# Eigenvalue proof help

1. Jan 12, 2008

### 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.

2. Jan 12, 2008

### HallsofIvy

Staff Emeritus
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].

3. Jan 12, 2008

### mathboy

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

4. Jan 12, 2008

### morphism

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

5. Jan 12, 2008

### mathboy

I don't know if you are asking rhetorically or not. But my textbook doesn't state the converse.

6. Jan 12, 2008