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

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.

• FYI: use \\ to make a new row in a matrix, and don't put it after the last row. – kccu May 14 at 20:16
• @kccu: thank you. I just couldn't make it show properly. – The way of life May 14 at 20:17

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}

• Thank you. I like the approach using trace. What happens if instead of $h=\begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix}$ we would take a positive trace orthogonal map? Say a rotation by $\frac{\pi}{3}$. Does the method fail or can it be generalized? If you think it is worthy of a new question I will open a new one. – The way of life May 14 at 20:37
• If your matrix has trace $t\neq$, then, by the same argument, the distance from a conjugate of such a matrix to $\operatorname{Id}$ will be at least $\sqrt2\left\lvert\frac t2-1\right\rvert>0$. – José Carlos Santos May 14 at 20:42
• I see, I proved your claim. This is actually very nice. I guess that a similar calculation works for any $n\times n$ matrix of trace not equal to $n$ by literally the same argument (after diagonalization, and with a little more algebra), am I right? – The way of life May 15 at 4:35
• I suppose so, but I didn't try it. – José Carlos Santos May 15 at 5:49