Isometry

Suppose that $V$ and $W$ are inner product spaces. An isometry is a linear map $T : V \to W$ such that $\forall v_{1}, v_{2} \in V$ :

⟨ T (v_{1}), T (v_{2}) ⟩_{W} = ⟨ v_{1}, v_{2} ⟩_{V}

If $| | \cdot | |$ is the induced norm, then an isometry satisfies:

| | T (v) | |_{W}^{2} = ⟨ T (v), T (v) ⟩_{W} = ⟨ v, v ⟩ = | | v | |^{2}

so $| | T (v) | | = | | v | |$

If $V$ is a finite inner product space and $T : V \to V$ is an isometry

$T$ is injective: $v \in V s.t T (v) = 0$
$⟨ T (v), T (v) ⟩ = 0 = ⟨ 0, 0 ⟩ = ⟨ v, v ⟩ ⟹ v = 0$
by rank nullity it is also surjective

$T : V \to V$ is an isometry where $V$ is finite dimensional

$T$ is an isometry
$\forall$ orthonormal basis $B of V$ such that $T (B)$ is also an orthonormal basis
$\exists$ and orthonormal basis $B$ of $V$ such that $T (B)$ is also an orthonormal basis
$\forall$ $v \in V$ $| | T (v) | | = | | v | |$

$1 \to 2$
$B = {b_{1}, \dots, b_{n}}$
$⟨ T (b_{i}), T (b_{j}) ⟩ = ⟨ b_{i}, b_{j} ⟩ = 0$ if $i = j, 1 otherwise$

$2 \to 3$ immediately follows
$3 \to 4$

\begin{aligned} ⟨ T (v), T (v) ⟩ = ⟨ \sum_{i = 1}^{n} T (c_{i} b_{i}), \sum_{j = 1}^{n} T (c_{j} b_{j}) ⟩ & = \sum_{i = 1}^{n} \sum_{i = j}^{n} c_{i} c_{j} b_{i} b_{j} \\ = \sum_{i = 1}^{n} c_{i}^{2} = | | T (v) | |^{2} \end{aligned}

\begin{array}{r} ⟨ v, v ⟩ = ⟨ \sum_{i = 1}^{n} c_{i} b_{i}, \sum_{i = j}^{n} c_{j} b_{j} ⟩ = \sum_{j = 1}^{n} \sum_{i = 1}^{n} c_{j} c_{i} b_{j} b_{i} = \sum_{i = 1}^{n} c_{i}^{2} = | | v | |^{2} \end{array}

$4 \to 1$

\begin{align} &\lvert \lvert T(v)+T(w) \rvert \rvert=\lvert \lvert T(v+w) \rvert \rvert=\lvert \lvert v+w \rvert \rvert \text{ by assumption} \\ \\ \langle T(v)+T(w),T(v)+T(w) \rangle &=\langle T(v),T(v)+T(w) \rangle +\langle T(w),T(v)+T(w) \rangle \\ &=\langle T(v),T(v) \rangle +\langle T(v),T(w) \rangle +\langle T(w),T(v) \rangle+\langle T(w),T(w) \rangle \\ &=\lvert \lvert T(v) \rvert \rvert {#2} +2\langle T(w),T(v) \rangle +\lvert \lvert T(w) \rvert \rvert {#2} \\ &=\lvert \lvert v \rvert \rvert {#2} +2\langle T(v),T(w) \rangle +\lvert \lvert w \rvert \rvert {#2} \\ \\ &=\langle v+w,v+w \rangle =\lvert \lvert v \rvert \rvert {#2} +2\langle v,w \rangle +\lvert \lvert w \rvert \rvert {#2} \\ &\implies \langle T(v),T(w) \rangle =\langle v,w \rangle \end{align}

If $(V, ⟨ \cdot, \cdot ⟩)$ is finite and $T : V \to V$ is linear the following are equivalent:

$1 \to 2$
since $B$ is orthonormal then $C_{B}$ is an isometry, we have

[T (b_{i})]_{B} \cdot [T (b_{j})]_{B} = ⟨ C_{B} (T (b_{i})), C_{B} (T (b_{j})) ⟩ = ⟨ T (b_{i}), T (b_{j}) ⟩ = ⟨ b_{i}, b_{j} ⟩

$2 \to 3$ trivial
$3 \to 1$

⟨ C_{B} (T (b_{i}), C_{B} (T (b_{j})) ⟩ = 0 or 1 ⟹ ⟨ T (b_{i}), T (b_{j}) ⟩ = 0 or 1

so orthonormal

If $B$ is orthonormal then the fact that $C_{B}$ is an isometry comes from:

⟨ w, v ⟩_{V} = ⟨ \sum_{i = 1}^{n} c_{i} b_{_{i}}, \sum_{j = 1}^{n} c_{j} b_{j} ⟩ = \sum_{i = 1}^{n} c_{i} ⟨ b_{i}, \sum_{i = j}^{n} c_{j} b_{j} ⟩ = \sum_{i = 1}^{n} c_{i} c_{j} = ⟨ C_{B} (w), C_{B} (v) ⟩_{E u c}