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 $\mathbb{R}^n$.

---

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.

Answer 2

I am slightly unsatisfied with my previous solution to this problem, so I give below another answer that is laid out in the same spirit as this answer by @JasonDeVito to Problem 1-8(b)^[1].

---

By Problem 1-8(a)^[1], we know that if $T \colon \mathbf{R}^n \to \mathbf{R}^n$ is norm preserving then it is angle preserving. Clearly, if $T$ is angle preserving, then so is $cT$ for any nonzero scalar $c$. We shall show that, conversely, for any angle preserving $T \colon \mathbf{R}^n \to \mathbf{R}^n$ there exists a nonzero scalar $c$ such that $cT$ is norm preserving. This will characterise angle preserving linear operators on $\mathbf{R}^n$.

Suppose that there exist unit vectors $\mathbf{x},\mathbf{y}$ with $\lVert T(\mathbf{x}) \rVert \neq \lVert T(\mathbf{y}) \rVert$. Then, on the one hand, we have $$ \langle \mathbf{x} + \mathbf{y}, \mathbf{x} - \mathbf{y} \rangle = \langle \mathbf{x}, \mathbf{x} \rangle - \langle \mathbf{y}, \mathbf{y} \rangle = 1 - 1 = 0, $$ so $\angle(\mathbf{x} + \mathbf{y},\mathbf{x} - \mathbf{y}) = \pi/2$. But, on the other hand, we have \begin{align*} \langle T(\mathbf{x} + \mathbf{y}), T(\mathbf{x} - \mathbf{y}) \rangle &= \langle T(\mathbf{x}), T(\mathbf{x}) \rangle - \langle T(\mathbf{y}), T(\mathbf{y})\rangle \\ &= \lVert T(\mathbf{x}) \rVert^2 - \lVert T(\mathbf{y}) \rVert^2\\ &\neq 0. \end{align*} Also note that $T(\mathbf{x} + \mathbf{y})$ and $T(\mathbf{x} - \mathbf{y})$ are nonzero vectors. Hence, we can talk about $$ \frac{\langle T(\mathbf{x} + \mathbf{y}), T(\mathbf{x} - \mathbf{y}) \rangle}{\lVert T(\mathbf{x} + \mathbf{y}) \rVert \cdot \lVert T(\mathbf{x} - \mathbf{y})\rVert}. $$ Since this is nonzero, it implies that $\angle(T(\mathbf{x} + \mathbf{y}),T(\mathbf{x} - \mathbf{y})) \neq \pi/2$. Thus, $T$ is not angle preserving. Hence, if $T$ is angle preserving, then $\lVert T(\mathbf{x}) \rVert$ equals a constant, say $r$, for all unit vectors $\mathbf{x}$. But then, if $\mathbf{x}$ is any nonzero vector, then $r = \lVert T(\mathbf{x}/\lVert \mathbf{x} \rVert) \rVert = \lVert T(\mathbf{x}) \rVert/\lVert \mathbf{x} \rVert$. Hence, for $c := r^{-1}$, $cT$ is norm preserving.

Thus, the angle preserving linear transformations $T \colon \mathbf{R}^n \to \mathbf{R}^n$ are precisely those for which there exists a scalar $c$ such that $cT \colon \mathbf{R}^n \to \mathbf{R}^n$ is norm preserving.

---

Notes:

1. Here is the complete statement of the problem for ease of reference (Calculus on Manifolds, Michael Spivak, pages 4–5):

1-8. $\quad$ If $\mathbf{x},\mathbf{y} \in \mathbb{R}^n$ are non-zero, the angle between $\mathbf{x}$ and $\mathbf{y}$, denoted $\angle(\mathbf{x},\mathbf{y})$, is defined as $\arccos\left(\langle \mathbf{x}, \mathbf{y} \rangle / \lVert \mathbf{x} \rVert \cdot \lVert \mathbf{y} \rVert \right)$, which makes sense by Theorem 1-1(2)^[2]. The linear transformation $T$ is angle preserving if $T$ is $1$-$1$, and for $\mathbf{x},\mathbf{y} \neq \mathbf{0}$ we have $\angle(T(\mathbf{x}),T(\mathbf{y})) = \angle(\mathbf{x},\mathbf{y})$.

$\quad$ (a) Prove that if $T$ is norm preserving, then $T$ is angle preserving.

$\quad$ (b) Suppose there is a basis $\mathbf{x}_1, \dotsc, \mathbf{x}_n$ of $\mathbf{R}^n$ and numbers $\lambda_1, \dotsc, \lambda_n$ such that $T(\mathbf{x}_i) = \lambda_i \mathbf{x}_i$. Prove that if $T$ is angle preserving, then all $\lvert \lambda_i \rvert$ are equal.

$\quad$ (c) What are all angle preserving $T\colon\mathbf{R}^n \to \mathbf{R}^n$?

2. Theorem 1-1(2) (on page 2) says that if $\mathbf{x}, \mathbf{y} \in \mathbf{R}^n$, then $\lvert \sum_{i=1}^n x^i y^i \rvert = \lvert \mathbf{x} \rvert \cdot \lvert \mathbf{y} \rvert$.