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HallsofIvy

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[tex]T= \left[\begin{array}{cc}0 & i \\i & 0\end{array}right][/tex]

Then T has eigenvalue i, with eigenvecor [a, a] and eigenvalue -i with eigenvector [1, -1].

[tex]T*= \left[\begin{array}{cc}0 & -i \\ -i & 0\end{array}\right][/tex]

which also eigenvalue i but now with eigenvectors [a, -a] and eigenvalue -i with eigenvectors [a, a].

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Ok, I just read that it is true if T is a normal operator. Thanks.

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morphism

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What about the converse? Namely, if T and T* share their eigenvectors, is T necessarily normal?

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I don't know if you are asking rhetorically or not. But my textbook doesn't state the converse.What about the converse? Namely, if T and T* share their eigenvectors, is T necessarily normal?

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morphism

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I'm just throwing it out there. It may be a good exercise to think about this.

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