Let $X = \left\lbrace \begin{pmatrix} a & b \\ c & d \end{pmatrix} \right\rbrace$ be the set of linear mappings $\mathbb{R}^2$ onto itself. The topology in it is given by its obvious identification with $\mathbb{R}^4$ . Equivalence relation: $A \sim B \iff A = LBL^{-1}$ , where $L$ is some invertible matrix. It is required to describe the quotient set $X/\!\sim$ and the quotient space. Is this quotient space Hausdorff ?
I can't find the quotient space here.
The condition does not say which topology is defined on $\mathbb{R}^4$ , I guess it's standard, metric space $(x_1,x_2,x_3,x_4) \mapsto (a,b,c,d)$.
No, the quotient is not Hausdorff. Consider the null matrix. It is a class all by it tself (that is, the only matrix similar to the null matrix is the null matrix itself). But as close as you wish to the null matrix you will find a matrix of the type $\left(\begin{smallmatrix}0&\lambda\\0&0\end{smallmatrix}\right)$ (with $\lambda\neq0$), which is similar to $\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$. So, in this quotient space the null matrix and $\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ are distinct elements, but every neighborhood of the former contains the later.