Conjugation of $\left(\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix}\right)$

Question

I'm interested in the following question.

Let $h=\begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix}$. This is an orthogonal map which is quite far away from the identity (say in the Frobenius norm, or max norm).

I was wondering how close to the identity can the map $f=g^{-1}hg$ be. First, when $g$ is in $SO(2,\mathbb{R})$ it is clear that $f$ is always far from $I$ since $SO(2,\mathbb{R})$ is abelian. So I was wondering what happens if we let $g$ be in a larger group. Say $g\in SL(2,\mathbb{R})$. Does it allow for $f$ to be arbitrarily close to $I$? I couldn't think of any counterexample. I tried doing the direct computation, and I think it might show that indeed $f$ cannot be close to $I$. But I was looking for a less direct way to approach this.

Answer 1

No, you can't make it arbitrarily close to $\operatorname{Id}$. If a $2\times2$ matrix is conjugate to $\left[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right]$, then its trace is $0$. It is therefore of the form $\left[\begin{smallmatrix}a&b\\c&-a\end{smallmatrix}\right]$. And\begin{align}\left\lVert\begin{bmatrix}a&b\\c&-a\end{bmatrix}-\operatorname{Id}\right\rVert_F&=\sqrt{(a-1)^2+b^2+c^2+(-a-1)^2}\\&\geqslant\sqrt{(a-1)^2+(a+1)^2}\\&\geqslant\sqrt2.\end{align}