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Math Help - Inner Product

  1. #1
    arg
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    Inner Product

    I'd be grateful for any assistance.


    V is a finite dimensional vector space over F (where F is R or C).

    Let V be a real vector space and let T be a linear operator on V. Prove that if V has a basis consisting of eigenvectors for T, then there is an inner product on V for which T is self-adjoint.
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    Quote Originally Posted by arg View Post
    I'd be grateful for any assistance.


    V is a finite dimensional vector space over F (where F is R or C).

    Let V be a real vector space and let T be a linear operator on V. Prove that if V has a basis consisting of eigenvectors for T, then there is an inner product on V for which T is self-adjoint.
    Let \{e_1,\ldots,e_n\} be a basis consisting of eigenvectors for T, and define an inner product by \langle x,y\rangle = \sum x_iy_i, where x = \sum x_ie_i, y = \sum y_ie_i. Note that this will make T selfadjoint (as you can easily check) provided that V be a real vector space. But the result is false over a complex vector space.
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    arg
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    Quote Originally Posted by Opalg View Post
    Let \{e_1,\ldots,e_n\} be a basis consisting of eigenvectors for T, and define an inner product by \langle x,y\rangle = \sum x_iy_i, where x = \sum x_ie_i, y = \sum y_ie_i. Note that this will make T selfadjoint (as you can easily check) provided that V be a real vector space. But the result is false over a complex vector space.

    Thank you. However, I can't see how to show that T is self-adjoint.
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    Quote Originally Posted by arg View Post
    Thank you. However, I can't see how to show that T is self-adjoint.
    If \lambda_i is the eigenvalue for the eigenvector e_i, then \langle Te_i,e_i\rangle = \langle e_i,Te_i\rangle = \lambda. If i≠j then \langle Te_i,e_j\rangle = \langle e_,Te_j\rangle = 0. It follows by linearity that \langle Tx,y\rangle = \langle x,Ty\rangle for all x and y.
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    arg
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    Quote Originally Posted by Opalg View Post
    If \lambda_i is the eigenvalue for the eigenvector e_i, then \langle Te_i,e_i\rangle = \langle e_i,Te_i\rangle = \lambda. If i≠j then \langle Te_i,e_j\rangle = \langle e_,Te_j\rangle = 0. It follows by linearity that \langle Tx,y\rangle = \langle x,Ty\rangle for all x and y.

    How do we know that \langle Te_i,e_j\rangle = \langle e_,Te_j\rangle = 0?
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    Quote Originally Posted by arg View Post
    How do we know that \langle Te_i,e_j\rangle = \langle e_,Te_j\rangle = 0?
    Because Te_i=\lambda e_i, and \langle e_i,e_j\rangle=0 by the way in which the inner product was defined.
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  7. #7
    arg
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    Thanks!
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