Linear-Algebra-Done-Right · Unit 20 · Video 1 · Interactive Practice

Isometries: Five Equivalent Ways to Preserve Length

IKey Formulas

Formula	Name	What it says
$\\|Sv\\| = \\|v\\|$ for every $v \in V$	Isometry	$S \in \mathcal{L}(V, W)$ , $V$ and $W$ inner product spaces; $W$ may differ from $V$ , and $S$ need not be onto: $(x, y) \mapsto (x, y, 0)$ is an isometry $\mathbf{R}^2 \to \mathbf{R}^3$
$Sv = 0 \Rightarrow \\|v\\| = \\|Sv\\| = 0 \Rightarrow v = 0$	Isometries are injective	For an isometry $S$ : $\operatorname{null} S = \{0\}$ , so $\dim V \le \dim W$
$Se_k = g_k \Rightarrow \\|Sv\\|^2 = \sum_{k=1}^n \|\langle v, e_k \rangle\|^2 = \\|v\\|^2$	Building isometries	$e_1, \dots, e_n$ orthonormal basis of $V$ and $g_1, \dots, g_n$ orthonormal list in $W$ : $S$ keeps the coefficients $\langle v, e_k \rangle$ , so $S$ is an isometry
(a) $S$ isometry $\iff$ (b) $S^*S = I$ $\iff$ (c) $\langle Su, Sv \rangle = \langle u, v \rangle$ for all $u, v \in V$ $\iff$ (d) $Se_1, \dots, Se_n$ orthonormal $\iff$ (e) columns of $\mathcal{M}\big(S, (e_1, \dots, e_n), (f_1, \dots, f_m)\big)$ orthonormal in $\mathbf{F}^m$	Characterizations of isometries	$V$ , $W$ finite-dimensional inner product spaces, $S \in \mathcal{L}(V, W)$ ; $e_1, \dots, e_n$ orthonormal basis of $V$ ; $f_1, \dots, f_m$ orthonormal basis of $W$ ; Euclidean inner product on $\mathbf{F}^m$

Key Insight: Unit-length images are not enough: (d) and (e) also demand orthogonality, and that is exactly what preserves the length of a sum such as $e_1 + e_2$ .

IIOrthonormal Columns Preserve Every Length

An operator on $\mathbf{R}^2$ preserves all lengths exactly when its columns $Se_1$ , $Se_2$ are orthonormal.

IIIFive Implications Prove All Twenty

Following the cycle (a) ⇒ (b) ⇒ (c) ⇒ (d) ⇒ (e) ⇒ (a) leads from any condition to any other.

from to

IV $A^{\mathrm{T}}A$ Records the Column Inner Products

For real $A$ and $v = (x, y)$ , $\|Av\|^2$ is assembled from the entries of $A^{\mathrm{T}}A$ , the inner products of the columns.

Matrix x 2 y −1

💡 With respect to orthonormal bases, $\mathcal{M}(S^*) = \mathcal{M}(S)^{\mathrm{T}}$ for real matrices, so $A^{\mathrm{T}}A = I$ is condition (b) and its entries are condition (e).

VQuiz Questions

Problem 1 · Length of an Image

Given: $S \in \mathcal{L}(\mathbf{R}^2, \mathbf{R}^3)$ has matrix $A = \begin{pmatrix} 3/5 & 0 \\ 4/5 & 0 \\ 0 & 1 \end{pmatrix}$ with respect to the standard bases (Euclidean inner products). Find $\|S(2, -1)\|$ .

✅ Correct!

S(2, -1) = (\tfrac{6}{5}, \tfrac{8}{5}, -1)

and

\tfrac{36}{25} + \tfrac{64}{25} + 1 = 5

, which is

\|(2, -1)\|^2

: the columns are orthonormal, so

S

is an isometry.

❌ Not quite. That is

\|(6, 8, -1)\|

: the factor

\tfrac{1}{5}

in the first column was dropped. The first column is

(\tfrac35, \tfrac45, 0)

, a unit vector.

❌ Not quite.

3 = 2\|c_1\| + 1 \cdot \|c_2\|

adds the lengths of the two pieces. Lengths of orthogonal vectors combine through squares:

\|2c_1 - c_2\|^2 = 4\|c_1\|^2 + \|c_2\|^2 = 5

❌ Not quite. This divides the third entry by

5

as well, giving

(\tfrac65, \tfrac85, -\tfrac15)

. The second column is

(0, 0, 1)

; only the first column carries the

\tfrac15

Show solution

The columns are $c_1 = (\tfrac35, \tfrac45, 0)$ and $c_2 = (0, 0, 1)$ , so

S(2, -1) = 2c_1 - c_2 = \left(\tfrac{6}{5}, \tfrac{8}{5}, -1\right), \qquad \|S(2, -1)\|^2 = \tfrac{36}{25} + \tfrac{64}{25} + 1 = 4 + 1 = 5.

So $\|S(2, -1)\| = \sqrt{5} = \|(2, -1)\|$ . This agrees with condition (e): $\|c_1\|^2 = \tfrac{9 + 16}{25} = 1$ , $\|c_2\| = 1$ and $\langle c_1, c_2 \rangle = 0$ , so $S$ is an isometry and every length is preserved.

Problem 2 · Which Matrix Gives an Isometry?

Given: each matrix is $\mathcal{M}(S)$ for some $S \in \mathcal{L}(\mathbf{R}^2, \mathbf{R}^3)$ with respect to the standard bases (Euclidean inner products). Which $S$ is an isometry?

✅ Correct!

\|c_2\|^2 = \tfrac{16 + 9 + 25}{50} = 1

\|c_1\| = 1

and

\langle c_1, c_2 \rangle = \tfrac{-12 + 12 + 0}{25\sqrt{2}} = 0

: orthonormal columns, so condition (e) holds.

❌ Not quite. Both columns are unit vectors, but

\langle c_1, c_2 \rangle = \tfrac{12 + 12}{25} = \tfrac{24}{25} \neq 0

. Then

S(1, -1) = c_1 - c_2 = (-\tfrac15, \tfrac15, 0)

has

\|S(1, -1)\|^2 = \tfrac{2}{25}

, while

\|(1, -1)\|^2 = 2

❌ Not quite. The columns are orthogonal, but

\|c_1\| = \|(3, 4, 0)\| = 5

, so

\|S(1, 0)\| = 5 \neq 1 = \|(1, 0)\|

. Orthonormal means orthogonal and unit length.

❌ Not quite. The columns are orthogonal (

\tfrac{12 - 12 + 0}{25} = 0

), but

\|c_2\|^2 = \tfrac{16 + 9 + 1}{25} = \tfrac{26}{25}

, so

\|S(0, 1)\| = \tfrac{\sqrt{26}}{5} \neq 1

Show solution

By (a) $\iff$ (e) with the standard (orthonormal) bases, $S$ is an isometry exactly when the two columns form an orthonormal list in $\mathbf{R}^3$ . Check each matrix's columns $c_1$ , $c_2$ :

(a): $\|c_1\| = \|c_2\| = 1$ , but $\langle c_1, c_2 \rangle = \tfrac{24}{25}$ — not orthogonal.
(b): $\langle c_1, c_2 \rangle = 0$ , but $\|c_1\| = 5$ — not unit.
(c): $\langle c_1, c_2 \rangle = 0$ , but $\|c_2\|^2 = \tfrac{26}{25}$ — not unit.
(d): $\|c_1\|^2 = \tfrac{9 + 16}{25} = 1$ , $\|c_2\|^2 = \tfrac{16}{50} + \tfrac{9}{50} + \tfrac{1}{2} = 1$ , $\langle c_1, c_2 \rangle = \tfrac{1}{5 \cdot 5\sqrt{2}}(-12 + 12) + 0 = 0$ — orthonormal.

Only (d) defines an isometry.

Problem 3 · Completing an Isometry

Given: $S \in \mathcal{L}(\mathbf{R}^2)$ has matrix $\begin{pmatrix} 3/5 & p \\ 4/5 & q \end{pmatrix}$ with respect to the standard basis (Euclidean inner product), where $p, q \in \mathbf{R}$ . Find all $(p, q)$ for which $S$ is an isometry.

✅ Correct! The second column must be a unit vector orthogonal to

(\tfrac35, \tfrac45)

, and the line orthogonal to it meets the unit circle in exactly two points: a rotation and a reflection.

❌ Not quite.

(-\tfrac45, \tfrac35)

works (a rotation), but so does its negative

(\tfrac45, -\tfrac35)

\tfrac35 \cdot \tfrac45 + \tfrac45 \cdot (-\tfrac35) = 0

and its length is

1

. That

S

is a reflection, and reflections preserve length too.

❌ Not quite.

(\tfrac45, \tfrac35)

is a unit vector, but

\langle (\tfrac35, \tfrac45), (\tfrac45, \tfrac35) \rangle = \tfrac{24}{25} \neq 0

, so the columns are not orthogonal and (e) fails.

❌ Not quite. Unit length alone is not enough. For

(p, q) = (0, 1)

the columns have inner product

\tfrac45 \neq 0

, and

\|S(1, -1)\|^2 = \|(\tfrac35, -\tfrac15)\|^2 = \tfrac{2}{5} \neq 2

Show solution

By (a) $\iff$ (e), $S$ is an isometry exactly when the columns are orthonormal. The first column $(\tfrac35, \tfrac45)$ already has length $1$ . The second must satisfy

\tfrac35 p + \tfrac45 q = 0 \quad\text{and}\quad p^2 + q^2 = 1.

The first equation gives $p = -\tfrac43 q$ ; substituting, $\tfrac{16}{9}q^2 + q^2 = \tfrac{25}{9}q^2 = 1$ , so $q = \pm\tfrac35$ and $p = \mp\tfrac45$ .

(p, q) = \left(-\tfrac45, \tfrac35\right) \quad\text{or}\quad (p, q) = \left(\tfrac45, -\tfrac35\right).

The first gives a rotation of $\mathbf{R}^2$ , the second a reflection; both are isometries.

Problem 4 · Inner Products Come Along for Free

Given: $S \in \mathcal{L}(\mathbf{R}^2, \mathbf{R}^3)$ is the isometry with matrix $\begin{pmatrix} 3/5 & 0 \\ 4/5 & 0 \\ 0 & 1 \end{pmatrix}$ with respect to the standard bases (Euclidean inner products). Find $\langle S(1, 2), S(3, -1) \rangle$ .

✅ Correct! By (a)

\Rightarrow

(c),

\langle S(1, 2), S(3, -1) \rangle = \langle (1, 2), (3, -1) \rangle = 3 - 2 = 1

; the direct computation

\tfrac{27}{25} + \tfrac{48}{25} - 2 = 1

agrees.

❌ Not quite.

\tfrac{27}{25} + \tfrac{48}{25} = 3

uses only the first two coordinates.

S(1, 2) = (\tfrac35, \tfrac45, 2)

and

S(3, -1) = (\tfrac95, \tfrac{12}{5}, -1)

, so the third coordinates add

2 \cdot (-1) = -2

❌ Not quite.

5\sqrt{2} = \sqrt{5}\,\sqrt{10} = \|(1, 2)\|\,\|(3, -1)\|

is the product of the lengths, which by Cauchy–Schwarz only bounds

|\langle u, v \rangle|

. Equality would need

(1, 2)

and

(3, -1)

to be parallel.

❌ Not quite.

\langle (1, 2), (3, -1) \rangle = 1 \cdot 3 + 2 \cdot (-1) = 3 - 2

; the second product is negative.

Show solution

The columns $(\tfrac35, \tfrac45, 0)$ and $(0, 0, 1)$ are orthonormal, so $S$ is an isometry, and by condition (c) it preserves inner products:

\langle S(1, 2), S(3, -1) \rangle = \langle (1, 2), (3, -1) \rangle = 3 - 2 = 1.

Check directly. $S(1, 2) = (\tfrac35, \tfrac45, 2)$ and $S(3, -1) = (\tfrac95, \tfrac{12}{5}, -1)$ , so

\langle S(1, 2), S(3, -1) \rangle = \tfrac{27}{25} + \tfrac{48}{25} - 2 = 3 - 2 = 1.

The angle between the two vectors is preserved as well, since both lengths and the inner product are.

Solved: 0 / 4

Isometries: Five Equivalent Ways to Preserve Length

IKey Formulas

IIOrthonormal Columns Preserve Every Length

IIIFive Implications Prove All Twenty

IVATAA^{\mathrm{T}}AATA Records the Column Inner Products

VQuiz Questions

IV $A^{\mathrm{T}}A$ Records the Column Inner Products