Linear-Algebra-Done-Right · Unit 22 · Video 1 · Interactive Practice

The Norm of a Linear Map Is Its Largest Singular Value

IKey Formulas

Formula	Name	Where it comes from
$\\|Tv\\| \le s_1\\|v\\|$	Upper bound for $\\|Tv\\|$	SVD formula and Bessel's inequality; equality at $v = e_1$
$\\|T\\| = \max\{\\|Tv\\| : v \in V,\ \\|v\\| \le 1\}$	Norm of $T \in \mathcal{L}(V, W)$	The maximum exists and equals $s_1$
$\\|\lambda T\\| = \|\lambda\|\,\\|T\\|$ , $\ \\|S + T\\| \le \\|S\\| + \\|T\\|$	Basic properties	Also $\\|T\\| \ge 0$ , and $\\|T\\| = 0 \iff T = 0$
$\big\|\,\\|S\\| - \\|T\\|\,\big\| \le \\|S - T\\|$	Reverse triangle inequality	Triangle inequality with $S = (S - T) + T$

Key Insight: $T$ sends the unit ball into the ball of radius $s_1$ , and $e_1$ reaches its edge, so the longest output from the unit ball has length exactly $s_1$ .

IIThe Farthest Image of a Unit Vector

Every unit vector $v$ lands inside the circle of radius $s_1$ , and only $\pm e_1$ land on it.

Map Direction of v 53°

IIIEach Map Peaks in Its Own Direction

The unit vector that maximizes $\|(S + T)v\|$ need not maximize $\|Sv\|$ or $\|Tv\|$ , so $\|S + T\|$ can fall short of $\|S\| + \|T\|$ .

Pair Angle θ of v = (cos θ, sin θ) 30°

IVQuiz Questions

Problem 1 · Norm of a Diagonal Operator

Given: $T \in \mathcal{L}(\mathbf{R}^3)$ defined by $T(x, y, z) = (2x, -5y, z)$ , with the usual inner product — find $\|T\|$ .

✅ Correct!

\|Tv\|^2 = 4x^2 + 25y^2 + z^2 \le 25\|v\|^2

, with equality at

v = (0, 1, 0)

❌ Not quite. The stretch along the second axis is

|-5| = 5

; the sign of a diagonal entry does not shorten the output.

❌ Not quite.

\sqrt{4 + 25 + 1}

adds up the stretches in all three directions at once, but no single unit vector gets all three:

\|Tv\| \le 5

for every unit

v

❌ Not quite. A norm is a maximum of lengths, so it is never negative:

\|T\| \ge 0

Show solution

For $v = (x, y, z)$ ,

\|Tv\|^2 = 4x^2 + 25y^2 + z^2 \le 25(x^2 + y^2 + z^2) = 25\|v\|^2,

so $\|Tv\| \le 5$ whenever $\|v\| \le 1$ . The unit vector $(0, 1, 0)$ gives $T(0, 1, 0) = (0, -5, 0)$ , of length $5$ . Hence $\|T\| = 5$ . (Equivalently: $T^*T(x, y, z) = (4x, 25y, z)$ , so the singular values are $5, 2, 1$ and $\|T\| = s_1 = 5$ .)

Problem 2 · The Norm of a Sum

Given: $S, T \in \mathcal{L}(\mathbf{R}^2)$ defined by $S(x, y) = (2x, 0)$ and $T(x, y) = (0, 3y)$ — find $\|S + T\|$ .

✅ Correct!

(S + T)(x, y) = (2x, 3y)

, whose largest stretch is

3

, reached at

(0, 1)

❌ Not quite.

\|S\| + \|T\| = 2 + 3 = 5

is only an upper bound.

S

peaks at

(1, 0)

and

T

(0, 1)

; no single unit vector does both.

❌ Not quite.

(S + T)(1, 1) = (2, 3)

has length

\sqrt{13}

, but

(1, 1)

has length

\sqrt{2}

, not

1

. Divide by

\sqrt{2}

\sqrt{13/2} \approx 2.55

❌ Not quite. Write

(S + T)(x, y) = (2x, 3y)

and bound its length on the unit circle.

Show solution

$(S + T)(x, y) = (2x, 3y)$ , so

\|(S + T)(x, y)\|^2 = 4x^2 + 9y^2 \le 9(x^2 + y^2).

Thus $\|(S + T)v\| \le 3$ for $\|v\| \le 1$ , with equality at $v = (0, 1)$ . So $\|S + T\| = 3$ , strictly less than $\|S\| + \|T\| = 2 + 3 = 5$ : the triangle inequality is strict here because $S$ and $T$ reach their norms at different unit vectors.

Problem 3 · A Rank-One Map

Given: $T \in \mathcal{L}(\mathbf{R}^3, \mathbf{R}^2)$ defined by $Tv = \langle v, u\rangle w$ , where $u = (1, 2, 2)$ and $w = (3, 4)$ — find $\|T\|$ .

✅ Correct!

\|Tv\| = |\langle v, u\rangle|\,\|w\| \le \|v\|\,\|u\|\,\|w\| = 15\|v\|

, with equality at

v = u/3

❌ Not quite.

5 = \|w\|

is the length of the output direction, but the coefficient

\langle v, u\rangle

can be as large as

\|u\| = 3

on a unit vector.

❌ Not quite.

45 = \|T u\|

, but

u

has length

3

, not

1

. Use the unit vector

u/\|u\|

❌ Not quite. Bound

|\langle v, u\rangle|

with the Cauchy–Schwarz inequality.

Show solution

Here $\|u\| = \sqrt{1 + 4 + 4} = 3$ and $\|w\| = \sqrt{9 + 16} = 5$ . For every $v$ , by the Cauchy–Schwarz inequality,

\|Tv\| = |\langle v, u\rangle|\,\|w\| \le \|v\|\,\|u\|\,\|w\| = 15\|v\|.

So $\|Tv\| \le 15$ on the unit ball. For $v = u/3$ (a unit vector), $\langle v, u\rangle = \|u\|^2/3 = 3$ and $\|Tv\| = 3 \cdot 5 = 15$ . Hence $\|T\| = 15$ . This is also $T$ 's only positive singular value: $Tv = 15\,\langle v, u/3\rangle\,(w/5)$ is a singular value decomposition.

Problem 4 · Close Maps Have Close Norms

Given: $S, T \in \mathcal{L}(V, W)$ with $\|S\| = 7$ and $\|S - T\| = 2$ — what do these facts force about $\|T\|$ ?

✅ Correct! The reverse triangle inequality gives

|7 - \|T\|| \le 2

, and both ends occur:

T = \tfrac57 S

and

T = \tfrac97 S

❌ Not quite. Distance

2

does not pin down

\|T\|

T = \tfrac57 S

gives

\|T\| = 5

and

T = \tfrac97 S

gives

\|T\| = 9

, both with

\|S - T\| = 2

❌ Not quite. The triangle inequality gives

\|T\| \le \|S\| + \|T - S\| = 9

, an upper bound, not a lower one.

Show solution

By the reverse triangle inequality, $\big|\,\|S\| - \|T\|\,\big| \le \|S - T\| = 2$ , so $|7 - \|T\|| \le 2$ , i.e.

5 \le \|T\| \le 9.

Nothing sharper holds. If $T = \tfrac57 S$ , then $\|T\| = 5$ and $\|S - T\| = \tfrac27\|S\| = 2$ . If $T = \tfrac97 S$ , then $\|T\| = 9$ and $\|S - T\| = \tfrac27\|S\| = 2$ .

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