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

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Homework Help Overview

The discussion revolves around the properties of eigenvalues and eigenvectors related to linear operators T and its adjoint T*. Participants are exploring the implications of these properties, particularly in the context of normal operators.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • Participants are questioning whether T* shares eigenvectors with T when T has a specific eigenvalue and eigenvector. There is also exploration of the converse situation regarding the normality of T based on shared eigenvectors with T*.

Discussion Status

The discussion is ongoing, with some participants providing examples and counterexamples to clarify the properties of T and T*. Others are raising questions about the implications of these properties, particularly concerning normal operators, without reaching a consensus.

Contextual Notes

There is mention of a textbook that does not provide certain proofs or statements regarding the relationships between T and T*, which may affect the participants' understanding and exploration of the topic.

mathboy
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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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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].
 
Ok, I just read that it is true if T is a normal operator. Thanks.
 
What about the converse? Namely, if T and T* share their eigenvectors, is T necessarily normal?
 
morphism said:
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.
 
I'm just throwing it out there. It may be a good exercise to think about this.
 

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