# Eigenvectors of $\begin{bmatrix}a&-b\\b&a\end{bmatrix}$

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

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 $Ax = \begin{bmatrix}a&-b\\b&a\end{bmatrix} \begin{bmatrix}1\\-i\end{bmatrix} = \begin{bmatrix}a+bi\\b-ai\end{bmatrix} = (a+bi) \begin{bmatrix}1\\-i\end{bmatrix}$.

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 $\begin{bmatrix}-i&-1\\1&-i\end{bmatrix}$. After multiplying the top equation by $i$, I got $\begin{bmatrix}1&-i\\1&-i\end{bmatrix}$ ~ $\begin{bmatrix}1&-i\\0&0\end{bmatrix}$. 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.

There are an infinity of eigenvectors related to a given eigenvalue (in fact a whole vector sub space). Let's check that

$$\begin{bmatrix} 1\\-i \end{bmatrix}=-i\cdot\begin{bmatrix}i \\1\end{bmatrix}$$

So "your" eigenvector is proportionate to the one "of the textbook"

Similarly for the other eigenvalue

There is none: $(-i,1)$ is a scalar multiple of $(1,i)$ that is:

$-i(1,i) = (-i,-i^2) = (-i,-(-1)) = (-i,1)$, and similarly for your other eigenvector.

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

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

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.