Linear-Algebra-Done-Right · Unit 14 · Video 2 · Interactive Practice

Norms, Orthogonality and the Pythagorean Theorem

IKey Formulas

Formula	Name	Hypotheses
$\\|v\\| = \sqrt{\langle v, v \rangle}$	Norm	$V$ an inner product space over $\mathbf{F}$ ( $\mathbf{F} = \mathbf{R}$ or $\mathbf{C}$ ), $v \in V$
$\\|(z_1, \dots, z_n)\\| = \sqrt{\|z_1\|^2 + \cdots + \|z_n\|^2}$	Euclidean norm	$\mathbf{F}^n$ with the Euclidean inner product
$\\|v\\| = 0 \iff v = 0$ , $\ \\|\lambda v\\| = \|\lambda\|\,\\|v\\|$	Basic properties of the norm	$V$ an inner product space, $v \in V$ , $\lambda \in \mathbf{F}$
$\langle u, v \rangle = 0 \;\Rightarrow\; \\|u + v\\|^2 = \\|u\\|^2 + \\|v\\|^2$	Pythagorean theorem	$V$ an inner product space, $u, v \in V$ orthogonal

Key Insight: Work with $\|v\|^2 = \langle v, v \rangle$ — both proofs expand an inner product instead of handling a square root.

IIScaling by a Complex Number

Multiplying $z = (1 + 2i, 2, 4i) \in \mathbf{C}^3$ by $\lambda \in \mathbf{C}$ multiplies every coordinate's modulus, and the norm, by $|\lambda|$ .

IIIOrthogonal Functions on $[-1, 1]$

With $\langle f, g \rangle = \int_{-1}^{1} fg$ , orthogonality means the signed area under $fg$ cancels to zero.

IVThe Pythagorean Theorem in $\mathbf{R}^2$

With the Euclidean inner product, the square on $u + v$ equals the other two squares exactly when $\langle u, v \rangle = 0$ .

💡 Challenge: with $u = (2, 1)$ fixed, find three different $v$ that make the squares balance, and say what those $v$ have in common.

VQuiz Questions

Problem 1 · A Norm in $\mathbf{C}^3$

Given: $w = (2 - i,\ 1 + 3i,\ 1) \in \mathbf{C}^3$ with the Euclidean inner product — find $\|w\|$ .

✅ Correct! The squared moduli are

5

10

and

1

, which add to

16

, so

\|w\| = 4

❌ Not quite. That uses only the real parts

2, 1, 1

. Each term is

|w_k|^2 = (\operatorname{Re} w_k)^2 + (\operatorname{Im} w_k)^2

, so the imaginary parts count too.

❌ Close, but…

16

\|w\|^2 = \langle w, w \rangle

. The norm is its square root.

❌ Not quite. The norm is not the sum of the moduli

|w_k|

— square the moduli, add, then take one square root.

❌ Not quite. Compute

|w_1|^2 + |w_2|^2 + |w_3|^2

first.

Show solution

For the Euclidean inner product, $\|w\| = \sqrt{|w_1|^2 + |w_2|^2 + |w_3|^2}$ . The squared moduli are

|2 - i|^2 = 4 + 1 = 5, \qquad |1 + 3i|^2 = 1 + 9 = 10, \qquad |1|^2 = 1.

\|w\|^2 = 5 + 10 + 1 = 16, \qquad \|w\| = 4.

The norm is $\mathbf{4}$ .

Problem 2 · Pulling Out a Complex Scalar

Given: $v = (1,\ 2i,\ 2) \in \mathbf{C}^3$ with the Euclidean inner product and $\lambda = -3 + 4i$ — find $\|\lambda v\|$ .

✅ Correct!

\|\lambda v\| = |\lambda|\,\|v\| = 5 \cdot 3 = 15

❌ Not quite. A norm is a nonnegative real number. The scalar comes out as

|\lambda|

, not as

\lambda

\|\lambda v\| = |\lambda|\,\|v\|

❌ Not quite.

|-3 + 4i|

\sqrt{(-3)^2 + 4^2} = 5

, not

3 + 4 = 7

❌ Close, but…

|\lambda|^2 = 25

belongs in

\|\lambda v\|^2 = |\lambda|^2 \|v\|^2

. Taking square roots of both sides gives

|\lambda|\,\|v\|

❌ Not quite. Find

\|v\|

and

|\lambda|

separately.

Show solution

\|v\|^2 = |1|^2 + |2i|^2 + |2|^2 = 1 + 4 + 4 = 9, \qquad \|v\| = 3.

|\lambda| = |-3 + 4i| = \sqrt{9 + 16} = 5.

\|\lambda v\| = |\lambda|\,\|v\| = 5 \cdot 3 = 15.

Direct check: $\lambda v = (-3 + 4i,\ -8 - 6i,\ -6 + 8i)$ , whose squared moduli are $25$ , $100$ , $100$ , summing to $225 = 15^2$ .

Problem 3 · Orthogonal Polynomials on $[0, 1]$

Given: $\mathcal{P}_1(\mathbf{R})$ , the real polynomials of degree at most $1$ , with the inner product $\langle p, q \rangle = \int_0^1 p q$ . Let $p(x) = x$ and $q(x) = x + c$ — find the $c \in \mathbf{R}$ that makes $p$ and $q$ orthogonal, then $\|p + q\|^2$ .

What is $c$ ?

What is $\|p + q\|^2$ ?

✅ Correct!

\langle x, x - \tfrac{2}{3} \rangle = 0

, so

\|p + q\|^2 = \|p\|^2 + \|q\|^2 = \tfrac{1}{3} + \tfrac{1}{9} = \tfrac{4}{9}

❌ Not quite. That solves

\int_0^1 (x + c) = 0

, which is

\langle 1, q \rangle = 0

. The condition is

\int_0^1 x(x + c) = 0

❌ Not quite.

c = 0

makes

q = p

, and

\langle x, x \rangle = \int_0^1 x^2 = \tfrac{1}{3} \ne 0

— a nonzero vector is never orthogonal to itself.

❌ Close, but… From

\tfrac{1}{3} + \tfrac{c}{2} = 0

you get

c = -\tfrac{2}{3}

; the fraction was inverted.

❌ Check $c$ . Expand

\int_0^1 (x^2 + c x)

and set it to zero.

❌ Not quite.

\tfrac{4}{3} = \|2x\|^2

— that drops

c

. Here

p + q = 2x - \tfrac{2}{3}

❌ Not quite. That is

(\|p\| + \|q\|)^2

. The Pythagorean theorem adds the squares of the norms:

\|p\|^2 + \|q\|^2

❌ Not quite.

\tfrac{2}{9} = \|p\|^2 - \|q\|^2 = \tfrac{1}{3} - \tfrac{1}{9}

. The Pythagorean theorem adds the squared norms.

❌ Check $\|p + q\|^2$ . Since

p \perp q

, add

\|p\|^2 = \int_0^1 x^2

and

\|q\|^2 = \int_0^1 (x - \tfrac{2}{3})^2

Show solution

Step 1: Orthogonality.

\langle p, q \rangle = \int_0^1 x(x + c)\,dx = \frac{1}{3} + \frac{c}{2} = 0 \;\Longrightarrow\; c = -\frac{2}{3}.

Step 2: The two norms squared.

\|p\|^2 = \int_0^1 x^2\,dx = \frac{1}{3}, \qquad \|q\|^2 = \int_0^1 \Big(x - \frac{2}{3}\Big)^2 dx = \frac{1}{3} - \frac{2}{3} + \frac{4}{9} = \frac{1}{9}.

Step 3: Pythagorean theorem. $\|p + q\|^2 = \frac{1}{3} + \frac{1}{9} = \frac{4}{9}$ .

Direct check: $p + q = 2x - \tfrac{2}{3}$ and $\int_0^1 (2x - \tfrac{2}{3})^2\,dx = \tfrac{4}{3} - \tfrac{4}{3} + \tfrac{4}{9} = \tfrac{4}{9}$ .

Problem 4 · Pythagoras in $\mathbf{C}^2$

Given: $u = (1, i)$ and $v = (i, 1)$ in $\mathbf{C}^2$ with the Euclidean inner product $\langle w, z \rangle = w_1 \overline{z_1} + w_2 \overline{z_2}$ — find $\langle u, v \rangle$ , then $\|u + v\|$ .

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

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

✅ Correct!

u \perp v

, so

\|u + v\|^2 = 2 + 2 = 4

and

\|u + v\| = 2

❌ Not quite.

1 \cdot i + i \cdot 1 = 2i

forgets the conjugate on the second slot:

\langle u, v \rangle = 1 \cdot \overline{i} + i \cdot \overline{1} = -i + i

❌ Not quite.

2 = |1|\,|i| + |i|\,|1|

uses moduli. The inner product keeps the complex numbers:

1 \cdot \overline{i} + i \cdot \overline{1} = -i + i

❌ Not quite. The inner product pairs each coordinate of

u

with the conjugate of the same coordinate of

v

1 \cdot \overline{i} + i \cdot \overline{1} = -i + i

❌ Check $\langle u, v \rangle$ . Conjugate the coordinates of

v

\overline{i} = -i

\overline{1} = 1

❌ Close, but…

4

\|u + v\|^2

. Take the square root.

❌ Not quite.

2\sqrt{2} = \|u\| + \|v\|

. The Pythagorean theorem adds squares:

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

❌ Not quite.

\sqrt{2}

\|u\|

alone. Both legs count:

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

❌ Check $\|u + v\|$ . Each of

u

v

has norm squared

|1|^2 + |i|^2 = 2

Show solution

Step 1: Inner product.

\langle u, v \rangle = 1 \cdot \overline{i} + i \cdot \overline{1} = -i + i = 0,

so $u$ and $v$ are orthogonal.

Step 2: Pythagorean theorem. $\|u\|^2 = |1|^2 + |i|^2 = 2$ and $\|v\|^2 = |i|^2 + |1|^2 = 2$ , so

\|u + v\|^2 = \|u\|^2 + \|v\|^2 = 4, \qquad \|u + v\| = 2.

Direct check: $u + v = (1 + i,\ 1 + i)$ and $|1 + i|^2 + |1 + i|^2 = 2 + 2 = 4$ .

Solved: 0 / 4

Norms, Orthogonality and the Pythagorean Theorem

IKey Formulas

IIScaling by a Complex Number

IIIOrthogonal Functions on [−1,1][-1, 1][−1,1]

IVThe Pythagorean Theorem in R2\mathbf{R}^2R2

VQuiz Questions

IIIOrthogonal Functions on $[-1, 1]$

IVThe Pythagorean Theorem in $\mathbf{R}^2$