(Last edited Apr 12, 2025.)

In my previous post on classifying isometries in $\mathbb{R}^2$, I proved that any linear isometry $T: \mathbb{R}^2\to\mathbb{R}^2$ satisfies \[\langle v, w\rangle = \langle T(v), T(w)\rangle\] for all $v, w\in\mathbb{R}^2$, where $\langle\cdot, \cdot\rangle$ is the dot product on $\mathbb{R}^2$. In other words, $T$ preserves the angle between any pair of vectors $v, w\in\mathbb{R}^2$.

I will now show that the same result holds if we replace $\mathbb{R}^2$ with any vector space $V$ that has an inner product. Let $V$ be an inner product space over $\mathbb{R}$ or $\mathbb{C}$, and let $T: V\to V$ be a linear isometry. Then for all $v, w\in V$, we have that \[||v + w||^2 = ||T(v + w)||^2 = ||T(v) + T(w)||^2.\] Expanding both sides gives \[||v||^2 + \langle v, w\rangle + \langle w, v\rangle + ||w||^2 = ||T(v)||^2 + \langle T(v), T(w)\rangle + \langle T(w), T(v)\rangle + ||T(w)||^2.\] Since $||v|| = ||T(v)||$ and $||w|| = ||T(w)||$, we have that \[\langle v, w\rangle + \langle w, v\rangle = \langle T(v), T(w)\rangle + \langle T(w), T(v)\rangle.\]

If $V$ is a vector space over $\mathbb{R}$, then the inner product is symmetric, so \[2\langle v, w\rangle = 2\langle T(v), T(w)\rangle\] and hence $\langle v, w\rangle = \langle T(v), T(w)\rangle$.

If $V$ is a vector space over $\mathbb{C}$, then the inner product is conjugate-symmetric, which means \[\langle v, w\rangle + \overline{\langle v, w\rangle} = \langle T(v), T(w)\rangle + \overline{\langle T(v), T(w)\rangle}.\] Hence, \[2\text{Re}(\langle v, w\rangle) = 2\text{Re}(\langle T(v), T(w)),\] so \[\text{Re}(\langle v, w\rangle) = \text{Re}(\langle T(v), T(w)\rangle).\] This equation holds for all vectors $v, w\in V$. Since $-iv$ is also an element of $V$, we can substitute $-iv$ for $v$ to get \[\text{Re}(\langle -iv, w\rangle) = \text{Re}(\langle T(-iv), T(w)\rangle).\] This simplifies to \[\text{Re}(-i\langle v, w\rangle) = \text{Re}(-i\langle T(v), T(w)\rangle).\] Since $\text{Re}(-iz) = \text{Im}(z)$ for any complex number $z$, it follows that \[\text{Im}(\langle v, w\rangle) = \text{Im}(\langle T(v), T(w)\rangle).\] Therefore, $\langle v, w\rangle = \langle T(v), T(w)\rangle$ since their real and imaginary parts are equal.