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Following an example from my textbook, we are looking for eigenvectors of a particular matrix with complex eigenvalues. The original matrix looks like $$\left(\begin{matrix} -1 & 2 \\ -1 & -3 \\ \end{matrix}\right)$$ with eigenvalues $$-2 \pm 2i$$ Inserting -$2+i$ into our $A- \lambda I$ version of the matrix, we get $$\left(\begin{matrix} 1-i & 2 \\ -1 & -1-i \\ \end{matrix}\right)$$ So far, I'm with them all the way. At this point, the eigenvector corresponding to this matrix is given to be $$\left(\begin{matrix} 2 \\ -1+i \\ \end{matrix}\right)$$
This is where they lose me. I calculate the eigenvector by solving the second row $[(-1)x_1 - (-1-i)x_2 = 0]$ for $x_2$ in terms of $x_1$ which gives us: $$x_2 = \frac{x_1}{-1-i}\tag{*}$$ and the eigenvector $$\left(\matrix{x_1\\x_2}\right)=\left(\begin{matrix} \frac{1}{-1-i} \\ 1 \\ \end{matrix}\right)\tag{**}$$ which is not a scalar multiple of their eigenvector. Multiplying this vector with $-1-i$ we get
$$\left(\begin{matrix} 1 \\ -1-i \\ \end{matrix}\right)$$ & multiplying their eigenvector with ${\frac{\overline{-1+i}}{2}}$ we get
$$\left(\begin{matrix} -1-i \\ 1 \\ \end{matrix}\right)$$

ie the "inverse" of what I get. This seems to be the general case, every time I calculate eigenvectors I get the mirror image of the result the book has. Since these vectors in general seem to be linearly independent (mine and theirs), they answers aren't equivalent. Am I missing something regarding these complex eigenvectors?

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    $\begingroup$ Calculation from $(*)$ to $(**)$ is wrong. $\endgroup$ – A.Γ. Sep 20 '15 at 17:08
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The equation $x_2=\frac{x_1}{-1-i}$ gives you the vector $(1, \frac{1}{-1-i}$, not the vector $(\frac{1}{-1-i}, 1)$. You set $x_1$ to $1$ and then $x_2$ becomes the fraction.

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  • $\begingroup$ Of course, thank you very much! Looking back on my notes, I made the exact same mistake of mixing up the parameters on two different problems, which is why I was "consistently" wrong. Again, thanks for your time! $\endgroup$ – Laplacinator Sep 20 '15 at 17:13

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