My first post on isometries showed that if $T$ is an isometry on $\mathbb{R}^2$ such that $T(0) = 0$, then $T$ an orthogonal linear transformation. We will now generalize this result to all inner product spaces.

I’ll briefly review the argument from the original post. We first proved this key lemma:

Lemma. If $v, w\in\mathbb{R}^2$ are nonzero and $||v + w|| = ||v|| + ||w||$, then there exists $c > 0$ such that $v = cw$.

Then, we proved that $T$ is linear—this portion of the argument only uses the lemma and the fact that $\mathbb{R}^2$ is a normed vector space. We could replace $\mathbb{R}^2$ with any normed vector space $V$ and the same argument would work (assuming that the Lemma holds if we replace $\mathbb{R}^2$ with $V$). Finally, we proved that every linear isometry on $\mathbb{R}^2$ is orthogonal, and we generalized this result to all inner product spaces in a follow-up post.

In fact, the only part of the proof that has not already been generalized to arbitrary inner product spaces is the Lemma. This post is dedicated to proving the generalized lemma.

Generalized Lemma. Let $V$ be an inner product space. Suppose $v, w\in V\setminus\{0\}$ satisfy $||v + w|| = ||v|| + ||w||$. Then there exists a positive scalar $c > 0$ such that $v = cw$.

Proof. Since $||v + w|| = ||v|| + ||w||$, we have

\begin{align*} ||v||^2 + 2\text{Re}(\langle v, w\rangle) + ||w||^2 &= ||v + w||^2 \\ &= (||v|| + ||w||)^2 \\ &= ||v||^2 + 2||v||\cdot ||w|| + ||w||^2, \end{align*}

so $\text{Re}(\langle v, w\rangle) = ||v||\cdot ||w||$. The Cauchy–Schwarz inequality for inner product spaces says that $|\langle v, w\rangle|\leq ||v||\cdot ||w||$, so $|\langle v, w\rangle|\leq \text{Re}(\langle v, w\rangle)$. But clearly

\begin{align*} |\langle v, w\rangle|^2 &= \text{Re}(\langle v, w\rangle)^2 + \text{Im}(\langle v, w\rangle)^2 \\ &\geq \text{Re}(\langle v, w\rangle)^2, \end{align*}

so $|\langle v, w\rangle| = \text{Re}(\langle v, w\rangle)$. Hence, $\text{Im}(\langle v, w\rangle) = 0$, so $\langle v, w\rangle = \text{Re}(\langle v, w\rangle) = ||v||\cdot ||w||$.

Since $v$ and $w$ are nonzero, we know that $||v||$ and $||w||$ are positive. Let $c = \frac{||v||}{||w||} > 0$. We will show that $v = cw$. Note that

\begin{align*} \langle cw, v - cw\rangle &= \langle cw, v\rangle - \langle cw, cw\rangle \\ &= c\langle w, v\rangle - c^2||w||^2 \\ &= c||v||\cdot ||w|| - c^2||w||^2 \\ &= c^2||w||^2 - c^2||w||^2\quad\text{(since $||v|| = c||w||$)}\\ &= 0. \end{align*}

Hence, $cw$ and $v - cw$ are orthogonal. The “Pythagorean Theorem” in inner product spaces says that if $\langle x, y\rangle = 0$, then $||x + y||^2 = ||x||^2 + ||y||^2$. Hence,

\begin{align*} ||v||^2 &= ||cw + (v - cw)||^2 \\ &= ||cw||^2 + ||v - cw||^2 \\ &= (c||w||)^2 + ||v - cw||^2 \\ &= ||v||^2 + ||v - cw||^2, \end{align*}

so $||v - cw||^2 = 0$. We conclude that $v - cw = 0$ and hence $v = cw$. QED.

This concludes the proof of our generalized theorem:

Theorem. Let $V$ be an inner product space and $f: V\to V$ be an isometry. Then the function $T: V\to V$ defined by $T(v) = f(v) - f(0)$ is an angle-preserving linear transformation.

In fact, we never use the assumption that $f$ maps $V$ to itself. This means we can generalize the theorem even further: if $V$ and $W$ are inner product spaces (over the same field $F\in\{\mathbb{R}, \mathbb{C}\}$) and $f: V\to W$ is an isometry, then $f-f(0)$ is linear and angle-preserving.