Proving Eigenvalues and Eigenvectors for T and T*: A Comprehensive Guide

[mathboy]

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.

 

[HallsofIvy]

There is no proof because it is not true. For example, if
$$T= \left[\begin{array}{cc}0 & i \\i & 0\end{array}right]$$
Then T has eigenvalue i, with eigenvecor [a, a] and eigenvalue -i with eigenvector [1, -1].
$$T*= \left[\begin{array}{cc}0 & -i \\ -i & 0\end{array}\right]$$
which also eigenvalue i but now with eigenvectors [a, -a] and eigenvalue -i with eigenvectors [a, a].

 

[mathboy]

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

 

[morphism]

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

 

[mathboy]

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.

 

[morphism]

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