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Let $$A=\left[\begin{matrix} 2 & -i \\ i & 2 \end{matrix}\right],$$

Show that $U_1 = \dfrac{1}{\sqrt{2}}(\Psi_1+i\Psi_2)$ and $U_2 = \dfrac{1}{\sqrt{2}}(\Psi_1-i\Psi_2)$ are orthonormal vectors of A and find the eigenvalues.

So I started off by stating that $Au=\lambda u$ So we should calculate:

$AU_1=\dfrac{1}{\sqrt{2}}(\Psi_1+i\Psi_2)$$\left[\begin{matrix} 2 & -i \\ i & 2 \end{matrix}\right],$ but I seem to be drawing a blank as how to do this? Can anyone please help me?

I think I can work from there, it's just this calculation I am struggling with.

Thanks

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  • $\begingroup$ What is meant by "orthonormal vector of A"? $\endgroup$ – mvw May 20 '14 at 17:38
  • $\begingroup$ I suppose that should be "orthonormal eigenvectors of $A$"? But what are $\Psi_1$ and $\Psi_2$? And are you familiar with the fact that the eigenvalues of a linear map (matrix) are precisely the roots of its characteristic polynomial? $\endgroup$ – Inactive - Objecting Extremism May 20 '14 at 18:11
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The characteristic polynomial is:

$(2-\lambda)^2-1=\lambda^2-4\lambda+4-1=\lambda^2-4\lambda+3=(\lambda-3)(\lambda-1)=0$

So $\lambda=3,1$ but the only case where the eigenvectors of a matrix, are the conjugates of each other, are when we have complex eigenvalues.

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  • $\begingroup$ Yes, I completely overlooked that fact! Thank you, I was making it far more complicated than it needed to be $\endgroup$ – sarahusher May 20 '14 at 18:21
  • $\begingroup$ No worries, happy to help :) $\endgroup$ – Ellya May 20 '14 at 18:35

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