0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.