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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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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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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