Eigenvectors of $\begin{bmatrix}a&-b\\b&a\end{bmatrix}$ I am currently confused as to what the correct eigenvectors are for $\begin{bmatrix}a&amp;-b\\b&amp;a\end{bmatrix}$.

I confirmed through my own calculations that the eigenvalues are a$\pm$bi. My textbook, Linear Algebra and its Applications, states that the corresponding eigenvectors are $\begin{bmatrix}1\\-i\end{bmatrix}$ and $\begin{bmatrix}1\\i\end{bmatrix}$.

This makes sense when verifying that $Ax = \lambda x$, as $Ax = \begin{bmatrix}a&amp;-b\\b&amp;a\end{bmatrix} \begin{bmatrix}1\\-i\end{bmatrix} = \begin{bmatrix}a+bi\\b-ai\end{bmatrix} = (a+bi) \begin{bmatrix}1\\-i\end{bmatrix}$. 

However, upon performing the calculations myself, I repeatedly found the eigenvectors to be $ \begin{bmatrix}-i\\1\end{bmatrix} $ and $ \begin{bmatrix}i\\1\end{bmatrix} $ rather than the given solution. Thinking that I could have made a calculation error, I plugged this into WolframAlpha  and got the same values. 

My calculation process was to solve for $ (A-(a+bi)I)x = 0 $, which I reduced down to $ \begin{bmatrix}-i&amp;-1\\1&amp;-i\end{bmatrix} $. After multiplying the top equation by $i$, I got $ \begin{bmatrix}1&amp;-i\\1&amp;-i\end{bmatrix} $ ~ $  \begin{bmatrix}1&amp;-i\\0&amp;0\end{bmatrix} $. Thus, $x_1 = ix_2 $ and $x_2$ is free. So $x = \begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix}i\\1\end{bmatrix} x_2 $, so the eigenvalue corresponding to $\lambda = a+bi$ is $\begin{bmatrix}i\\1\end{bmatrix}$. Similarly, the eigenvalue corresponding to $\lambda = a-bi$ is $\begin{bmatrix}-i\\1\end{bmatrix}$.

Where is the discrepancy between my calculated answer and the one given in the textbook?
Eigenvectors are unique upto some (non-zero) constant. Note that the two eigenvectors that are listed as answers are just $\pm i$ times the eigenvectors you found.
Let $r$ be the module and $\theta$ the argument of complex number $a+bi$, thus with

$$a=r \cos \theta \ \ \ \text{and} \ \ \  b=r \sin \theta$$

No answer mentions the geometrical interpretation of matrix

$$S=\begin{bmatrix}a&amp;-b\\b&amp;a\end{bmatrix}=r\begin{bmatrix}\cos \theta&amp;-\sin \theta\\\sin \theta&amp;\cos \theta\end{bmatrix}$$

that helps to understand why such matrices are bound to have non real eigenvalues.

The interpretation of $S$ as a geometrical transformation is clearly a rotation followed (or preceded) by an homothetic transform, i.e., a similitude. It is clear that (unless $\theta=0$!) no real vector can be transformed by $S$ into a real multiple of it. Therefore, the eigenvalues "have to be" complex.