Linear-Algebra-Done-Right · Unit 14 · Video 3 · Interactive Practice

Cauchy-Schwarz and the Triangle Inequality

IKey Formulas

Formula	Name	Hypotheses and equality case
$c = \dfrac{\langle u, v\rangle}{\\|v\\|^2}, \ w = u - cv \implies u = cv + w, \ \langle w, v\rangle = 0$	Orthogonal decomposition	$V$ an inner product space over $\mathbf{F}$ ( $\mathbf{F} = \mathbf{R}$ or $\mathbf{C}$ ), $u, v \in V$ , $v \ne 0$ ; the numerator is $\langle u, v\rangle$ , not $\langle v, u\rangle$
$\|\langle u, v\rangle\| \le \\|u\\|\,\\|v\\|$	Cauchy–Schwarz inequality	$u, v \in V$ ; equality $\iff$ one of $u, v$ is a scalar multiple of the other
$\\|u + v\\| \le \\|u\\| + \\|v\\|$	Triangle inequality	$u, v \in V$ ; equality $\iff$ one of $u, v$ is a nonnegative real multiple of the other
$\\|u + v\\|^2 + \\|u - v\\|^2 = 2\bigl(\\|u\\|^2 + \\|v\\|^2\bigr)$	Parallelogram equality	$u, v \in V$ ; always an equation

Key Insight: Cauchy–Schwarz is the Pythagorean theorem applied to the decomposition: $\|u\|^2 = \|cv\|^2 + \|w\|^2 \ge \|cv\|^2 = \dfrac{|\langle u, v\rangle|^2}{\|v\|^2}$ , with equality exactly when $w = 0$ .

IISplitting u Along v

The scalar $c = \langle u, v\rangle / \|v\|^2$ makes $w = u - cv$ orthogonal to $v$ , and $\|u\| \ge \|cv\|$ is Cauchy–Schwarz.

💡 In a complex inner product space the numerator must be $\langle u, v\rangle$ , with $u$ in the first slot: using $\langle v, u\rangle = \overline{\langle u, v\rangle}$ instead leaves a leftover that is in general not orthogonal to $v$ .

IIIWhich Step Loses Equality

With $\|u\| = 3$ and $\|v\| = 2$ in $\mathbf{R}^2$ , $\|u + v\| = \|u\| + \|v\|$ only when $v$ points the same way as $u$ .

Angle θ from u to v 45°

IVDiagonals Against Sides

The squared diagonals sum to twice the squared sides, because the cross terms $\pm 2\langle u, v\rangle$ cancel.

VQuiz Questions

Problem 1 · Decompose u Along v in R⁴

Given: in $\mathbf{R}^4$ with the Euclidean inner product, $u = (4, 1, 3, 0)$ and $v = (1, 1, 1, 1)$ — find $c$ and $w$ with $u = cv + w$ and $\langle w, v\rangle = 0$ .

✅ Correct!

c = 8/4 = 2

w = u - 2v = (2, -1, 1, -2)

, and

\langle w, v\rangle = 2 - 1 + 1 - 2 = 0

❌ Not quite. That divides by

\|v\| = 2

instead of

\|v\|^2 = 4

. Then

\langle w, v\rangle = 0 - 3 - 1 - 4 = -8 \ne 0

❌ Not quite. This

w

cv - u

. It is orthogonal to

v

, but

cv + w = (0, 3, 1, 4) \ne u

; the leftover must be

w = u - cv

❌ Not quite. That is

\|v\|^2 / \langle u, v\rangle

, the fraction upside down. Then

\langle w, v\rangle = \tfrac{7}{2} + \tfrac{1}{2} + \tfrac{5}{2} - \tfrac{1}{2} = 6 \ne 0

❌ Not quite. Compute

c = \langle u, v\rangle / \|v\|^2

, then

w = u - cv

Show solution

Here $v \ne 0$ , so the orthogonal decomposition applies.

\langle u, v\rangle = 4 + 1 + 3 + 0 = 8, \qquad \|v\|^2 = 1 + 1 + 1 + 1 = 4, \qquad c = \frac{\langle u, v\rangle}{\|v\|^2} = \frac{8}{4} = 2.

w = u - cv = (4, 1, 3, 0) - (2, 2, 2, 2) = (2, -1, 1, -2).

Check: $\langle w, v\rangle = 2 - 1 + 1 - 2 = 0$ and $cv + w = (2, 2, 2, 2) + (2, -1, 1, -2) = (4, 1, 3, 0) = u$ .

Problem 2 · A Negative Multiple

Given: in $\mathbf{R}^3$ with the Euclidean inner product, $u = (1, -2, 2)$ and $v = (-2, 4, -4)$ — find the true statement about Cauchy–Schwarz and the triangle inequality for this pair.

✅ Correct!

v = -2u

is a scalar multiple of

u

, which is all Cauchy–Schwarz equality asks, but

-2

is negative, so the triangle inequality is strict.

❌ Not quite. The triangle inequality needs a nonnegative real multiple. Here

u + v = (-1, 2, -2)

has norm

3

, while

\|u\| + \|v\| = 3 + 6 = 9

❌ Not quite. Cauchy–Schwarz equality allows any scalar multiple, negative ones included:

|\langle u, v\rangle| = |-18| = 18 = 3 \cdot 6

❌ Not quite. Triangle equality would force Cauchy–Schwarz equality, since the triangle proof passes through

|\langle u, v\rangle| \le \|u\|\,\|v\|

. Compute both sides of each.

❌ Not quite. Compute

\langle u, v\rangle

\|u\|

\|v\|

and

\|u + v\|

Show solution

$\|u\| = \sqrt{1 + 4 + 4} = 3$ , $\ \|v\| = \sqrt{4 + 16 + 16} = 6$ , and $\langle u, v\rangle = -2 - 8 - 8 = -18$ .

Cauchy–Schwarz: $|\langle u, v\rangle| = 18 = 3 \cdot 6 = \|u\|\,\|v\|$ , an equality, as it must be because $v = -2u$ is a scalar multiple of $u$ .

Triangle inequality: $u + v = (-1, 2, -2)$ , so $\|u + v\| = 3 \lt 9 = \|u\| + \|v\|$ . Equality would need one vector to be a nonnegative real multiple of the other; $u = -\tfrac{1}{2}v$ and $v = -2u$ only use negative scalars. In the proof, the real-part step fails: $\mathrm{Re}\,\langle u, v\rangle = -18 \lt 18 = |\langle u, v\rangle|$ .

Problem 3 · Cauchy–Schwarz for Integrals

Given: on the continuous real-valued functions on $[-1, 1]$ with $\langle f, g\rangle = \int_{-1}^{1} fg$ , take $f(x) = x$ and $g(x) = x^3$ . Cauchy–Schwarz, squared, reads $\Bigl|\int_{-1}^{1} fg\Bigr|^2 \le \Bigl(\int_{-1}^{1} f^2\Bigr)\Bigl(\int_{-1}^{1} g^2\Bigr)$ — find both sides.

Left side $\Bigl|\int_{-1}^{1} fg\Bigr|^2$

Right side $\Bigl(\int_{-1}^{1} f^2\Bigr)\Bigl(\int_{-1}^{1} g^2\Bigr)$

✅ Correct!

\tfrac{4}{25} \lt \tfrac{4}{21}

, strict because neither of

x

x^3

is a constant multiple of the other.

❌ Not quite.

\int_{-1}^{1} x^4\,dx = \tfrac{2}{5}

\langle f, g\rangle

itself; the left side squares it.

❌ Not quite. That is

\bigl(\int f\bigr)\bigl(\int g\bigr)

, which is not the inner product. Integrate the product:

fg = x^4

❌ Not quite.

\int_{-1}^{1} f^2 g^2 = \int_{-1}^{1} x^8 = \tfrac{2}{9}

appears in neither side: the left side is the square of

\int fg

, the right side the product of two separate integrals.

❌ Not quite. That is

\bigl(\int x^2\bigr)^2

; the second factor is

\|g\|^2 = \int_{-1}^{1} x^6 = \tfrac{2}{7}

, not

\int x^2

❌ Not quite.

\int_{-1}^{1} x^2 = \tfrac{2}{3}

(not

\tfrac{1}{3}

): integrate over the whole interval

[-1, 1]

❌ Check the left side. Integrate

fg = x^4

over

[-1, 1]

, then square.

❌ Check the right side. Multiply

\|f\|^2 = \int_{-1}^{1} x^2

\|g\|^2 = \int_{-1}^{1} x^6

Show solution

Left side: $\langle f, g\rangle = \int_{-1}^{1} x \cdot x^3\,dx = \int_{-1}^{1} x^4\,dx = \Bigl[\tfrac{x^5}{5}\Bigr]_{-1}^{1} = \tfrac{2}{5}$ , so $|\langle f, g\rangle|^2 = \tfrac{4}{25}$ .

Right side: $\|f\|^2 = \int_{-1}^{1} x^2\,dx = \tfrac{2}{3}$ and $\|g\|^2 = \int_{-1}^{1} x^6\,dx = \tfrac{2}{7}$ , so the product is $\tfrac{4}{21}$ .

\frac{4}{25} = 0.16 \lt \frac{4}{21} \approx 0.190.

The inequality is strict: equality would need one function to be a scalar multiple of the other, and $x^3 = \lambda x$ for all $x \in [-1, 1]$ is impossible.

Problem 4 · The Other Diagonal

Given: in a real inner product space, $\|u\| = 3$ , $\|v\| = 5$ and $\|u + v\| = 7$ — find $\|u - v\|$ and $\langle u, v\rangle$ .

What is $\|u - v\|$ ?

What is $\langle u, v\rangle$ ?

✅ Correct!

\|u - v\|^2 = 2(9 + 25) - 49 = 19

, and

49 = 9 + 25 + 2\langle u, v\rangle

gives

\langle u, v\rangle = \tfrac{15}{2}

, consistent with Cauchy–Schwarz since

\tfrac{15}{2} \le 3 \cdot 5

❌ Not quite. The parallelogram equality uses

\|u + v\|^2 = 49

, not

7

\|u - v\|^2 = 68 - 49

❌ Not quite.

2(3 + 5) - 7

uses lengths where the equality needs squared lengths:

\|u - v\|^2 = 2(3^2 + 5^2) - 7^2

❌ Not quite.

19

\|u - v\|^2

; take the square root.

❌ Not quite. The cross term is

2\langle u, v\rangle = 15

; divide by

2

❌ Not quite.

\|u + v\|^2 = 49

exceeds

\|u\|^2 + \|v\|^2 = 34

, so

2\langle u, v\rangle = 49 - 34

is positive.

❌ Not quite.

\tfrac{19}{2}

is half of

\|u - v\|^2

. Expand

\|u + v\|^2 = \|u\|^2 + \|v\|^2 + 2\langle u, v\rangle

instead.

❌ Check the diagonal. Use

\|u + v\|^2 + \|u - v\|^2 = 2\bigl(\|u\|^2 + \|v\|^2\bigr)

with squared lengths.

❌ Check the inner product. In a real space

\|u + v\|^2 = \|u\|^2 + \|v\|^2 + 2\langle u, v\rangle

Show solution

Parallelogram equality:

\|u - v\|^2 = 2\bigl(\|u\|^2 + \|v\|^2\bigr) - \|u + v\|^2 = 2(9 + 25) - 49 = 19, \qquad \|u - v\| = \sqrt{19}.

Expansion: in a real inner product space $\mathrm{Re}\,\langle u, v\rangle = \langle u, v\rangle$ , so

49 = \|u + v\|^2 = \|u\|^2 + \|v\|^2 + 2\langle u, v\rangle = 34 + 2\langle u, v\rangle, \qquad \langle u, v\rangle = \frac{15}{2}.

Check with the other diagonal: $\|u - v\|^2 = 34 - 2 \cdot \tfrac{15}{2} = 19$ . Both inequalities hold strictly: $|\langle u, v\rangle| = \tfrac{15}{2} \lt 15 = \|u\|\,\|v\|$ and $\|u + v\| = 7 \lt 8 = \|u\| + \|v\|$ .

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