Linear-Algebra-Done-Right · Unit 29 · Video 4 · Interactive Practice

|det T| Is Volume: Unitary, Positive and Singular Values

IKey Formulas

Formula	Name	Hypotheses
$\|\det S\| = 1$	Unitary operators	$S$ unitary on a finite-dimensional inner product space ( $\mathbf{F} = \mathbf{R}$ or $\mathbf{C}$ )
$\det T \ge 0$	Positive operators	$T$ positive
$\|\det T\| = \sqrt{\det(T^*T)} = s_1 \cdots s_n$	Singular values	$s_1, \dots, s_n$ the singular values of $T \in \mathcal{L}(V)$
$\operatorname{volume} T(\Omega) = \|\det T\|\,\operatorname{volume} \Omega$	Volume scaling	$T \in \mathcal{L}(\mathbf{R}^n)$ , $\Omega \subseteq \mathbf{R}^n$

Key Insight: $T = S\sqrt{T^*T}$ with $S$ unitary: $S$ moves volume without changing it, and $\sqrt{T^*T}$ stretches along orthonormal axes by the singular values.

IIThe Unit Circle Becomes an Ellipse

$T$ maps the unit circle onto an ellipse with semi-axes $s_1, s_2$ , so every area is multiplied by $s_1 s_2 = |\det T|$ .

IIIUnitary and Positive Operators

Rotations and reflections preserve area ( $|\det| = 1$ ); a positive operator stretches along orthogonal axes by nonnegative factors ( $\det \ge 0$ ).

Operator θ 30° λ₁ 2 λ₂ 0.5

IVQuiz Questions

Problem 1 · Volume from Singular Values

Given: $T \in \mathcal{L}(\mathbf{R}^3)$ has singular values $3$ , $2$ and $\tfrac{1}{2}$ . Find the volume of $T(\text{unit cube})$ .

✅ Correct!

|\det T| = 3 \cdot 2 \cdot \tfrac{1}{2} = 3

, and the cube has volume

1

❌ Not quite. Volume factors multiply: use the product of the singular values, not their sum.

❌ Not quite.

9 = \det(T^*T)

, the product of the squares;

|\det T|

is its square root.

❌ Not quite. That is the factor for

T^{-1}

;

T

multiplies volume by

s_1s_2s_3

Show solution

|\det T| = s_1 s_2 s_3 = 3 \cdot 2 \cdot \tfrac{1}{2} = 3, \qquad \operatorname{volume} T(\text{cube}) = |\det T| \cdot 1 = 3.

Problem 2 · Possible Determinants of a Unitary Operator

Given: $S$ is a unitary operator on $\mathbf{C}^2$ with the standard inner product. Which numbers can $\det S$ be?

✅ Correct!

|\det S| = 1

, and every such number occurs:

\operatorname{diag}(e^{i\theta}, 1)

is unitary with determinant

e^{i\theta}

❌ Not quite. That is the real case. Over

\mathbf{C}

\operatorname{diag}(i, 1)

is unitary with determinant

i

❌ Not quite.

1 = \det(S^*S) = |\det S|^2

forces

|\det S| = 1

❌ Not quite. A positive real other than

1

has absolute value

\neq 1

, and

\operatorname{diag}(i, 1)

is unitary with a non-real determinant.

Show solution

$S^*S = I$ , so $1 = \det(S^*S) = \overline{\det S}\,\det S = |\det S|^2$ , giving $|\det S| = 1$ .

Conversely, for any $\theta$ , $\operatorname{diag}(e^{i\theta}, 1)$ preserves norms (its columns are orthonormal) and has determinant $e^{i\theta}$ . So the possible values are exactly the complex numbers of absolute value $1$ .

Problem 3 · Area of the Image of a Disk

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ with matrix $\begin{pmatrix} 3 & 1 \\ 1 & -2 \end{pmatrix}$ and $\Omega$ the disk of radius $2$ centred at $0$ . Find the area of $T(\Omega)$ .

✅ Correct!

|\det T| = |-6 - 1| = 7

and the disk has area

4\pi

❌ Not quite.

7\pi

is the image of the unit disk; this disk has radius

2

, so area

4\pi

❌ Check the sign. Areas are nonnegative: the factor is

|\det T| = 7

, not

\det T = -7

❌ Not quite. The area of a disk of radius

2

\pi \cdot 2^2 = 4\pi

, not

2\pi

Show solution

\det T = 3 \cdot (-2) - 1 \cdot 1 = -7, \qquad \operatorname{area} T(\Omega) = |\det T| \cdot \operatorname{area} \Omega = 7 \cdot 4\pi = 28\pi.

Problem 4 · The Determinant of a Positive Square Root

Given: $T$ is a positive operator on $\mathbf{R}^2$ whose square $T^2$ has matrix $\begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix}$ . Find $\det T$ .

✅ Correct!

(\det T)^2 = \det(T^2) = 9

and

\det T \ge 0

❌ Not quite.

(-3)^2 = 9

as well, but a positive operator has

\det T \ge 0

❌ Not quite.

9

\det(T^2)

;

\det T

is a number whose square is

9

❌ Not quite.

\det(T^2) = 25 - 16 = 9

, and

\det T = \sqrt{9}

Show solution

By the product rule, $(\det T)^2 = \det(T^2) = 5 \cdot 5 - 4 \cdot 4 = 9$ , so $\det T = \pm 3$ . A positive operator has an orthonormal eigenbasis with nonnegative eigenvalues, so $\det T \ge 0$ . Hence $\det T = 3$ .

Check: $T^2$ has eigenvalues $9$ and $1$ (eigenvectors $(1, 1)$ and $(1, -1)$ ), so $T$ has eigenvalues $3$ and $1$ , and $\det T = 3$ .

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