Linear transformation with no real eigenvalue Let $f:\Bbb R^2\to\Bbb R^2$ a linear transformation with no real eigenvalue. The question is to prove that for some basis of $\Bbb R^2$ the matrix of $f$ has the form
$$\begin{pmatrix}\alpha&-\beta\\
\beta&\alpha\end{pmatrix}$$
where $\alpha\in\Bbb R$ and $\beta\in\Bbb R\setminus\{0\}$. I don'tknow how begin the answer, I tried with characteristic polynomial of $f$ which is a quadratic polynomial $x^2+bx+c$ with negative discriminant but no success. Any suggestion?
 A: To begin with, we note that $f$ has the complex eigenvalues $\alpha \pm \beta i$.
Take $g = f - \alpha \,I$ (where $I$ denotes the identity transformation).  It now suffices to find a basis such that the matrix of $g$ has the form
$$
M = \pmatrix{0&-\beta\\\beta & 0}
$$
Note that $g$ has characteristic polynomial $x^2 + \beta^2$.  Thus, we have $g^2  = - \beta^2 I$, by the Cayley-Hamilton theorem.  Select any non-zero $x_1 \in \Bbb R^2$ to be the first element of our basis.  Take $x_2 = \frac 1 \beta g(x_1)$ to be the second element of our basis.  We find that
$$
g(x_1) = \beta x_2\\
g(x_2) = g \left( \frac 1 \beta  g(x_1)\right) = \frac{g^2(x_1)}{\beta} = \frac{-\beta^2 x_1}{\beta} = -\beta x_1
$$
Thus, we indeed find that the matrix of $g$ with respect to $\{x_1,x_2\}$ is $M$.  Thus, the matrix of $f = g + \alpha I$ is
$$
M + \alpha I = \pmatrix{0&-\beta\\\beta & 0} + \alpha \pmatrix{1&0\\0&1} = \pmatrix{\alpha& - \beta \\ \beta & \alpha}
$$

Note: rather than characterizing $\alpha,\beta$ as the complex eigenvalues of $f$, we could say that $\alpha = \frac 12 Tr(f)$, and $\beta$ is such that $(f - \alpha I)^2 = - \beta^2 I$.
Or, note that any quadratic with negative descriminant can be rewritten by completing the square to give
$$
x^2 + bx + c = (x-\alpha)^2 - \beta^2
$$
