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HallsofIvy · #4 November 6th, 2009, 08:13 PM

You "rref" with complex number exactly like with real numbers! Just remember to multiply and divide correctly!

But I would bother with "rref" here. Just use the definition of "eigenvalue": If $\lambda$ is an eigenvalue of A, then there exist a non-zero vector, v, such that $Av= \lambda v$ and the eigenvectors are the vectors satisfying that.

Since i is an eigenvalue, you must have $\begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= i\begin{bmatrix} x \\ y \end{bmatrix}$

$\begin{bmatrix}y \\-x\end{bmatrix}= \begin{bmatrix}ix \\ iy\end{bmatrix}$ .

That gives the two equations y= ix and -x= iy which, since 1/i= -i, are really the same equation. From y= ix, we can write $\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}x \\ ix\end{bmatrix}= x\begin{bmatrix}1 \\ i\end{bmatrix}$ .

The eigenvectors corresponding to eigenvalue i are multiples of $\begin{bmatrix}1 \\ i\end{bmatrix}$ .

Now you find the eigenvectors corresponding to eigenvalue -i.