Linear-Algebra-Done-Right · Unit 21 · Video 2 · Interactive Practice

Singular Value Decomposition: Rank, Isometries, Proof

IKey Formulas

Formula	Name	What it decides
$T$ injective $\iff 0$ is not a singular value of $T$	Injectivity test	Look for a zero in the list
$\dim \operatorname{range} T = \#\{k : s_k > 0\}$	Rank count	$T$ is surjective $\iff$ this count $= \dim W$
$S$ isometry $\iff$ every singular value of $S$ equals $1$	Isometry test	Every entry counts, zeros included
$Tv = s_1\langle v, e_1\rangle f_1 + \cdots + s_m\langle v, e_m\rangle f_m$	Singular value decomposition	$T^*Te_k = s_k^2 e_k$ and $f_k = Te_k / s_k$

Key Insight: every linear map is a sum of simple pieces $s_k\langle v, e_k\rangle f_k$ : measure $v$ along $e_k$ , scale by $s_k$ , point along $f_k$ .

IIVisualization 1 — What the List Decides

The list of singular values alone settles the rank, injectivity, surjectivity and whether $T$ is an isometry.

Map dim V dim W

IIIVisualization 2 — The Unit Circle Becomes an Ellipse

The SVD sends the unit circle onto an ellipse whose half-axes are $s_1 f_1$ and $s_2 f_2$ .

θ 60° s₂ 1

IVVisualization 3 — Building the Two Lists

Both orthonormal lists come from $T^*T$ : its eigenvectors are the $e_k$ , and $f_k = Te_k / s_k$ .

Map Step 1

Step 1 · Eigenvectors of $T^*T$

T^*T = \begin{pmatrix} 3 & 4 \\ 0 & 5 \end{pmatrix}\begin{pmatrix} 3 & 0 \\ 4 & 5 \end{pmatrix} = \begin{pmatrix} 25 & 20 \\ 20 & 25 \end{pmatrix}

T^*Te_1 = 45\,e_1, \quad T^*Te_2 = 5\,e_2, \qquad e_1 = \tfrac{1}{\sqrt{2}}(1, 1), \ e_2 = \tfrac{1}{\sqrt{2}}(1, -1)

Step 2 · Singular values

s_1 = \sqrt{45} = 3\sqrt{5}, \qquad s_2 = \sqrt{5}

Both are positive, so $m = 2 = \dim W$ and $0$ is not in the list: $T$ is injective and surjective.

Step 3 · Define $f_k = Te_k / s_k$

f_1 = \frac{Te_1}{s_1} = \frac{1}{3\sqrt{5}} \cdot \frac{1}{\sqrt{2}}(3, 9) = \frac{1}{\sqrt{10}}(1, 3)

f_2 = \frac{Te_2}{s_2} = \frac{1}{\sqrt{5}} \cdot \frac{1}{\sqrt{2}}(3, -1) = \frac{1}{\sqrt{10}}(3, -1)

Step 4 · The $f_k$ are orthonormal

\langle f_j, f_k\rangle = \frac{1}{s_j s_k}\langle e_j, T^*Te_k\rangle = \frac{s_k}{s_j}\langle e_j, e_k\rangle

\langle f_1, f_2\rangle = \tfrac{1}{10}(3 - 3) = 0, \qquad \|f_1\|^2 = \|f_2\|^2 = \tfrac{1 + 9}{10} = 1

Step 5 · Reassemble $Tv$ for $v = (1, 0)$ , where $\langle v, e_1\rangle = \langle v, e_2\rangle = \tfrac{1}{\sqrt{2}}$ :

s_1\langle v, e_1\rangle f_1 + s_2\langle v, e_2\rangle f_2 = \tfrac{3}{2}(1, 3) + \tfrac{1}{2}(3, -1) = (3, 4) = Tv

Step 1 · Eigenvectors of $T^*T$

T^*T = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} 5 & 10 \\ 10 & 20 \end{pmatrix}

T^*Te_1 = 25\,e_1, \quad T^*Te_2 = 0, \qquad e_1 = \tfrac{1}{\sqrt{5}}(1, 2), \ e_2 = \tfrac{1}{\sqrt{5}}(2, -1)

Step 2 · Singular values

s_1 = \sqrt{25} = 5, \qquad s_2 = \sqrt{0} = 0

Only one is positive, so $m = 1$ : $0$ is in the list, so $T$ is not injective, and $1 \lt 2 = \dim W$ , so $T$ is not surjective.

Step 3 · Define $f_k = Te_k / s_k$ for $k \le m = 1$ only:

f_1 = \frac{Te_1}{s_1} = \frac{1}{5} \cdot \frac{1}{\sqrt{5}}(5, 10) = \frac{1}{\sqrt{5}}(1, 2)

No $f_2$ is defined: dividing by $s_2 = 0$ is not allowed.

Step 4 · The zero singular value sends $e_2$ to $0$

k = 2 > m: \quad T^*Te_2 = 0 \ \Rightarrow\ e_2 \in \operatorname{null} T^*T = \operatorname{null} T

Indeed $T(2, -1) = (0, 0)$ , and the one-vector list $f_1$ is orthonormal: $\|f_1\|^2 = \tfrac{1 + 4}{5} = 1$ .

Step 5 · Reassemble $Tv$ for $v = (1, 0)$ , where $\langle v, e_1\rangle = \tfrac{1}{\sqrt{5}}$ ; the $e_2$ term vanishes:

s_1\langle v, e_1\rangle f_1 = 5 \cdot \tfrac{1}{\sqrt{5}} \cdot \tfrac{1}{\sqrt{5}}(1, 2) = (1, 2) = Tv

VQuiz Questions

Problem 1 · Read the List

Given: $T \in \mathcal{L}(\mathbf{F}^5, \mathbf{F}^3)$ has singular values $4$ , $2$ , $2$ , $0$ , $0$ . Which statement is true?

✅ Correct! Three positive singular values give

\dim \operatorname{range} T = 3 = \dim \mathbf{F}^3

, and the zeros mean

T

is not injective.

❌ Not quite.

0

appears in the list, so

T

is not injective:

\operatorname{null} T^*T = \operatorname{null} T

is nonzero.

❌ Close, but… surjectivity compares the number of positive singular values,

3

, with

\dim W = 3

, not with the length of the list.

❌ Not quite. Count the positive singular values and compare with

\dim W

; look for a zero to test injectivity.

Show solution

Injectivity: $0$ is a singular value of $T$ , so $T$ is not injective. In fact $\dim \operatorname{null} T = 5 - 3 = 2$ , matching the two zeros.

Surjectivity: the positive singular values are $4, 2, 2$ , so

\dim \operatorname{range} T = \#\{k : s_k > 0\} = 3 = \dim \mathbf{F}^3,

and a subspace of $\mathbf{F}^3$ of dimension $3$ is all of $\mathbf{F}^3$ . So $T$ is surjective but not injective.

Problem 2 · Every Entry Counts

Given: $P$ is the orthogonal projection of $\mathbf{R}^3$ onto a plane through the origin. Its singular values are $1$ , $1$ , $0$ . Is $P$ an isometry?

✅ Correct! An isometry needs all singular values equal to

1

; the

0

fails that, and a nonzero vector

u

with

Pu = 0

has

\|Pu\| = 0 \ne \|u\|

❌ Close, but… the theorem asks for all singular values to equal

1

, zeros included. The

0

means

P

kills a nonzero vector.

❌ Not quite. The theorem says exactly that the singular values decide it:

S

is an isometry

\iff

every singular value equals

1

❌ Not quite. Being an isometry means

P^*P = I

, not merely that

P^*P

is positive. Here

P^*P = P \ne I

Show solution

$S$ is an isometry $\iff S^*S = I \iff$ every eigenvalue of $S^*S$ equals $1 \iff$ every singular value of $S$ equals $1$ .

For the projection, $P^*P = P^2 = P$ has eigenvalues $1, 1, 0$ , so the singular values are $1, 1, 0$ . The last one is not $1$ , so $P$ is not an isometry. Concretely, a unit normal $u$ to the plane has $Pu = 0$ , so $\|Pu\| = 0 \ne 1 = \|u\|$ .

Problem 3 · Build the Decomposition

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ , $T(x, y) = (2x + 2y,\ x - y)$ . Find its singular values and the vector $f_1$ of its SVD.

What are the singular values?

With $e_1 = \tfrac{1}{\sqrt{2}}(1, 1)$ , what is $f_1$ ?

✅ Correct!

s_1 = 2\sqrt{2}

s_2 = \sqrt{2}

f_1 = (1, 0)

and

f_2 = (0, 1)

, so

Tv = 2\sqrt{2}\langle v, e_1\rangle f_1 + \sqrt{2}\langle v, e_2\rangle f_2

❌ Close, but…

8

and

2

are the eigenvalues of

T^*T

. The singular values are their square roots.

❌ Not quite.

\sqrt{5}

is the length of each column of the matrix, the square root of a diagonal entry of

T^*T

, not of an eigenvalue.

❌ Not quite. Those are the absolute values of the eigenvalues of

T

itself. Singular values come from the eigenvalues of

T^*T

❌ Not quite. Compute

T^*T

, find its eigenvalues and take square roots.

❌ Not quite.

f_1

lives in the target and is usually not

e_1

: it is

Te_1 / s_1

❌ Close, but… that is

Te_1

itself, of length

s_1 = 2\sqrt{2}

. Divide by

s_1

to get a unit vector.

❌ Not quite.

(0, 1)

f_2 = Te_2 / s_2

. Use

f_1 = Te_1 / s_1

Show solution

Step 1: with the matrix $\begin{pmatrix} 2 & 2 \\ 1 & -1 \end{pmatrix}$ ,

T^*T = \begin{pmatrix} 2 & 1 \\ 2 & -1 \end{pmatrix}\begin{pmatrix} 2 & 2 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 5 & 3 \\ 3 & 5 \end{pmatrix}.

Step 2: $(1, 1)$ is an eigenvector with eigenvalue $8$ and $(1, -1)$ one with eigenvalue $2$ , so $e_1 = \tfrac{1}{\sqrt{2}}(1, 1)$ , $e_2 = \tfrac{1}{\sqrt{2}}(1, -1)$ and

s_1 = \sqrt{8} = 2\sqrt{2}, \qquad s_2 = \sqrt{2}.

Step 3: $Te_1 = \tfrac{1}{\sqrt{2}}(4, 0)$ , so

f_1 = \frac{Te_1}{s_1} = \frac{1}{2\sqrt{2}} \cdot \frac{1}{\sqrt{2}}(4, 0) = (1, 0).

Likewise $Te_2 = \tfrac{1}{\sqrt{2}}(0, 2)$ gives $f_2 = (0, 1)$ , and $f_1, f_2$ is orthonormal, as the proof guarantees.

Problem 4 · From the Decomposition Back to the List

Given: $T \in \mathcal{L}(\mathbf{F}^6, \mathbf{F}^4)$ satisfies $Tv = 7\langle v, e_1\rangle f_1 + 3\langle v, e_2\rangle f_2 + \langle v, e_3\rangle f_3$ for all $v$ , with $e_1, e_2, e_3$ orthonormal in $\mathbf{F}^6$ and $f_1, f_2, f_3$ orthonormal in $\mathbf{F}^4$ . Find the list of singular values of $T$ and $\dim \operatorname{null} T$ .

✅ Correct! The positive singular values are

7, 3, 1

; the list has length

\dim \mathbf{F}^6 = 6

, and

\dim \operatorname{null} T = 6 - 3 = 3

. So

T

is neither injective nor surjective.

❌ Not quite. The list of singular values has length

\dim V = 6

, the dimension of the domain, not

\dim W = 4

❌ Close, but… the SVD shows only the positive singular values. Zeros fill the list up to length

\dim V = 6

, and they record the null space.

❌ Close, but…

4 - 3

uses

\dim W

. Rank–nullity uses the domain:

\dim \operatorname{null} T = \dim V - \dim \operatorname{range} T

Show solution

In an SVD the coefficients $s_k$ are the positive singular values, so they are $7, 3, 1$ and $m = 3 = \dim \operatorname{range} T$ .

The full list has length $\dim \mathbf{F}^6 = 6$ , so three zeros follow: $7, 3, 1, 0, 0, 0$ .

By rank–nullity, $\dim \operatorname{null} T = 6 - 3 = 3$ , matching the three zeros. Since $3 \lt 4 = \dim W$ , $T$ is not surjective, and since $0$ is in the list, $T$ is not injective.

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