# 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

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.