Linear-Algebra-Done-Right · Unit 23 · Video 3 · Interactive Practice

Volume Scales by the Product of the Singular Values

IKey Formulas

Formula	Name	Where it comes from
$\operatorname{volume}\big(u + P(r_1 e_1, \dots, r_n e_n)\big) = r_1 \times \cdots \times r_n$	Volume of a box	Definition; other regions are approximated by disjoint boxes ( $\mathbf{F} = \mathbf{R}$ )
$\operatorname{volume} T(\Omega) = (s_1 \times \cdots \times s_n)\,\operatorname{volume}\,\Omega$	Volume changes by the product of the singular values	$T$ invertible: boxes along the $e$ 's go to boxes along the $f$ 's with edges $r_k s_k$
$\mathcal{M}(T) = \operatorname{diag}(\lambda_1, \dots, \lambda_n)$ , $\ \mathcal{M}(T^*) = \operatorname{diag}(\bar\lambda_1, \dots, \bar\lambda_n)$	Normal $T$ , $\mathbf{F} = \mathbf{C}$	Complex spectral theorem: an orthonormal basis of eigenvectors
$\\|T\\| = \max_k \|\lambda_k\|$	Norm of a normal operator	Singular values of a normal $T$ are $\|\lambda_1\|, \dots, \|\lambda_n\|$

Key Insight: Every operator has singular values, even an operator on a real space with no eigenvalues; their product is the single factor by which $T$ multiplies every volume.

IIOne Factor for Every Box

Each box along $e_1, e_2$ goes to a box along $f_1, f_2$ with area multiplied by $s_1 s_2$ , so the whole approximation scales by $s_1 s_2$ .

Operator Strips n 5

IIIReading Properties off the Eigenvalues

For a normal operator with eigenvalues $\lambda_1, \lambda_2$ , each property in the table holds exactly when both eigenvalues lie in its subset of $\mathbf{C}$ .

💡 The table needs normality: $N(z_1, z_2) = (z_2, 0)$ has only the eigenvalue $0$ , yet it is not self-adjoint, not a projection, and has $\|N\| = 1$ .

IVQuiz Questions

Problem 1 · Area of an Image Disk

Given: $T(x, y) = (2x + y,\ x + 2y)$ on $\mathbf{R}^2$ and $B$ the unit disk — find the area of $T(B)$ .

✅ Correct!

s_1 = 3

s_2 = 1

, so the area is

3 \cdot 1 \cdot \pi

❌ Multiply, don't add. The factor is

s_1 s_2 = 3 \cdot 1

, not

s_1 + s_2

❌ $9$ and $1$ are the eigenvalues of $T^*T$ . The singular values are their square roots.

❌ $T$ does change area. It stretches the direction

(1, 1)

by 3.

Show solution

The matrix $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ is symmetric with eigenvalues $3$ (eigenvector $(1, 1)$ ) and $1$ (eigenvector $(-1, 1)$ ). So $T^*T = T^2$ has eigenvalues $9$ and $1$ , and the singular values are $s_1 = 3$ , $s_2 = 1$ .

\operatorname{area} T(B) = s_1 s_2 \operatorname{area} B = 3 \cdot 1 \cdot \pi = 3\pi.

Indeed $T(B)$ is the ellipse with semi-axes $3$ along $(1, 1)$ and $1$ along $(-1, 1)$ .

Problem 2 · No Real Eigenvalues, Still a Factor

Given: $T(x, y) = (x - 2y,\ 2x + y)$ on $\mathbf{R}^2$ , which has no real eigenvalues — find the factor by which $T$ multiplies every area.

✅ Correct!

T^*T = 5I

, so

s_1 = s_2 = \sqrt5

and

s_1 s_2 = 5

❌ $T$ is a rotation times a stretch. It turns the plane, but it also scales every length by

\sqrt5

❌ That is one singular value. Both equal

\sqrt5

, and area takes their product.

❌ The theorem uses singular values, not eigenvalues. Every operator has singular values.

Show solution

With $A = \begin{pmatrix} 1 & -2 \\ 2 & 1 \end{pmatrix}$ , $A^{\mathsf T}A = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix}$ . So both singular values are $\sqrt5$ and

\operatorname{area} T(\Omega) = \sqrt5 \cdot \sqrt5 \operatorname{area}\Omega = 5\operatorname{area}\Omega .

The characteristic polynomial $\lambda^2 - 2\lambda + 5$ has roots $1 \pm 2i$ , so there are no real eigenvalues; the singular values do not need them.

Problem 3 · A Cube in R³

Given: $T(x, y, z) = (y,\ 2z,\ 3x)$ on $\mathbf{R}^3$ — find the volume of $T\big(P(e_1, e_2, e_3)\big)$ , the image of the open unit cube.

✅ Correct! The singular values are

3, 1, 2

, and

3 \cdot 1 \cdot 2 = 6

❌ $T$ does more than shuffle coordinates. It also multiplies them by

1, 2, 3

❌ $36$ is the product of the eigenvalues of $T^*T$ . Take square roots first.

❌ $9 + 1 + 4$ adds the eigenvalues of $T^*T$ . Volume multiplies the singular values.

Show solution

$Te_1 = (0, 0, 3)$ , $Te_2 = (1, 0, 0)$ , $Te_3 = (0, 2, 0)$ : orthogonal vectors of lengths $3, 1, 2$ . So $T^*T = \operatorname{diag}(9, 1, 4)$ and the singular values are $3, 1, 2$ .

The cube is the box along $e_1, e_2, e_3$ with $r_1 = r_2 = r_3 = 1$ , so its image is the box $P\big((0, 0, 3), (1, 0, 0), (0, 2, 0)\big)$ with volume

3 \times 1 \times 2 = 6 = (s_1 s_2 s_3)\operatorname{volume} P(e_1, e_2, e_3).

Problem 4 · Reading the Table

Given: $T$ a normal operator on $\mathbf{C}^2$ with eigenvalues $\tfrac{i}{2}$ and $-\tfrac{i}{2}$ — which list contains exactly the properties of $T$ from the table?

✅ Correct!

0

is not an eigenvalue, both eigenvalues have real part

0

, and

\|T\| = \max|\lambda_k| = \tfrac12

❌ A conjugate pair is not real. Self-adjoint needs every eigenvalue in

\mathbf{R}

, and

\tfrac{i}{2} \notin \mathbf{R}

❌ Unitary needs $|\lambda_k| = 1$ . Here

|\lambda_k| = \tfrac12

❌ Positive needs eigenvalues in $[0, \infty)$ .

\tfrac{i}{2}

is not a nonnegative real number.

Show solution

In an orthonormal basis of eigenvectors, $\mathcal{M}(T) = \operatorname{diag}\big(\tfrac{i}{2}, -\tfrac{i}{2}\big)$ . Check each row:

invertible: $0$ is not an eigenvalue — yes;
self-adjoint: needs $\lambda_k \in \mathbf{R}$ — no; positive: needs $\lambda_k \in [0, \infty)$ — no;
skew: $\operatorname{Re}\lambda_k = 0$ for both — yes (indeed $\mathcal{M}(T^*) = \operatorname{diag}\big(-\tfrac{i}{2}, \tfrac{i}{2}\big) = -\mathcal{M}(T)$ );
orthogonal projection: needs $\lambda_k \in \{0, 1\}$ — no; unitary: needs $|\lambda_k| = 1$ — no;
$\|T\| = \max_k|\lambda_k| = \tfrac12 < 1$ — yes.

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