Can you please give me a "good" reason that the following two matrices are not conjugate in $SL(2, \mathbb{R})$? I'm sure I could prove it with a computation but I'd like to know why they're not conjugate. For instance, I hope there is some sort of conjugate-invariant property of one that is not a property of the other. Thank you.
$$\left(\begin{matrix} 1&1\\ 0&1 \end{matrix}\right)$$
$$\left(\begin{matrix} 1&-1\\ 0&1\\ \end{matrix}\right) $$
Subtract the identity matrix from your two matrices, you are essentially asking why the $2\times2$ nilpotent Jordan block is not similar to its negative in $SL(2,\mathbb R)$. Here is one answer: for real $2\times2$ nilpotent matrices, the sign of the difference between the two antidiagonal elements is preserved by conjugation in $SL(2,\mathbb R)$.
Proof. Suppose $A\ne0$ is a real $2\times2$ nilpotent matrix. Since $SL(2,\mathbb R)$ is path-connected, if the difference between the two antidiagonal elements is not preserved by conjugation, by intermediate value theorem, the difference must be vanish by an appropriate conjugation in $SL(2,\mathbb R)$. As $A$ is traceless, it follows that for some $S$, $$SAS^{-1}=\pmatrix{x&y\\ y&-x}.$$ Yet $\det(A)=0$ because $A$ is nilpotent. Therefore $x=y=0$ and $A=0$, which is a contradiction.
The computation is of course the real proof, but as to the "why": These skew translation differ by the fact that one has a bit of a "left turn" in it and the other a "right turn". Such is not invariant under conjugation with general matrices, but it is if we conjugate with orientation preserving matrices (i.e with positive determinant, here specifically determinant $+1$).
