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.

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

  • $\begingroup$ 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. $\endgroup$ – The way of life May 14 at 20:37
  • $\begingroup$ 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 $\sqrt2\left\lvert\frac t2-1\right\rvert>0$. $\endgroup$ – José Carlos Santos May 14 at 20:42
  • $\begingroup$ 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? $\endgroup$ – The way of life May 15 at 4:35
  • $\begingroup$ I suppose so, but I didn't try it. $\endgroup$ – José Carlos Santos May 15 at 5:49

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