assert that any self-adjoint linear transformation on a vector space over the real numbers has eigenvalues.
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  4. #4
    Guy
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    Quote Originally Posted by HallsofIvy View Post
    Since the set of real numbers is a subset of the complex numbers, even though <u, v> is defined to be a real number, you can think of it as a function from pairs of vectors to a complex number.

    For example, R^2 with inner product <(a, b), (c, d)>= ac+ bd has inner product always real but the matrix \begin{bmatrix}0 & 1\\ -1 & 0\end{bmatrix} has characteristic equation \lambda^2+ 1= 0 which has i and -i as roots. Since we can't have Av= iv in R^2, the corresponding linear transformation does NOT have any eigevalues. But we can, temporarily switch to "over C^n so that any linear transformation has eigenvalues, show, as you did, that, for a self-adjoint linear transformation, they must be real and so assert that any self-adjoint linear transformation on a vector space over the real numbers has eigenvalues.
    Yeah, I get that. I guess my concern for this particular problem is that it's part of the build up to the Spectral Theorem. We need (counting multiplicities and including 0) n eigenvalues for our self-adjoint matrix to make things work I think, and so if you have an inner product over R^n you want to show that this is the case for self adjoint matricies. If you can extend the inner product to C^n, like you can with the dot product, then it's fine. I just worry that there is an inner product that you can't "switch over" to C^n in the same way and that I wouldn't be able to show the Spectral Theorem if we are working in R^n with an inner product that can't be extended to C^n.

    EDIT: I think I've shown that you can extend any inner product over R^n to one over C^n by taking
    ComplexInnerProduct(a + bi, c + di) = <a, c> + <b, d> + i[<b, c> - <a, d>]
    so I guess it's irrelevant.
    Last edited by Guy; September 5th 2010 at 07:10 PM.
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