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.
$$\left(\begin{matrix}
1&1\\
0&1
\end{matrix}\right)$$
$$\left(\begin{matrix}
1&-1\\
0&1\\
\end{matrix}\right)
$$
Best Answer
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.