I am working with $\Gamma$, a positive definite matrix of dimension 2, which is not necessarily symmetric. I assume that $\Gamma$ has real eigenvalues and that the eigenvalues are positive (since $\Gamma$ is positive definite).
Up to similarity, I think that there are two families of $\Gamma$:
- \begin{pmatrix}\lambda_1 & 0\\0&\lambda_2\end{pmatrix} where $\lambda_1$ and $\lambda_2$ are positive scalars.
- \begin{pmatrix}\lambda & 1\\0&\lambda\end{pmatrix} i.e. $\Gamma$ is similar to a Jordan canonical form where $\lambda$ is positive.
Do you think that I am correct or there exists another family for $\Gamma$ (up to similarity)?
Assuming everything is in $\mathbb{R}$.
You are wrong. $A>0$ does not imply that $P^{-1}AP>0$. Choose for example $A=diag(1,2)$ and $P^{-1}AP=\begin{pmatrix}-1&-6\\1&4\end{pmatrix}$.
In the sequel, I assume that the matrices are real.
On the other hand, for every $x\not= 0$, $x^TAx>0$ IFF for every $x\not= 0$, $x^TP^TAPx>0$ (where $P$ is invertible). Then we must consider congruence classes and not similarity classes.
There are at least $2$ congruence classes for your $>0$ matrices that have $>0$ eigenvalues. That of $I_2$ and for example that of $B=\begin{pmatrix}2&2\\3&4\end{pmatrix}$. The classification is known over $M_n(\mathbb{C})$ (Horn, Sergeichuk) but, over $M_n(\mathbb{R})$, I don't know.
EDIT. About the OP's question, the answer is YES, because any considered matrix (with $>0$ eigenvalues) in $M_n(\mathbb{R})$ is similar to one of the presented two models. Conversely, the first model is $>0$ and the second one is similar to $\begin{pmatrix}\lambda&a\\0&\lambda\end{pmatrix}$, where $4\lambda^2>a^2>0$, which is $>0$.