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Math Help - T normal operator has splitting characteristic polynomial - eigenvectors form basis?

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    T normal operator has splitting characteristic polynomial - eigenvectors form basis?

    Question: T is a normal operator on a finite-dimensional real inner product space V with a characteristic polynomial that splits. Prove that V has an orthonormal basis of eigenvectors of T.

    By Schur's theorem, because the characteristic polynomial of T splits, there exists an orthonormal basis \beta for V such that [T]_{\beta} is upper triangular. How does the fact that the inner product space being real ensure that such a basis is composed of the eigenvectors of T? Thanks!
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    Quote Originally Posted by Last_Singularity View Post
    Question: T is a normal operator on a finite-dimensional real inner product space V with a characteristic polynomial that splits. Prove that V has an orthonormal basis of eigenvectors of T.

    By Schur's theorem, because the characteristic polynomial of T splits, there exists an orthonormal basis \beta for V such that [T]_{\beta} is upper triangular. How does the fact that the inner product space being real ensure that such a basis is composed of the eigenvectors of T? Thanks!

    Schur's theorem tells us that U^{*}TU=A , where A is upper triangular and U is unitary.
    As all the eigenvalues of T are real (otherwise its char. pol. wouldn't split over the reals!), T is self-adjoint or Hermitian (i.e., T=T^{*}), and then:

    A^{*}=\left(U^{*}TU\right)^{*}=U^{*}T^{*}U=U^{*}TU  =A\Longrightarrow\,A=A^{*} , and since A is upper triangular then A^{*} is lower triangular, thus A is in fact diagonal and we're done.

    Tonio
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