Conjugate classes for 2x2 matrices I've read an unproved claim that every real 2 by 2 matrix is similar to a matrix in exactly one of the following categories:
$\alpha \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$
$\alpha \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} + \beta \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}$
$\alpha \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} + \beta \begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix}$
$\alpha \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} + \beta \begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}$
where $\alpha, \beta \in \mathbb{R}$ and $\beta \neq 0$.
What is a general procedure for finding the appropriate class and the appropriate values of $\alpha$ and $\beta$ for an arbitrary real 2 by 2 matrix? 
I presume that one starts by computing the Jordan normal form of the matrix, which looks like
$P\begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2\end{bmatrix}P^{-1} \quad \quad $ or $\quad \quad P\begin{bmatrix} \lambda & 1 \\ 0 & \lambda\end{bmatrix}P^{-1}$.
In the latter case, it's pretty clear how to express (a conjugate of) the matrix in the 4th form above. In the former case (the diagonalizable case) when $\lambda_1 == \lambda_2$, it's even easier to express the matrix in the 1st form above. The hard part (the part I need help with) is how to express a diagonalizable matrix with distinct eigenvalues in either the 2nd or 3rd form above. 
In addition, I'd like an explanation of how one could generalize this result to larger matrices. 
Thanks
 A: The similarity type of a $2\times2$ matrix with distinct eigenvalues is determined precisely by those eigenvalues. Additionally, if the matrix is real, its eigenvalues will either both be real or will be a conjugate pair of complex numbers.
Note that
$\begin{bmatrix} \alpha & \beta \\ \beta & \alpha\end{bmatrix}$
has $\alpha + \beta$ and $\alpha - \beta$ as eigenvalues, with eigenvectors $\left(\begin{array}{c}1 \\ 1\end{array}\right)$ and $\left(\begin{array}{c}1 \\ -1\end{array}\right)$. So if your original matrix has real eigenvalues $\lambda_1$ and $\lambda_2$, it is similar to a matrix of the second type, with $\alpha=\frac{\lambda_1+\lambda_2}{2}$, $\beta=\frac{\lambda_1-\lambda_2}{2}$.
Similarly, $\begin{bmatrix} \alpha & -\beta \\ \beta & \alpha\end{bmatrix}$ has $\alpha+\beta i$ and $\alpha-\beta i$ as eigenvalues, with respective eigenvectors $\left(\begin{array}{c}i \\ 1\end{array}\right)$ and $\left(\begin{array}{c}1 \\ i\end{array}\right)$. So if your original matrix has complex eigenvalues $\lambda_{1,2}=\alpha \pm \beta i$, it is similar to a matrix of the third type.
It seems difficult to generalize this usefully to larger matrices, except insofar as the Jordan form itself is a kind of generalization. For larger matrices, real eigenvalues no longer come naturally in pairs, complex eigenvalues need no longer be simple, and so on.
