Equivalence classes of similar $2\times 2$ matrices How can we describe the equivalence classes under the similarity relation for $2 \times 2$ matrices with respect to the field of real numbers, $\mathbb{R}$? How would the equivalence classes change if the field is $\mathbb{C}$? 
I know that for ${Mat} _{1\times1}(\mathbb{R})$ each matrix has its own equivalence class, and I know that for $2\times 2$ matrices, the identity and zero matrices have their own equivalence class. But how can we describe the rest of the equivalence classes with respect to transformations and bases? 

I will answer part of your question.
You can determine the equivalence class of an $n\times n$ matrix by looking at its real Jordan form. In the $2\times2$ case, the real Jordan form of a matrix may take one of the following forms:


*

*$\begin{pmatrix}\lambda&0\\0&\mu\end{pmatrix}$ for some $\lambda\ge\mu$.

*$\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}$ for some $\lambda\in\mathbb{R}$.

*$\begin{pmatrix}a&-b\\ b&a\end{pmatrix}$ for some real number $a$ and nonzero real number $b$. Over $\mathbb{C}$, this real Jordan form is similar to the complex Jordan form $\begin{pmatrix}a+bi&0\\0&a-bi\end{pmatrix}$.


In general, if $J$ is the real Jordan form of an $n\times n$ matrix $A$, the equivalence class of $A$ can be written as $\{S^{-1}AS: S\in GL_n(\mathbb{R})\}=\{S^{-1}JS: S\in GL_n(\mathbb{R})\}$. Two real matrices are similar if and only if they have the same real Jordan form. That is, they are $\mathbb{R}$-similar if and only if they are $\mathbb{C}$-similar. So, the equivalence class of a real matrix (or strictly speaking, the intersection of its equivalence class with $M_n(\mathbb{R})$) remain unchanged if you change the field from $\mathbb{R}$ to $\mathbb{C}$.
