What are all angle preserving linear operators on $\mathbb R^n$?

Question

I´m working on Spivak's Calculus on Manifolds and I met this exercise. My immediate answer was 'all the rotations' but I can't explain why. Am I right? Can you give a hint or something to be able to answer this question in a more formal way?

Answer 1

In $\mathbb{R}^2$ we can see that rotations preserve angles, but then so do reflections and uniform expansions. Note that the rotation transformation $T(x,y)=(x \cos \theta, y \sin \theta)$ is an orthogonal transformation and that the expansion and reflection transformations essentially boil down to $T(v_1,v_2)=(\lambda_1 v_1, \lambda_2 v_2)$, where $\{ v_1, v_2 \}$ is a basis of $\mathbb{R}^2$ and $|\lambda_1|=|\lambda_2|$. Hence we can conjecture that a transformation $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is angle-preserving precisely when $kT:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is orthogonal, for some $k \in \mathbb{R}$. Note that $\mathbb{R}^2$ is not playing any special role in this conjecture any more, so we can go ahead and attempt a proof for $\mathbb{R}^n$.

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Proof:

Let $T:\mathbb{R}^n \rightarrow \mathbb{R}^n$ and $k \in \mathbb{R}$ such that $kT$ is orthogonal. Let $x, y \in \mathbb{R}^n$. Then, $$ \arccos\left(\frac{\big\langle T(x), T(y) \big\rangle}{\|T(x)\| \, \|T(y)\|}\right) = \arccos\left(\frac{\big\langle kT(x), kT(y) \big\rangle}{\|kT(x)\| \, \|kT(y)\|}\right) = \arccos\left(\frac{\big\langle x, y \big\rangle}{\|x\| \, \|y\|}\right) $$ because $\|kT(x)\|^2=\big\langle kT(x), kT(x) \big\rangle = \langle x, x\rangle = \|x\|^2$, so $\|kT(x)\|=\|x\|$ for all $x$. Hence, $T$ is angle-preserving.

Conversely, suppose that $T$ is angle-preserving. Let $\{ e_1, \dots, e_n \}$ be the standard basis for $\mathbb{R}^n$. Then, $$ \arccos\left(\frac{\big\langle T(e_i), T(e_j) \big\rangle}{\|T(e_i)\| \, \|T(e_j)\|} \right) =\arccos\left(\frac{\big\langle e_i, e_j \big\rangle}{\|e_i\| \, \|e_j\|} \right) $$ $$ \Rightarrow \big\langle T(e_i), T(e_j) \big\rangle = \|T(e_i)\| \, \|T(e_j)\| \, \big\langle e_i, e_j \big\rangle = \|T(e_i)\| \, \|T(e_j)\| \, \delta_{ij} $$

Now, let $x=(x_1,\dots,x_n)$ and $y=(y_1,\dots,y_n)$. Then, $$ \begin{align} \big\langle T(x), T(y) \big\rangle &= \Big\langle \sum_{i=1}^{n} x_i T(e_i), \sum_{j=1}^{n} y_j T(e_j) \Big\rangle \\ &= \sum_{i=1}^{n} \sum_{j=1}^{n} x_i y_j \big\langle T(e_i), T(e_j) \big\rangle \\ &= \sum_{i=1}^{n} \sum_{j=1}^{n} x_i y_j \|T(e_i)\| \, \|T(e_j)\| \, \delta_{ij} \\ &= \sum_{i=1}^{n} x_i y_i \| T(e_i) \|^2 \end{align} $$

Now, taking a hint from rschwieb's answer to a similar question: Action of angle-preserving linear transformation on basis vectors, we can conclude that $\|T(e_i)\|=\|T(e_j)\|=\lambda$ (say). If $k=1/\lambda$, then $\big\langle kT(x), kT(y) \big\rangle = \big\langle x, y \big\rangle.$

Hence, proved.