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

José Carlos Santos
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Metso
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2 Answers2

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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.

José Carlos Santos
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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?

Emilio Novati
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  • That third matrix isn't in JNF: it isn't triangular. – DonAntonio Feb 03 '18 at 14:05
  • @Emilio Novati: not really:( – Metso Feb 03 '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 03 '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 03 '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 : https://people.math.osu.edu/costin.10/5101/Eigenvalues%20p20-30.pdf – Emilio Novati Feb 03 '18 at 14:21
  • You can also see: https://math.stackexchange.com/questions/338277/equivalence-classes-of-similar-2-times-2-matrices – Emilio Novati Feb 03 '18 at 14:25
  • @EmilioNovati Even the link you sent says that third matrix is not in JCF: "For example, if M is a matrix in Jordan normal form, then it is block diagonal..." . It is the beginning of page 21 (second page in that paper). And real JCF is, as far as I know, a real matrix. It doesn't fit to complex valued matrices. – DonAntonio Feb 03 '18 at 16:19
  • Maybe that the name of that matrix is not ''Jordan'' ( but I've seen also this name used), but it is the ''canonical'' real form for matrices that are not triangularizable. – Emilio Novati Feb 03 '18 at 16:43
  • @EmilioNovati No, it is not. If the matrix isn't triangularizable then its form is like the second matrix in your post. Of course, for matrices $;2\times2;$ , the only such possibility involves a double eigenvalue with geometric multiplicity of 1. – DonAntonio Feb 03 '18 at 21:11