Chapter 10 · Vector Algebra

Example 19: Cauchy-Schwarz inequality $|\vec a\cdot\vec b|\le|\vec a||\vec b|$

ExamplesScalar product and |cos θ| ≤ 1 · Section 10.6.1

The question

For any two vectors $\vec a$ and $\vec b$ ⁠, we always have $|\vec a\cdot\vec b| \le |\vec a||\vec b|$ (Cauchy-Schwartz inequality).

Exam answer

Case 1. If $\vec a = \vec 0$ or $\vec b = \vec 0$ ⁠, then $\vec a\cdot\vec b = 0$ and $|\vec a||\vec b| = 0$ ⁠, so $|\vec a\cdot\vec b| = 0 = |\vec a||\vec b|$ and the result holds.

Case 2. Let $\vec a \ne \vec 0$ and $\vec b \ne \vec 0$ ⁠, and let $\theta$ be the angle between them. Then

\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta

\begin{aligned} \Rightarrow |\vec a\cdot\vec b| &= |\vec a||\vec b||\cos\theta| \\ &\le |\vec a||\vec b| \end{aligned}

since $|\cos\theta| \le 1$ and $|\vec a||\vec b| > 0$ ⁠.

$\therefore\ |\vec a\cdot\vec b| \le |\vec a||\vec b|$ for any two vectors. Hence proved.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The scalar product is $|\vec a||\vec b|\cos\theta$ ⁠, and a cosine is never bigger than 1 in size. So the scalar product can never exceed the product of the lengths:

|\vec a\cdot\vec b| = |\vec a||\vec b|\,|\cos\theta| \le |\vec a||\vec b|.

In pictures: $|\vec b||\cos\theta|$ is the length of the shadow of $\vec b$ on the line of $\vec a$ ⁠, and a shadow is never longer than the stick.

Given, to find, and the plan

Given: Any two vectors $\vec a$ and $\vec b$ ⁠.
To show: $|\vec a\cdot\vec b| \le |\vec a||\vec b|$ ⁠.
Plan: (1) Deal with a zero vector separately (no angle is defined). (2) For nonzero vectors, write $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$ and take absolute values. (3) Use $|\cos\theta| \le 1$ ⁠. (4) See when equality holds.

Basics you need

1The scalar product

For nonzero vectors, $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$ ⁠, $0 \le \theta \le \pi$ ⁠; if $\vec a = \vec 0$ or $\vec b = \vec 0$ ⁠, the angle is not defined and $\vec a\cdot\vec b = 0$ (Section 10.6.1, Definition 2). Lesson: the dot product

2Bounds on cosine

$-1 \le \cos\theta \le 1$ ⁠, that is $|\cos\theta| \le 1$ ⁠, with $|\cos\theta| = 1$ only for $\theta = 0$ or $\pi$ in $[0, \pi]$ ⁠. Basics: cos θ is a shadow

3Absolute value of a product

$|pq| = |p||q|$ for real numbers, and $|p| = p$ when $p \ge 0$ ⁠. Lesson: Examples 13–21

Solution

The case of a zero vector

If $\vec a = \vec 0$ or $\vec b = \vec 0$ ⁠, then $\vec a\cdot\vec b = 0$ by definition, and one of the lengths is 0, so $|\vec a||\vec b| = 0$ ⁠. Hence $|\vec a\cdot\vec b| = 0 = |\vec a||\vec b|$ ⁠, and the inequality holds (with equality).

This case must be handled first because the angle $\theta$ between $\vec 0$ and another vector is not defined, so the formula with $\cos\theta$ cannot be used. The book calls this case "trivial".

"If either $\vec a = \vec 0$ or $\vec b = \vec 0$ ⁠, then $\theta$ is not defined, and in this case, we define $\vec a\cdot\vec b = 0$ " (Section 10.6.1, Definition 2). Lesson: the dot product
Nonzero vectors: take absolute values

Now let $\vec a \ne \vec 0$ and $\vec b \ne \vec 0$ ⁠, with angle $\theta$ between them. Then $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$ ⁠, and so
$|\vec a\cdot\vec b| = |\vec a||\vec b|\,|\cos\theta|.$
The absolute value of a product is the product of the absolute values, and $|\vec a|$ ⁠, $|\vec b|$ are already non-negative, so only $\cos\theta$ needs the bars. The book writes the same fact divided through: $\frac{|\vec a\cdot\vec b|}{|\vec a||\vec b|} = |\cos\theta|$ ⁠, which needs $|\vec a||\vec b| \ne 0$ ⁠, true in this case.

$\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$ (Section 10.6.1, Definition 2); $|pq| = |p||q|$ ⁠. Lesson: Examples 13–21
Use $|\cos\theta| \le 1$
$\begin{aligned} |\vec a\cdot\vec b| &= |\vec a||\vec b|\,|\cos\theta| \\ &\le |\vec a||\vec b| \times 1 = |\vec a||\vec b|. \end{aligned}$
Together with Step 1, $|\vec a\cdot\vec b| \le |\vec a||\vec b|$ for any two vectors.

Drawn to scale for a sample pair, one scale on both axes: $|\vec a| = 4$ ⁠, $|\vec b| = 3$ ⁠, $\theta = 120^\circ$ ⁠. The shadow of $\vec b$ on the line of $\vec a$ falls behind O and has length $|\vec b||\cos\theta| = 1.5$ ⁠, shorter than $|\vec b| = 3$ ⁠. So $|\vec a\cdot\vec b| = 4 \times 1.5 = 6$ ⁠, less than $|\vec a||\vec b| = 12$ ⁠.

Multiplying the inequality $|\cos\theta| \le 1$ by the positive number $|\vec a||\vec b|$ keeps its direction. The bars around $\vec a\cdot\vec b$ matter: when $\theta$ is obtuse, $\vec a\cdot\vec b$ is negative, and the inequality is about its size.

$|\cos\theta| \le 1$ ⁠. Basics: cos θ is a shadow
When does equality hold?

For nonzero vectors, $|\vec a\cdot\vec b| = |\vec a||\vec b|$ exactly when $|\cos\theta| = 1$ ⁠, that is $\theta = 0$ or $\theta = \pi$ ⁠: the vectors point the same way or opposite ways, so they are collinear. With Step 1, equality holds exactly when $\vec a$ and $\vec b$ are collinear (a zero vector counts as collinear with every vector).

In every other case $0 < \theta < \pi$ ⁠, so $|\cos\theta| < 1$ and the inequality is strict. In the shadow picture: the shadow is as long as the stick only when the stick lies along the line.

$\theta = 0$ or $\pi$ gives $\vec a\cdot\vec b = \pm|\vec a||\vec b|$ (Section 10.6.1, Observations 3 and 4); collinear vectors (Section 10.5). Lesson: the dot product

Result

For any two vectors, $|\vec a\cdot\vec b| \le |\vec a||\vec b|$ ⁠, with equality exactly when $\vec a$ and $\vec b$ are collinear.

(Textbook answer: the inequality holds trivially when $\vec a = \vec 0$ or $\vec b = \vec 0$ ⁠; otherwise $\frac{|\vec a\cdot\vec b|}{|\vec a||\vec b|} = |\cos\theta| \le 1$ ⁠, therefore $|\vec a\cdot\vec b| \le |\vec a||\vec b|$ ⁠. The book spells the name "Cauchy-Schwartz"; the mathematician is Hermann Amandus Schwarz, so it is usually written Cauchy-Schwarz.)

Check it with numbers

$\vec a = 2\hat i + 3\hat j + 2\hat k$ ⁠, $\vec b = \hat i + 2\hat j + \hat k$ ⁠: $\vec a\cdot\vec b = 2 + 6 + 2 = 10$ and $|\vec a||\vec b| = \sqrt{17}\sqrt6 = \sqrt{102} \approx 10.099505$ ⁠. So $10 \le 10.099505$ ⁠, a close call because the angle is small (about $8.05^\circ$ ⁠).

$\vec a = \hat i - 2\hat j + 2\hat k$ ⁠, $\vec b = -2\vec a = -2\hat i + 4\hat j - 4\hat k$ (collinear, opposite): $\vec a\cdot\vec b = -2 - 8 - 8 = -18$ ⁠, and $|\vec a||\vec b| = 3 \times 6 = 18$ ⁠. Equality, $|-18| = 18$ ⁠, as Step 4 predicts.

Common mistakes

Skipping the zero-vector case. For $\vec a = \vec 0$ there is no angle, so the $\cos\theta$ argument does not apply; the case needs its own line.
Dividing by $|\vec a||\vec b|$ without saying it is nonzero. The book's form $\frac{|\vec a\cdot\vec b|}{|\vec a||\vec b|}$ is only allowed once both vectors are nonzero.
Dropping the absolute value. $\vec a\cdot\vec b \le |\vec a||\vec b|$ is true but weaker; the statement bounds the size $|\vec a\cdot\vec b|$ ⁠, so a large negative scalar product is ruled out too.

What to take away

$|\vec a\cdot\vec b| = |\vec a||\vec b||\cos\theta| \le |\vec a||\vec b|$ ⁠: a shadow is never longer than the stick.
Handle the zero vector first, since no angle is defined for it.
Equality holds exactly for collinear vectors.

Every fact used, and where it comes from

Fact	Used in	Source
$\vec a\cdot\vec b = \|\vec a\|\|\vec b\|\cos\theta$ ⁠; $\vec a\cdot\vec b = 0$ if a vector is zero	Steps 1, 2	Section 10.6.1, Definition 2 (Lesson: the dot product)
$\|\cos\theta\| \le 1$ ⁠, equal to 1 only at $0$ and $\pi$	Steps 3, 4	Trigonometry (Basics: cos θ is a shadow)
$\theta = 0$ or $\pi$ gives $\vec a\cdot\vec b = \pm\|\vec a\|\|\vec b\|$	Step 4	Section 10.6.1, Observations 3 and 4
Collinear vectors	Step 4	Section 10.5
The book's working and answer	Result	Example 19, Section 10.6.2 (Lesson: Examples 13–21)

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Example 19: Cauchy-Schwarz inequality ∣a⃗⋅b⃗∣≤∣a⃗∣∣b⃗∣|\vec a\cdot\vec b|\le|\vec a||\vec b|∣a⋅b∣≤∣a∣∣b∣

Basics you need

Solution

The case of a zero vector

Nonzero vectors: take absolute values

Use ∣cos⁡θ∣≤1|\cos\theta| \le 1∣cosθ∣≤1

When does equality hold?

Every fact used, and where it comes from

Example 19: Cauchy-Schwarz inequality $|\vec a\cdot\vec b|\le|\vec a||\vec b|$

Use $|\cos\theta| \le 1$