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Angle-Preserving Transformations

Recall that for any field \mathbb K, to construct the complex numbers \mathbb C_{\mathbb K}, we needed to define i = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} =: \sqrt{-1}. Geometrically, this means for any u_1,u_2 \in \mathbb K, i \begin{bmatrix} u_1 \\ u_2\end{bmatrix} = \begin{bmatrix} u_2 \\ -u_1 \end{bmatrix}.

Something curious happens in the case \mathbb K = \mathbb R. For any \mathbf u := \begin{bmatrix} u_1 \\ u_2 \end{bmatrix}, \mathbf v := \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} \in \mathbb R^2, we have

\begin{aligned} \langle i \mathbf u, i \mathbf v\rangle &= \left\langle \begin{bmatrix} -u_2 \\ u_1 \end{bmatrix}, \begin{bmatrix} -v_2 \\ v_1 \end{bmatrix} \right\rangle \\ &= (-u_2)(-v_2) + u_1 v_1 \\ &= u_1 v_1 + u_2 v_2 \\ &= \left\langle \begin{bmatrix} u_1 \\ u_2 \end{bmatrix}, \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} \right\rangle = \langle \mathbf u, \mathbf v\rangle. \end{aligned}

In other words, i : \mathbb R^2 \to \mathbb R^2 preserves angles. We generalise to other linear transformations.

Let V be an inner product space over \mathbb K, where \mathbb K equals \mathbb R or \mathbb C.

Definition 1. We say that a linear transformation T : V \to V, also known as a linear operator on V, is:

  • angle-preserving if for any \mathbf u, \mathbf v \in V, \langle T(\mathbf u), T(\mathbf v) \rangle = \langle \mathbf u, \mathbf v \rangle,
  • norm-preserving if for any \mathbf v \in V, \|T(\mathbf v)\| = \|\mathbf v\|.

It is clear that if T is angle-preserving, then it is norm-preserving. Does that hold in the opposite direction? More is true, in fact, as long as V has an orthonormal basis.

Theorem 1. Suppose V has an orthonormal basis \{\mathbf e_\alpha : \alpha \in I\}. The following are equivalent:

  • T is angle-preserving,
  • T is norm-preserving,
  • \{T (\mathbf e_\alpha) : \alpha \in I\} is orthonormal,
  • T^* = T^{-1}, i.e. T is unitary.

Proof. Suppose T is norm preserving. Fix \gamma \in \mathbb K to be tuned. Consider the expression

\| T(\mathbf e_\alpha + \gamma \mathbf e_\beta)\|^2 = \| \mathbf e_\alpha + \gamma \mathbf e_\beta\|^2.

Writing the norm as an inner product, the left-hand simplifies to

\begin{aligned} \|T(\mathbf e_\alpha)\|^2 + \bar{\gamma} \langle T(\mathbf e_\alpha), T(\mathbf e_\beta) \rangle +\gamma \overline{\langle T(\mathbf e_\alpha), T(\mathbf e_\beta) \rangle} + |\gamma|^2 \|T(\mathbf e_\beta)\|^2. \end{aligned}

Since \langle \mathbf e_\alpha, \mathbf e_\beta \rangle = 0, the right-hand side simplifies to

\begin{aligned} \|\mathbf e_\alpha\|^2 + |\gamma|^2 \|\mathbf e_\beta \|^2. \end{aligned}

Since T is norm-preserving, equating the expressions yields

\bar{\gamma} \langle T(\mathbf e_\alpha), T(\mathbf e_\beta) \rangle +\gamma \overline{\langle T(\mathbf e_\alpha), T(\mathbf e_\beta) \rangle} = 0.

By algebra, setting \gamma = \langle T(\mathbf e_\alpha), T(\mathbf e_\beta) \rangle and simplify

| \langle T(\mathbf e_\alpha), T(\mathbf e_\beta) \rangle |^2 = |\gamma|^2 = 0

so that \langle T(\mathbf e_\alpha), T(\mathbf e_\beta) \rangle = 0, as required. Furthermore,

\begin{aligned} (T^* \circ T)(\mathbf e_\alpha) &= \sum_{\beta} \langle (T^* \circ T)(\mathbf e_\alpha), \mathbf e_\beta\rangle \mathbf e_\beta \\ &= \sum_{\beta} \langle T^*(T(\mathbf e_\alpha)), \mathbf e_\beta\rangle \mathbf e_\beta \\ &= \sum_{\beta} \langle T(\mathbf e_\alpha), T(\mathbf e_\beta) \rangle \mathbf e_\beta \\ &= 1 \cdot \mathbf e_\alpha + \sum_{\alpha \neq \beta} 0 \cdot \mathbf e_\beta = \mathbf e_\alpha. \end{aligned}

Therefore, T^* = T^{-1}. Finally, under this assumption, for any \mathbf u, \mathbf v \in V,

\begin{aligned} \langle T(\mathbf u),T(\mathbf v)\rangle &= \langle \mathbf u, T^* ( T(\mathbf v) )\rangle \\ &= \langle \mathbf u, T^{-1} ( T(\mathbf v) ) \rangle \\ &= \langle \mathbf u, \mathbf v \rangle. \end{aligned}

In particular, T^* \circ T = I_V = T \circ T^*. In this case, we call T a normal operator.

Lemma 1. Unitary operators are normal. Furthermore, if T^* = T (i.e. T is self-adjoint), then T is normal.

Theorem 2. Let T : V \to V be a normal operator. The following propositions hold:

  • For any \mathbf u, \mathbf v, \langle T(\mathbf u),T(\mathbf v)\rangle = \langle T^*(\mathbf u),T^*(\mathbf v)\rangle.
  • If T(\mathbf v) = \lambda \mathbf v, then T^*(\mathbf v) = \bar{\lambda} \mathbf v.
  • If T(\mathbf u) = \lambda \mathbf u and T(\mathbf v) = \mu \mathbf v and \lambda \neq \mu, then \mathbf u \perp \mathbf v.

Proof. If T \circ T^* = T^* \circ T, then for any \mathbf u, \mathbf v,

\begin{aligned} \langle T(\mathbf u),T(\mathbf v)\rangle &= \langle \mathbf u,(T^* \circ T)(\mathbf v)\rangle \\ &= \langle \mathbf u,(T \circ T^*)(\mathbf v)\rangle \\ &= \langle T^*(\mathbf u), T^*(\mathbf v)\rangle .\end{aligned}

We leave it as an exercise to check that T - \lambda is normal. Then

\begin{aligned} \langle (T^* - \bar\lambda)(\mathbf v), (T^* - \bar\lambda)(\mathbf v)\rangle &= \langle (T - \lambda)^* (\mathbf v), (T -\lambda)^*(\mathbf v)\rangle \\ &= \langle (T - \lambda) (\mathbf v), (T -\lambda)(\mathbf v)\rangle = 0. \end{aligned}

Finally,

\lambda \langle \mathbf u, \mathbf v \rangle = \langle T(\mathbf u), \mathbf v\rangle = \langle \mathbf u, T^*(\mathbf v)\rangle =\mu \langle \mathbf u, \mathbf v \rangle

yields the desired result since \lambda - \mu \neq 0.

You may now be wondering: what on earth are we doing, and why? We have started from the notion of angle-preserving transformations, and ended up at normal operators—for what? Well, we want to prove a rather elegant result: any real symmetric matrix \mathbf A can be diagonalised! Furthermore, the matrix we use to diagonalise \mathbf A can be comprised of our favorite type of basis—an orthonormal basis.

—Joel Kindiak, 15 Mar 25, 1554H

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