Eigenvectors from complex eigenvalues

zfolwick
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how does one systematically find the eigenvectors of a 2x2 (or higher) Real matrix given complex eigenvalues?
 
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The same way you do it when you have real eigenvalues.
 
For example, the eigenvalues of the matrix
\begin{bmatrix}0 & -1 \\ 1 & 0 \end{bmatrix}
are i and - i.

If < x, y> is an eigenvector corresponding to eigenvalue i then we must have
\begin{bmatrix}0 &amp; -1 \\ 1 &amp; 0 \end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}- y \\ x\end{bmatrix}= \begin{bmatrix} ix \\ iy\end{bmatrix}

So we must have -y= ix and x= iy. Since 1/i= -i, those are equivalent. Any such eigenvector is of the form < x, y>= <iy, y>= y<i , 1>.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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