The quotient of M_n(R) by conjugation is not Hausdorff

Idea. The orbit of the Jordan block is not closed: its closure contains the orbit $\{I\}$ of the identity. In the quotient, any neighbourhood of $[I]$ then contains $[J]$.

Step 1: $J_t$ is conjugate to $J$. For $t\ne0$, let $D_t=\operatorname{diag}(t,1,\dots,1)$. Then
$$D_tJD_t^{-1}=\begin{pmatrix}1&t\\0&1\end{pmatrix}\oplus I_{n-2}=J_t.$$
(Check in the $2\times2$ block: $\begin{pmatrix}t&0\\0&1\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}1/t&0\\0&1\end{pmatrix}=\begin{pmatrix}t&t\\0&1\end{pmatrix}\begin{pmatrix}1/t&0\\0&1\end{pmatrix}=\begin{pmatrix}1&t\\0&1\end{pmatrix}$.) Also $J_t\to I$ as $t\to0$, and $J\ne I$ (and $J$ is not conjugate to $I$, since $PIP^{-1}=I$).

Step 2: neighbourhoods of $[I]$ contain $[J]$. Let $\pi:M_n\to M_n/GL_n$ be the quotient map and $W$ an open set of the quotient containing $[I]$. Then $\pi^{-1}(W)$ is open, contains $I$, and is a union of orbits. As $J_t\to I$, $J_t\in\pi^{-1}(W)$ for small $t\ne0$; since $J_t$ is conjugate to $J$ and $\pi^{-1}(W)$ is a union of orbits, $J\in\pi^{-1}(W)$, i.e. $[J]\in W$.

Step 3: conclusion. Every open neighbourhood of $[I]$ contains the different point $[J]$. So no open sets separate $[I]$ and $[J]$ (not Hausdorff), and there is no neighbourhood of $[I]$ missing $[J]$ (not $T_1$: the point $[J]$ is not closed, because every neighbourhood of $[I]$ meets $\{[J]\}$).

Remark. The quotient is well behaved on the open set of matrices with $n$ distinct real eigenvalues, where conjugacy classes are classified by the eigenvalues (and are closed). The "bad" behaviour comes from non-diagonalisable matrices: the closure of a conjugacy class contains matrices with the same characteristic polynomial but "more diagonal" Jordan form.

Lesson. Closures of orbits that are not closed destroy separation axioms in the quotient.