Eigenvalue proof help

  • Thread starter mathboy
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  • #1
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I know that if T has eigenvalue k, then T* has eigenvalue k bar. But if T has eigenvector x, does T* also have eigenvector x? If so, how do you prove it? I don't see that in my textbook.
 

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  • #2
HallsofIvy
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There is no proof because it is not true. For example, if
[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].
 
  • #3
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Ok, I just read that it is true if T is a normal operator. Thanks.
 
  • #4
morphism
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What about the converse? Namely, if T and T* share their eigenvectors, is T necessarily normal?
 
  • #5
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What about the converse? Namely, if T and T* share their eigenvectors, is T necessarily normal?
I don't know if you are asking rhetorically or not. But my textbook doesn't state the converse.
 
  • #6
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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