Linear-Algebra-Done-Right · Unit 14 · Video 1 · Interactive Practice

Inner Products: Why Complex Spaces Need a Conjugate

IKey Formulas

Formula	Statement	Hypotheses
$\langle w, z \rangle = w_1\overline{z_1} + \cdots + w_n\overline{z_n}$	The Euclidean inner product; on $\mathbf{R}^n$ the conjugates do nothing and it is the dot product	$\mathbf{F} = \mathbf{R}$ or $\mathbf{C}$ ; $w = (w_1, \ldots, w_n)$ and $z = (z_1, \ldots, z_n)$ in $\mathbf{F}^n$
$\langle w, z \rangle = c_1 w_1\overline{z_1} + \cdots + c_n w_n\overline{z_n}$	A weighted inner product, with $\langle w, w \rangle = c_1\|w_1\|^2 + \cdots + c_n\|w_n\|^2$	$\mathbf{F} = \mathbf{R}$ or $\mathbf{C}$ ; weights $c_1, \ldots, c_n > 0$ (with $c = (1, -1)$ , $\langle (1,1), (1,1) \rangle = 0$ )
$\langle u, v \rangle = \overline{\langle v, u \rangle}$	Conjugate symmetry; with $u = v$ it forces $\langle v, v \rangle \in \mathbf{R}$ , and on a real space it reads $\langle u, v \rangle = \langle v, u \rangle$	$\langle \cdot, \cdot \rangle$ an inner product on a vector space $V$ over $\mathbf{F}$ ; $u, v \in V$
$\langle u, v + w \rangle = \langle u, v \rangle + \langle u, w \rangle$ , $\ \langle u, \lambda v \rangle = \bar{\lambda}\langle u, v \rangle$	The second slot is additive and scalars come out conjugated (the first slot gives $\langle \lambda u, v \rangle = \lambda\langle u, v \rangle$ )	$V$ an inner product space over $\mathbf{F}$ ; $u, v, w \in V$ , $\lambda \in \mathbf{F}$

Key Insight: For $z = (1, i) \in \mathbf{C}^2$ , squaring and adding gives $1^2 + i^2 = 0$ although $z \neq 0$ . Multiplying each coordinate by its conjugate gives $1 \cdot \overline{1} + i \cdot \overline{i} = 1 + 1 = 2$ , a positive real number, and that is why the conjugate sits in the definition.

IISquares Versus Conjugates in C²

Squaring and adding the coordinates of a nonzero $z \in \mathbf{C}^2$ can give $0$ or worse; $z_1\overline{z_1} + z_2\overline{z_2}$ never does.

vector z

IIITesting a Candidate Formula on R²

A formula on $\mathbf{R}^2$ is an inner product only if all five properties hold for every $u$ , $v$ and $\lambda$ .

⟨x, y⟩ = λ 2

💡 One choice of vectors that breaks a property disproves it, but passing at one choice proves nothing: $2x_1y_1 + x_2y_2$ passes everywhere because it is linear in $x$ , symmetric, and $\langle x, x \rangle = 2x_1^2 + x_2^2$ .

IVScalars Leave the Second Slot Conjugated

Multiplying $\langle u, v \rangle$ by $\lambda$ rotates it one way in $\mathbf{C}$ ; multiplying by $\bar{\lambda}$ rotates it the opposite way.

u, v

VQuiz Questions

Problem 1 · A Weighted Inner Product on R³

Given: the inner product on $\mathbf{R}^3$ defined by $\langle x, y \rangle = 2x_1y_1 + x_2y_2 + 3x_3y_3$ , with $x = (1, -2, 2)$ and $y = (3, 1, -1)$ . Find $\langle x, y \rangle$ .

✅ Correct!

2(1)(3) + (-2)(1) + 3(2)(-1) = 6 - 2 - 6 = -2

❌ That is the dot product.

3 - 2 - 2 = -1

ignores the weights

2

1

3

; each product

x_ky_k

must be multiplied by its weight

c_k

❌ Weights in the wrong order.

3(1)(3) + (-2)(1) + 2(2)(-1) = 3

uses the weights reversed;

c_1 = 2

multiplies

x_1y_1

and

c_3 = 3

multiplies

x_3y_3

❌ The sign is real. Positivity constrains only

\langle v, v \rangle

; the inner product of two different vectors can be negative, and here it is

-2

Show solution

The weights are $c_1 = 2$ , $c_2 = 1$ , $c_3 = 3$ , all positive, so this is an inner product on $\mathbf{R}^3$ (on $\mathbf{R}$ the conjugates do nothing). Multiply coordinate by coordinate, weight, and add:

\langle x, y \rangle = 2(1)(3) + (-2)(1) + 3(2)(-1) = 6 - 2 - 6 = -2.

A negative value is allowed: positivity says only that $\langle v, v \rangle \geq 0$ . For instance $\langle x, x \rangle = 2(1)^2 + (-2)^2 + 3(2)^2 = 2 + 4 + 12 = 18 \geq 0$ .

Problem 2 · Which Property Breaks?

Given: the formula $\langle x, y \rangle = |x_1y_1| + |x_2y_2|$ on $\mathbf{R}^2$ . Exactly one of the properties listed below fails for it. Which one?

✅ Correct! The absolute values swallow the sign of

\lambda

: with

\lambda = -1

u = (1, 2)

v = (3, -1)

we get

\langle -u, v \rangle = 5

but

-\langle u, v \rangle = -5

❌ Positivity holds.

\langle v, v \rangle = |v_1^2| + |v_2^2| = v_1^2 + v_2^2 \geq 0

❌ Definiteness holds.

v_1^2 + v_2^2 = 0

forces

v_1 = v_2 = 0

❌ Symmetry holds.

|u_1v_1| = |v_1u_1|

and

|u_2v_2| = |v_2u_2|

, so swapping

u

and

v

changes nothing; on

\mathbf{R}

this is conjugate symmetry.

Show solution

Check each listed property.

Positivity and definiteness: $\langle v, v \rangle = v_1^2 + v_2^2$ , which is $\geq 0$ and is $0$ only for $v = 0$ . Both hold.

Symmetry: each term $|x_ky_k|$ is unchanged when $x$ and $y$ swap. It holds.

Homogeneity: $\langle \lambda u, v \rangle = |\lambda|\,\langle u, v \rangle$ , which differs from $\lambda\langle u, v \rangle$ whenever $\lambda < 0$ and $\langle u, v \rangle \neq 0$ . With $u = (1, 2)$ , $v = (3, -1)$ , $\lambda = -1$ :

\langle -u, v \rangle = |(-1)(3)| + |(-2)(-1)| = 3 + 2 = 5 \neq -5 = -\langle u, v \rangle.

So the formula is not an inner product. (Additivity in the first slot, which is not listed, fails too: $u = (1, 0)$ , $v = w = (-2, 0)$ give $\langle u + v, w \rangle = 2$ but $\langle u, w \rangle + \langle v, w \rangle = 2 + 4 = 6$ .)

Problem 3 · The Euclidean Inner Product on C²

Given: $\mathbf{C}^2$ with the Euclidean inner product, $u = (1 + i,\ 2)$ and $v = (i,\ 1 - i)$ . Find $\langle u, v \rangle$ and $\langle v, u \rangle$ .

What is ⟨u, v⟩?

What is ⟨v, u⟩?

✅ Correct!

\langle u, v \rangle = 3 + i

and

\langle v, u \rangle = 3 - i = \overline{3 + i}

, exactly as conjugate symmetry demands.

❌ Check ⟨u, v⟩. Conjugate the coordinates of the second vector:

\overline{i} = -i

and

\overline{1 - i} = 1 + i

❌ The conjugate is missing.

(1 + i)i + 2(1 - i) = 1 - i

is the plain coordinate product; the Euclidean inner product uses

w_k\overline{z_k}

❌ Wrong slot conjugated. Conjugating

u

instead of

v

gives

\overline{1 + i}\,i + 2(1 - i) = 3 - i

, which is

\langle v, u \rangle

. In

\langle u, v \rangle

the conjugates go on the second vector.

❌ Too many conjugates.

\overline{1 + i}\,\overline{i} + \overline{2}\,\overline{(1 - i)} = 1 + i

conjugates both vectors, which gives

\overline{1 - i}

, the conjugate of the plain coordinate product. Only the second vector's coordinates are conjugated.

❌ The conjugate is missing.

i(1 + i) + (1 - i)(2) = 1 - i

is the plain coordinate product, which is the same in either order;

\langle v, u \rangle

conjugates the coordinates of

u

❌ Check ⟨v, u⟩. Either compute

i\,\overline{(1 + i)} + (1 - i)\,\overline{2}

, or conjugate your answer for

\langle u, v \rangle

❌ Not symmetric over C. Swapping the slots conjugates the value:

\langle v, u \rangle = \overline{\langle u, v \rangle} = \overline{3 + i}

❌ Conjugate, not negate.

\overline{3 + i} = 3 - i

; only the imaginary part changes sign.

Show solution

Step 1: ⟨u, v⟩. Multiply each coordinate of $u$ by the conjugate of the matching coordinate of $v$ :

\langle u, v \rangle = (1 + i)\,\overline{i} + 2\,\overline{(1 - i)} = (1 + i)(-i) + 2(1 + i) = (1 - i) + (2 + 2i) = 3 + i.

Step 2: ⟨v, u⟩. Directly:

\langle v, u \rangle = i\,\overline{(1 + i)} + (1 - i)\,\overline{2} = i(1 - i) + (2 - 2i) = (1 + i) + (2 - 2i) = 3 - i.

Check: conjugate symmetry predicts $\langle v, u \rangle = \overline{\langle u, v \rangle} = \overline{3 + i} = 3 - i$ . ✓

Problem 4 · Scalars in Both Slots

Given: $V$ an inner product space over $\mathbf{C}$ and $u, v \in V$ with $\langle u, v \rangle = 2 - i$ . Find $\langle (1 + i)u,\ 3i\,v \rangle$ .

✅ Correct!

(1 + i)\,\overline{3i}\,(2 - i) = (1 + i)(-3i)(2 - i) = (3 - 3i)(2 - i) = 3 - 9i

❌ The second-slot scalar must be conjugated.

(1 + i)(3i)(2 - i) = -3 + 9i

treats the inner product as linear in both slots;

\langle u, \lambda v \rangle = \bar{\lambda}\langle u, v \rangle

, so

3i

comes out as

-3i

❌ That is the other convention. Conjugating the first-slot scalar instead gives

(1 - i)(3i)(2 - i) = 9 + 3i

, the physicists' convention. Here homogeneity holds in the first slot, so

1 + i

comes out unchanged.

❌ Only one scalar is conjugated.

(1 - i)(-3i)(2 - i) = -9 - 3i

conjugates both; a scalar leaves the first slot unchanged.

Show solution

Step 1: first slot. Homogeneity in the first slot: $\langle (1 + i)u, 3i\,v \rangle = (1 + i)\,\langle u, 3i\,v \rangle$ .

Step 2: second slot. Scalars come out conjugated: $\langle u, 3i\,v \rangle = \overline{3i}\,\langle u, v \rangle = -3i\,(2 - i)$ .

Step 3: multiply.

(1 + i)(-3i)(2 - i) = (-3i - 3i^2)(2 - i) = (3 - 3i)(2 - i) = 6 - 3i - 6i + 3i^2 = 3 - 9i.

The derivation of Step 2 is the theorem from the video: $\langle u, \lambda v \rangle = \overline{\langle \lambda v, u \rangle} = \overline{\lambda\langle v, u \rangle} = \bar{\lambda}\,\overline{\langle v, u \rangle} = \bar{\lambda}\langle u, v \rangle$ .

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