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How do you find the eigenvalues (hence the eigenvectors too) of a matrix with complex bits like this:

$$\hat{H}=\epsilon \begin{vmatrix} 0&i&0 \\\\ -i&0&0 \\\\ 0&0&-i \end{vmatrix}$$

With $\epsilon$ real. How do you find the eigenvalues (hence the eigenvectors too) of a matrix with complex bits like this:

$$\hat{H}=\epsilon \begin{vmatrix} 0&i&0 \\\\ -i&0&0 \\\\ 0&0&-i \end{vmatrix}$$

With $\epsilon$ real.

I get so far as this:

$$|\hat{H}-\lambda{}I | = \epsilon \begin{vmatrix} -\lambda&i&0 \\\\ -i&0-\lambda&0 \\\\ 0&0&-i-\lambda \end{vmatrix} =0$$

$$|\hat{H}-\lambda{}I | =\epsilon(-\lambda(-\lambda(i+\lambda))+i(0-i(i+\lambda))$$

$$=\epsilon(\lambda^2 (i+\lambda)+(i+\lambda))=0$$

Then I am not sure what to do next. I think the $\epsilon$ cancels and I can do this:

$$= \lambda^3+\lambda^2i+i+\lambda=0$$

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2 Answers 2

up vote 1 down vote accepted

You have a factor $\lambda +i$, which gives $-1$ as eigenvalue. Then you have to solve $\lambda^2+1=0$, which gives $i$ again and $-i$.

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$$|xI-H|=\epsilon(x+i)(x^2-1)=\epsilon(x+i)(x-1)(x+1)$$

So the eignevalues are $\,-i\,,\,\pm 1\,$. The matrix is thus diagonalizable

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+1 for the extra information about the matrix being diagonalizable. That is right. I am not sure how you deduced it though! –  Magpie Nov 4 '12 at 14:59
1  
Since a matrix is diagonalizable iff there exists a basis of its eigenvectors, and since eigenvectors belonging to different eigenvalues are linearly independent, a $\,3\times 3\,$ matrix with 3 different eigenvalues is diagonalizable. –  DonAntonio Nov 4 '12 at 17:11

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