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

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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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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