We’ve studied special classes of linear maps over an inner product space to itself like normal maps and self-adjoint maps. Today we will study a new special class

Definition

$T:V \rightarrow V$ is an isometry if $$ \begin{align*} \langle T(x), T(y) \rangle = \langle x, y \rangle \end{align*} $$

For example, in $\mathbf{R}^3$, this means that isometries preserve lengths and angles.
When $V$ is over $\mathbf{C}$, an isometry is sometimes called unitary while if $V$ is over $\mathbf{R}$, $T$ is called orthogonal.

Example 1

Let $T = L_A$ where

$$ \begin{align*} A = \begin{pmatrix} \cos\theta & - \sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \end{align*} $$

Remember from the previous lecture, $A$ was normal but not self-adjoint. We claim that $A$ is an isometry. We can verify this by evaluating

$$ \begin{align*} L_A \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} x\cos\theta - y\sin\theta \\ x\sin\theta + y\cos\theta \end{pmatrix} \end{align*} $$

while,

$$ \begin{align*} \left\langle L_A \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}, L_A \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} \right\rangle &= (x_1\cos\theta - y_1\sin\theta)(x_2\sin\theta + y_2\cos\theta) \\ & = x_1y_1 + x_2y_2 \end{align*} $$

Example 2

Let

$$ \begin{align*} V &= C^0([0,1]), \mathbf{C}) \\ &= \{ F(t) = f(t) + ig(t) \ | \ f, g \in C^0([0,1)]\} \\ \langle F, G \rangle &= \int_0^1 F(t)\overline{G(t)} dt \end{align*} $$

Suppose that $H \in V$ such that $|H(t)| = 1$. For real numbers, we have only two functions 1 and -1 but for complex numbers, we have many of these functions. For example $H(t) = e^{ih(t)}$.

Now we’re going to use this function to define an isometry. Specifically,

$$ \begin{align*} T \ : \ &V \rightarrow V &F \rightarrow FH \end{align*} $$

This is an isometry. $H$ is kind of rotating by some angle depending on $t$. To verify this,

$$ \begin{align*} \langle T(F), T(G) \rangle &= \int_0^1 (F(t)H(t))(\overline{G}(t)\overline{H}(t)) \ dt \\ &= \int_0^1 F(t)\overline{G}(t) \ dt \\ &= \langle F, G \rangle \end{align*} $$

Equivalent Conditions for Isometry

What can we say about Isometries? What does being an Isometry imply?

Theorem

The following are equivalent

$T: V \rightarrow V$ is an isometry
$TT^* = T^*T = I_V $ (an isometry is a normal map)
If $\beta$ is an orthonormal basis, then so is $T(\beta)$. [In other words, $T$ is an Isometry if it takes one orthonormal basis to another orthonormal basis].
$T(\beta)$ is orthonormal for some orthonormal $\beta$. [This is a weaker statement. $T$ needs to take one orthonormal basis to another at a minimum once (not every)].
$\Vert T(x) \Vert = \Vert x \Vert \ \forall x \in V $. So $T$ preserves lengths.

Why is $(e)$ true? to see why, we need the following lemma

Lemma

Suppose $S:V \rightarrow V$ is self-adjoint. If $\langle S(x), x \rangle = 0$ for all $x \in V$, then $S = T_{0_V}$ (the zero map)

Proof

Let $\beta = \{v_1,...,v_n\}$ be an orthonormal basis consisting of eigenvectors of $S$ (this is possible because $S$ is self adjoint). So $S(v_j) = \lambda_j v_j$ for $j = 1,...,n$. Then

$$ \begin{align*} 0 &= \langle S(v_j), v_j \rangle \quad \text{(by the given assumption)} \\ 0 &= \langle \lambda_j v_j, v_j \rangle \\ 0 &= \lambda_j \langle v_j, v_j \rangle \\ 0 &= \lambda_j \Vert v_j \Vert^2 \\ \end{align*} $$

$v_j$ is a basis vector so it’s not zero. Therefore, we must have $\lambda_j = 0$ for $j = 1,...,n$. This implies that $S$ is the zero map because all the eigenvalues are zero but we have a basis of eigenvectors. So every vector $x$ can be written as a linear combination of the basis vectors and when we apply $S$, we get a linear combination of $\lambda_j v_j$ where all the $\lambda_j$’s are zero so $S$ must be the zero vector.

Proof of Theorem

We’re ready to prove the theorem.

$(a) \implies (b):$
We’re given that $T$ is an isometry so we know by definition that $T$ preserves the inner product so $\langle T(x), T(y) \rangle = \langle x, y \rangle$. We want to show that $TT^* = T^*T$. We first observe that $T^*T$ is also self-adjoint (which means it’s diagonalizable). To see this, see that the adjoint of $(T^*T)$ is $(T^*T)^* = T^*(T^*)^* = T^*T$. Since $T^*T$ is self-adjoint, then we can add to it any multiple of the identity map and the new map is still self adjoint (proof is in last lecture?). So $T^*T - I_V$ is self adjoint. Now observe

$$ \begin{align*} \langle (T^*T - I_V)(x), x \rangle &= \langle T^*T(x) - x, x \rangle \\ &= \langle T^*T(x), x \rangle - \langle x, x \rangle \\ &= \langle T(x), T(x) \rangle - \langle x, x \rangle \quad \text{(by definition of adjoint)}\\ &= 0 \text{(Since $T$ is an Isometry)}\\ \end{align*} $$

So we showed that $T^*T - I_V$ is a self adjoint map where for any vector $x$, we have $\langle (T^*T - I_V)(x), x \rangle - 0$. So this satisfies the lemma we proved previously. This implies that $T^*T - I_V$ is the zero map. So

$$ \begin{align*} T^*T - I_V = T_0 \\ T^*T = I_V \\ T^* = T^{-1} \end{align*} $$

And we are done.

$(b) \implies (c):$
Let $\beta = \{v_1,...,v_n\}$ be an orthonormal basis where $\langle v_i, v_j \rangle = \delta_{ij}$. We know $T(\beta) = \{T(v_1),...,T(v_n)\}$. We need to show that $T(\beta)$ is an orthonormal basis which means that $\langle T(v_i), T(v_j) \rangle = \delta_{ij}$. Now observe

$$ \begin{align*} \langle (T(v_i), T(v_j) \rangle &= \langle v_i, T^*(T(v_j)) \rangle \\ &= \langle v_i, v_j \rangle \quad \text{(by (b), $T^*T = I_V$)} \\ &= \delta_{ij} \end{align*} $$

$(c) \implies (d):$ This is trivial since $c$ is a stronger statement

$(d) \implies (e):$
We need to show that if there is some orthonormal basis that can be taken to another orthonormal basis, then this means that $T$ preserves all lengths.

Let $\beta = \{v_1,...,v_n\}$ be an orthonormal basis with $T(\beta) = \{T(v_1),...,T(v_n)\} = \{w_1, ..., w_n\}$ also an orthonormal basis where $\langle v_i, v_j \rangle = \delta_{ij} = \langle w_i, w_j \rangle$. All these assumptions are by (d). Now we know to prove that $\Vert T(x) \Vert = \Vert x \Vert$ for any $x \in V$. Observe that

$$ \begin{align*} x &= \sum_j a_jv_j \\ T(x) &= \sum_j a_j T(v_j) = \sum_j a_j w_j \end{align*} $$

So now we use this to evaluate

$$ \begin{align*} \Vert T(x) \Vert^2 &= \langle T(x), T(x) \rangle \\ &= \langle \sum_i a_i w_i, \sum_j a_j w_j \rangle \\ &= \sum_{i,j} a_i \bar{a_j} \langle w_i, w_j \rangle \quad \text{(the sums and constants come out)} \\ &= \sum_{i,j} a_i \bar{a_j} (\delta_{ij}) \\ &= \sum_{i} |a_i|^2 \quad \text{(because $\delta_{ij} = 0$ when $i \neq j$)} \\ &= \langle x,x \rangle \end{align*} $$

In fact if we had started with $\Vert T(x) \Vert^2$, we would arrive at the same result $\sum_{i} |a_i|^2$. So we’re done.

$(e) \implies (a):$
We want to show that if we preserve lengths, then we actually preserve all inner products so $\Vert T(x) \Vert = \Vert x \Vert \implies \langle T(x), T(x) \rangle = \langle x, x \rangle$. To show this, we can use the following trick

$$ \begin{align*} \Vert x + y \Vert^2 &= \langle x+y, x+y \rangle \\ &= \Vert x\Vert^2 + \langle x, y \rangle + \langle y, x \rangle + \Vert y \Vert^2 \\ &= \Vert x\Vert^2 + \langle x, y \rangle + \overline{\langle x, y \rangle} + \Vert y \Vert^2 \end{align*} $$

What is $\langle x, y \rangle + \overline{\langle x, y \rangle}$? It’s 2 times the real part of $\langle x, y \rangle + \overline{\langle x, y \rangle}$. I don’t understand why? [TODO]

$$ \begin{align*} 2Re\langle x, y \rangle &= \frac{1}{2}(\Vert x + y\Vert^2 - \Vert x\Vert^2 - \Vert y\Vert^2) \\ Im\langle x, y \rangle &= -\frac{1}{2}(\Vert ix + y\Vert^2 - \Vert x\Vert^2 - \Vert y\Vert^2) \end{align*} $$

Isometries Are Rare

From research, we know that Isometries are actually rare. One fact is that $T$ is an isometry of $\mathbf{R}^2$ if and only if

$$ \begin{align*} T = L_A \text{ for } A = \begin{pmatrix} \cos\theta & - \sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \text{ or } A = \begin{pmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{pmatrix} \end{align*} $$

So these are the only possibilities for isometries for $\mathbf{R}^2$.

References

Math416 by Ely Kerman