# Quotient space and quotient set for $\mathrm{End}(\mathbb{R}^2)$

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 $(x_1,x_2,x_3,x_4) \mapsto (a,b,c,d)$.

• You could maybe try to use this for checking whether the quotient is Hausdorff or not. – Watson Feb 3 '18 at 13:34
• @Watson: Thank you for your respond but my problem that I can't define the quotient space.This is the main problem. – Metso Feb 3 '18 at 13:36
• It is the set of similarity classes in $\mathcal M_2(\mathbf R)$. – Bernard Feb 3 '18 at 13:48

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.

• Great answer. +1 – DonAntonio Feb 3 '18 at 14:07
• @DonAntonio Thank you. Coming from you, this means a lot. – José Carlos Santos Feb 3 '18 at 14:32

Hint:

the possible real Jordan forms of matrices in $M_2(\mathbb{R})$ are: $$\begin{bmatrix}a&0\\ 0&b \end{bmatrix} \quad \begin{bmatrix}a&1\\ 0&a \end{bmatrix} \quad \begin{bmatrix}a&-b\\ b&a \end{bmatrix}$$ and:

two real matrices are similar iff they have the same real Jordan form.

Can you find the quotient space from this?

• That third matrix isn't in JNF: it isn't triangular. – DonAntonio Feb 3 '18 at 14:05
• @Emilio Novati: not really:( – Metso Feb 3 '18 at 14:11
• Since by condition the matrix it is a reversible, and therefore non-degenerate, it can be regarded as the matrix of a transition from one basis in $\mathbb{R}^2$ to another. Since all such matrices are considered, I assume that the matrices of the same operator in all possible bases will be in the equivalence class. Thus, each class of equivalence is associated with an operator. As a result, we obtain a half-plane bounded by the line $y=x$ – Metso Feb 3 '18 at 14:15
• At the decision I have designated an axis abscissa $\sigma_1$, ordinate axis - $\sigma_2$ then I have a half-plane $S_1 = \left\lbrace\ -\infty < \sigma_1 < +\infty, \sigma_1 \leqslant \sigma_2 < +\infty \right\rbrace$ – Metso Feb 3 '18 at 14:18
• Well, this is the form of a matrix that has complex eigenvalues, so that it cannot be similar to a real triangular matrix. As far as I know also this form is called ''real Jordan form''. You can see the link in my answer or also : people.math.osu.edu/costin.10/5101/Eigenvalues%20p20-30.pdf – Emilio Novati Feb 3 '18 at 14:21