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I'm experiencing some confusion regarding self-adjoint operators. As background for my question, I give the following 3 results (all from Linear Algebra, 3rd ed. by Friedberg, Insel, and Spence):

Theorem 6.17: "Let $T$ be a linear operator on a finite-dimensional real inner product space $V$. Then $T$ is self-adjoint if and only if there exists an orthonormal basis $\beta$ consisting of eigenvectors of $T$."

Theorem 6.14 (Schur's Theorem): "Let $T$ be a linear operator on a finite-dimensional inner product space $V$. Suppose that the characteristic polynomial of $T$ splits. Then there exists an orthonormal basis $\beta$ for $V$ such that the matrix $[T]_\beta$ is upper triangular."

Lemma: "Let $T$ be a self-adjoint operator on a finite-dimensional inner product space $V$. Then (a) Every eigenvalue of $T$ is real. (b) Suppose that $V$ is a real inner product space. Then the characteristic polynomial of $T$ splits."

From these results I want to say this: Let $T$ be a self-adjoint linear operator on a finite-dimensional real vector space $V$. Then by the lemma, the characteristic polynomial splits, and by Schur's theorem there exists an orthonormal basis $\beta$ for $V$ so that $[T]_\beta$ is upper triangular. Let $A=[T]_\beta$. Then since $T$ is self-adjoint we have $A^*=[T]_\beta^*=[T^*]_\beta=[T]_\beta=A$ (this comes from the proof of Theorem 6.17) so that $A$ must be a diagonal matrix. So this seems to imply that every self-adjoint linear operator on a finite-dimensional real vector space is diagonal, but this is not the case since the self-adjoint matrices clearly include any real symmetric matrix. What am I not seeing?

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Being diagonal is not a property of an operator but of a matrix. That there exists an orthonormal basis $\beta$ in which $T$ is represented by a diagonal matrix doesn't imply that it is represented by a diagonal matrix in all orthonormal bases, so there's no contradiction here.

With respect to any orthonormal basis a self-adjoint operator is represented by a Hermitian (or self-adjoint) matrix, and the fact that there exists a basis $\beta$ in which it is represented by a diagonal matrix corresponds to the fact that every Hermitian matrix is diagonalizable by a unitary matrix.

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  • $\begingroup$ So every self-adjoint operator IS diagonalizable? But that means that there is some (unique?) basis in which it is represented by a diagonalizable matrix, but is not in every basis (I did forgot this and it was the source of my confusion). $\endgroup$
    – Alex Petzke
    Jan 18, 2012 at 0:58
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    $\begingroup$ @Alex: I think you meant "diagonal matrix", not "diagonalizable matrix"? It is always represented by a diagonalizable matrix, but only sometimes by a diagonal matrix, namely whenever the basis is a combination of bases of its eigenspaces. Thus there is a unique basis (up to sign flips) in which it is represented by a diagonal matrix iff its eigenvalues are all distinct. $\endgroup$
    – joriki
    Jan 18, 2012 at 1:21
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    $\begingroup$ If there are repeated eigenvalues, the diagonalizing basis may not be unique. $\endgroup$
    – GEdgar
    Jan 18, 2012 at 1:22
  • $\begingroup$ @joriki: Yes, diagonal matrix. GEdgar: Could we obtain two different bases simply by choosing two different bases for an eigenspace corresponding to one of the repeated eigenvalues, and combining it with the rest? $\endgroup$
    – Alex Petzke
    Jan 18, 2012 at 2:21
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    $\begingroup$ @Alex: Yes, that's basically what I was trying to say in my previous comment. If an eigenvalue is repeated, there are different bases for its eigenspace, and you can combine any bases for the eigenspaces into a basis that diagonalizes the matrix for the entire operator. $\endgroup$
    – joriki
    Jan 18, 2012 at 2:55

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