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I need to find eigenvectors and eigenvalues of $\begin{pmatrix} 1 & i \\ -i & 1 \end{pmatrix}$.

Attempt: When I find the equation which I have to solve for the eigenvalues I get $(\lambda -1)^2 +i=0$. Solving for $\lambda$ I get $\lambda =\pm \frac{1-i}{\sqrt{2}}+1$ using $\sqrt{-i}=\frac{1-i}{\sqrt{2}}$. However, my book lists the following answers: $\lambda =0;2$. Could you explain how to get to these answers. Thank you.

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You have a mistake in your characteristic polynomial. It should be $(\lambda-1)^2-1$. – 1015 Mar 5 '13 at 4:50
Thanks. I see it now. – Koba Mar 5 '13 at 4:54
up vote 3 down vote accepted

The characterisitic polynomial is $|A - \lambda I| = \begin{pmatrix} 1-\lambda & i \\ -i & 1-\lambda \end{pmatrix} = 0$.

This gives: $(1-\lambda^2) + i^2 = (1-\lambda^2) - 1 = \lambda (\lambda - 2) = 0$.

You should get an Eigensystem as follows:

$$\lambda_1 = 2, v_1 = (i, 1)$$

$$\lambda_2 = 0, v_2 = (-i, 1)$$

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Yeah my book says the same thing. But am I on the right track? Why my lambdas contain complex numbers? – Koba Mar 5 '13 at 4:50
Quick question. For the eigenvector corresponding to the second lambda can I have (i,-1). The equation will still be equal to zero. – Koba Mar 5 '13 at 5:02
Well-done, as usual ;-) +1 – amWhy Apr 25 '13 at 0:25

You've miscalculated. You should be taking the determinant of $$\left(\begin{array}{cc}1-\lambda & i\\-i & 1-\lambda\end{array}\right),$$ which is $$(1-\lambda)^2-(i)(-i)=(1-\lambda)^2-1=\lambda^2-2\lambda.$$ Setting that equal to $0$ will do the trick.

As an aside, you will occasionally encounter complex eigenvalues. It isn't (necessarly) anything to be concerned about, so long as they're correctly obtained. We'll deal with them in much the same way as with real eigenvalues.

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