6.1 Inner products and norms

In $ℝ^{2}$ the dot product $𝐯 \cdot$ $𝐰 =$ $𝑣_{1} 𝑤_{1} +$ $𝑣_{2} 𝑤_{2}$ knows the length of every vector and the angle between any two. This section takes the algebraic properties of the dot product that are responsible for this and makes them axioms. A vector space over $ℝ$ or $ℂ$ equipped with a product satisfying them, an inner product, acquires lengths and angles, even when its vectors are functions or matrices and there is no picture to look at.

6.1.1Length and angle in $ℝ^{2}$

For $𝐯 = (𝑣_{1}, 𝑣_{2})$ and $𝐰 = (𝑤_{1}, 𝑤_{2})$ in $ℝ^{2}$ , the dot product is

𝐯 \cdot 𝐰 = 𝑣_{1} 𝑤_{1} + 𝑣_{2} 𝑤_{2} \in ℝ .

Draw $𝐯$ as an arrow from the origin. Its length $‖ 𝐯 ‖$ is the hypotenuse of a right triangle with legs $| 𝑣_{1} |$ and $| 𝑣_{2} |$ , so by Pythagoras

{‖ 𝐯 ‖}^{2} = 𝑣_{1}^{2} + 𝑣_{2}^{2} = 𝐯 \cdot 𝐯 .

Now let $𝐯$ and $𝐰$ be nonzero, and let $𝜃 \in [0, 𝜋]$ be the angle between them. The arrows $𝐯$ and $𝐰$ , together with the arrow $𝐯 - 𝐰$ from the tip of $𝐰$ to the tip of $𝐯$ , form a triangle (Figure 6.1.1), and the law of cosines says

{‖ 𝐯 - 𝐰 ‖}^{2} = {‖ 𝐯 ‖}^{2} + {‖ 𝐰 ‖}^{2} - 2 ‖ 𝐯 ‖ ‖ 𝐰 ‖ cos 𝜃 .

We can also expand the left side algebraically. The dot product is symmetric, $𝐯 \cdot 𝐰 = 𝐰 \cdot 𝐯$ , and it is linear in each slot, so

{‖ 𝐯 - 𝐰 ‖}^{2} = (𝐯 - 𝐰) \cdot (𝐯 - 𝐰) = 𝐯 \cdot 𝐯 - 2 𝐯 \cdot 𝐰 + 𝐰 \cdot 𝐰 .

Comparing the two expressions, and using ${‖ 𝐯 ‖}^{2} = 𝐯 \cdot 𝐯$ and ${‖ 𝐰 ‖}^{2} = 𝐰 \cdot 𝐰$ , we get

𝐯 \cdot 𝐰 = ‖ 𝐯 ‖ ‖ 𝐰 ‖ cos 𝜃 .

(6.1.1)

So lengths and angles can both be read off from the dot product. In particular, $𝐯$ and $𝐰$ are perpendicular exactly when $𝐯 \cdot 𝐰 = 0$ , which is where Section 6.3 starts.

Figure 6.1.1. The triangle behind (6.1.1). Its sides have lengths

‖ 𝐯 ‖

‖ 𝐰 ‖

, and

‖ 𝐯 - 𝐰 ‖

, and the angle between the first two is

𝜃

Look at what the computation used: symmetry, linearity in each slot, and the fact that $𝐯 \cdot 𝐯$ is a nonnegative real number that vanishes only for $𝐯 = 𝟎$ , so that $‖ 𝐯 ‖ = \sqrt{𝐯 \cdot 𝐯}$ makes sense and is positive when $𝐯 \neq 𝟎$ . Turned around, any product with these properties, on any vector space, lets us define the length of a vector as $\sqrt{𝐯 \cdot 𝐯}$ , and (6.1.1) suggests a definition of the angle between two vectors. (Section 6.2 shows that the resulting value of $cos 𝜃$ lies in $[- 1, 1]$ .)

Which scalars?

The axioms will compare $𝐯 \cdot 𝐯$ with $0$ , and lengths are square roots. So the scalars need an order and square roots of their nonnegative elements. $ℝ$ has both, and the lack of either causes real trouble.

In $ℚ^{2}$ , the vector $(1, 1)$ would have length $\sqrt{2} \notin ℚ$ .
In $𝔽_{2}^{2}$ , the same formula gives $(1, 1) \cdot (1, 1) = 1 + 1 = 0$ , so a nonzero vector would have length $0$ . Ordering $𝔽_{𝑝}$ cannot help: an order compatible with the field operations would force $0 <$ $1 <$ $1 +$ $1 <$ $\dots <$ $𝑝 \cdot$ $1 =$ $0$ .
In $ℂ^{2}$ , $(1, 𝑖) \cdot$ $(1, 𝑖) =$ $1 +$ $𝑖^{2} =$ $0$ , and $(1, 1 + 𝑖) \cdot$ $(1, 1 + 𝑖) =$ $1 +$ $2 𝑖$ is not even real.

$ℂ$ has no useful order either, but it contains $ℝ$ , and the case $𝑛 = 1$ shows how to land in $ℝ$ . Viewing $𝑧 = 𝑎 + 𝑏 𝑖$ as the point $(𝑎, 𝑏)$ of the plane, its length is $\sqrt{𝑎^{2} + 𝑏^{2}}$ , and $𝑎^{2} + 𝑏^{2} = 𝑧 𝑧$ , not $𝑧^{2}$ . So on $ℂ^{𝑛}$ we conjugate the second factor and use $\sum_{𝑘} 𝑣_{𝑘} 𝑤_{𝑘}$ . Then $𝐯 \cdot 𝐯 = \sum_{𝑘} {| 𝑣_{𝑘} |}^{2}$ is real and nonnegative, and the price is that symmetry becomes “conjugate symmetry”. In this chapter the scalars are always real or complex.

Notation 6.1.1 (Real and complex scalars). Throughout Chapter 6, $𝔽$ denotes $ℝ$ or $ℂ$ . For $𝑧 = 𝑎 + 𝑏 𝑖 \in ℂ$ with $𝑎, 𝑏 \in ℝ$ , the conjugate of $𝑧$ is $𝑧 = 𝑎 - 𝑏 𝑖$ , its modulus is $| 𝑧 | = \sqrt{𝑎^{2} + 𝑏^{2}}$ , and its real and imaginary parts are $Re 𝑧 = 𝑎$ and $Im 𝑧 = 𝑏$ . For all $𝑧, 𝑤 \in ℂ$ ,

\begin{matrix} 𝑧 + 𝑤 = 𝑧 + 𝑤, 𝑧 𝑤 = 𝑧 𝑤, 𝑧 = 𝑧, \\ 𝑧 𝑧 = {| 𝑧 |}^{2}, 𝑧 + 𝑧 = 2 Re 𝑧, Re (- 𝑖 𝑧) = Im 𝑧, \end{matrix}

$| Re 𝑧 | \leq | 𝑧 |$ , and $𝑧 = 𝑧$ if and only if $𝑧 \in ℝ$ . Each of these is checked by writing $𝑧$ and $𝑤$ in coordinates; Section B.1 develops conjugation and the modulus further. For real $𝑧$ , $𝑧 = 𝑧$ and $| 𝑧 |$ is the absolute value, so every formula in this chapter specializes to $ℝ$ by erasing the bars.

6.1.2Inner products

Definition 6.1.2 (Inner product). Let $𝑉$ be a vector space over $𝔽$ . An inner product on $𝑉$ is a function $𝑉 \times 𝑉 \to 𝔽$ , written $(𝐯, 𝐰) \mapsto ⟨ 𝐯, 𝐰 ⟩$ , such that for all $𝐯, 𝐰, 𝐱 \in 𝑉$ and all $𝑎, 𝑏 \in 𝔽$ :

Conjugate symmetry. $⟨ 𝐯, 𝐰 ⟩ = ⟨ 𝐰, 𝐯 ⟩$ .
Linearity in the first slot. $⟨ 𝑎 𝐯 + 𝑏 𝐰, 𝐱 ⟩ =$ $𝑎 ⟨ 𝐯, 𝐱 ⟩ +$ $𝑏 ⟨ 𝐰, 𝐱 ⟩$ .
Positivity. $⟨ 𝐯, 𝐯 ⟩ \geq 0$ .
Definiteness (or non-degeneracy). $⟨ 𝐯, 𝐯 ⟩ = 0$ if and only if $𝐯 = 𝟎$ .

An inner product space is a vector space over $𝔽$ together with a specified inner product on it.

Remark 6.1.3. Taking $𝐰 = 𝐯$ in conjugate symmetry gives $⟨ 𝐯, 𝐯 ⟩ = ⟨ 𝐯, 𝐯 ⟩$ , so $⟨ 𝐯, 𝐯 ⟩$ is real for every $𝐯$ , and positivity makes sense even when $𝔽 = ℂ$ . Over $ℝ$ conjugation does nothing, and conjugate symmetry is plain symmetry.

Half of definiteness is automatic: linearity gives $⟨ 𝟎, 𝟎 ⟩ =$ $⟨ 0 𝟎, 𝟎 ⟩ =$ $0 ⟨ 𝟎, 𝟎 ⟩ =$ $0$ . So to verify definiteness one only has to show that $⟨ 𝐯, 𝐯 ⟩ = 0$ forces $𝐯 = 𝟎$ .

A vector space usually carries many inner products, and lengths and angles depend on the choice (Example 6.1.21). So we always say which inner product we use.

Linearity is required only in the first slot. The second slot inherits a twisted version of it from conjugate symmetry.

Proposition 6.1.4 (Sesquilinearity). Let $𝑉$ be an inner product space. For all $𝐯, 𝐰, 𝐱 \in 𝑉$ and $𝑎, 𝑏 \in 𝔽$ :

(a) $⟨ 𝐯, 𝑎 𝐰 + 𝑏 𝐱 ⟩ =$ $𝑎 ⟨ 𝐯, 𝐰 ⟩ +$ $𝑏 ⟨ 𝐯, 𝐱 ⟩$ ;
(b) $⟨ 𝟎, 𝐯 ⟩ = ⟨ 𝐯, 𝟎 ⟩ = 0$ .

Proof.routine

Uses. Definition 6.1.2 (Inner product), Notation 6.1.1 (Real and complex scalars)

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. The only axiom that says anything about the second slot is conjugate symmetry. Use it to move the second slot into the first.

Hint 2. Flip with conjugate symmetry, expand with linearity in the first slot, and flip back. Conjugation turns a sum into a sum and a product into a product.

Hint 3. For (b), write

𝟎 = 𝟎 + 𝟎

in the first slot.

Sketch. Conjugate symmetry turns

⟨ 𝐯, 𝑎 𝐰 + 𝑏 𝐱 ⟩

into the conjugate of

𝑎 ⟨ 𝐰, 𝐯 ⟩ +

𝑏 ⟨ 𝐱, 𝐯 ⟩

, which is

𝑎 ⟨ 𝐯, 𝐰 ⟩ +

𝑏 ⟨ 𝐯, 𝐱 ⟩

. For (b),

⟨ 𝟎, 𝐯 ⟩ =

⟨ 𝟎 + 𝟎, 𝐯 ⟩ =

2 ⟨ 𝟎, 𝐯 ⟩

, and conjugate symmetry handles

⟨ 𝐯, 𝟎 ⟩

(a) We compute

\begin{aligned} ⟨ 𝐯, 𝑎 𝐰 + 𝑏 𝐱 ⟩ & = ⟨ 𝑎 𝐰 + 𝑏 𝐱, 𝐯 ⟩ & (conjugate symmetry) \\ = 𝑎 ⟨ 𝐰, 𝐯 ⟩ + 𝑏 ⟨ 𝐱, 𝐯 ⟩ & (linearity in the first slot) \\ = 𝑎 ⟨ 𝐰, 𝐯 ⟩ + 𝑏 ⟨ 𝐱, 𝐯 ⟩ & (conjugation respects + and \cdot) \\ = 𝑎 ⟨ 𝐯, 𝐰 ⟩ + 𝑏 ⟨ 𝐯, 𝐱 ⟩ & (conjugate symmetry), \end{aligned}

as desired.

(b) By linearity in the first slot, $⟨ 𝟎, 𝐯 ⟩ =$ $⟨ 𝟎 + 𝟎, 𝐯 ⟩ =$ $⟨ 𝟎, 𝐯 ⟩ +$ $⟨ 𝟎, 𝐯 ⟩$ . Subtracting $⟨ 𝟎, 𝐯 ⟩$ from both sides gives $⟨ 𝟎, 𝐯 ⟩ = 0$ . Then $⟨ 𝐯, 𝟎 ⟩ =$ $⟨ 𝟎, 𝐯 ⟩ =$ $0 =$ $0$ . □

A function of two vectors that is linear in the first slot and satisfies (a) in the second is called sesquilinear (“one and a half times linear”). Over $ℝ$ the bars disappear: a real inner product is linear in each slot, that is, bilinear, and symmetric. Over $ℂ$ , scalars leave the second slot conjugated, as in $⟨ 𝐯, 𝑖 𝐰 ⟩ =$ $- 𝑖 ⟨ 𝐯, 𝐰 ⟩$ . Forgetting the bar is the most common error in computations with complex inner products.

Example 6.1.5 (Expanding an inner product). Let $𝐱, 𝐲$ be vectors in a complex inner product space with $⟨ 𝐱, 𝐱 ⟩ = 1$ , $⟨ 𝐲, 𝐲 ⟩ = 2$ , and $⟨ 𝐱, 𝐲 ⟩ = 𝑖$ , so that $⟨ 𝐲, 𝐱 ⟩ = 𝑖 = - 𝑖$ . To compute the inner product below, expand each slot, pulling scalars out of the first slot as they are and out of the second slot conjugated:

\begin{aligned} ⟨ (1 + 𝑖) 𝐱 + 𝐲, 2 𝐱 - 𝑖 𝐲 ⟩ \\ = (1 + 𝑖) \cdot 2 ⟨ 𝐱, 𝐱 ⟩ + (1 + 𝑖) - 𝑖 ⟨ 𝐱, 𝐲 ⟩ \\ + 2 ⟨ 𝐲, 𝐱 ⟩ + - 𝑖 ⟨ 𝐲, 𝐲 ⟩ \\ = 2 (1 + 𝑖) + (1 + 𝑖) 𝑖 \cdot 𝑖 + 2 (- 𝑖) + 𝑖 \cdot 2 \\ = (2 + 2 𝑖) + (- 1 - 𝑖) - 2 𝑖 + 2 𝑖 = 1 + 𝑖 . \end{aligned}

Here $(1 + 𝑖) 𝑖 \cdot$ $𝑖 =$ $(1 + 𝑖) (- 1)$ . This is Technique 5.1.7 for a sesquilinear function: each slot is split into its terms, and the four resulting values are known.

Over $ℂ$ the conjugation is forced on us, because positivity and definiteness together are incompatible with linearity in both slots.

Proposition 6.1.6 (Bilinear forms over

ℂ

are not positive definite). Let

𝑉 \neq {𝟎}

be a complex vector space, and let

𝐵 : 𝑉 \times 𝑉 \to ℂ

be linear in each slot when the other is fixed. If

𝐵 (𝐯, 𝐯)

is real and nonnegative for every

𝐯 \in 𝑉

, then

𝐵 (𝐯, 𝐯) = 0

for every

𝐯 \in 𝑉

. In particular,

𝐵

cannot satisfy both positivity and definiteness.

Proof.routine

Uses. Definition 6.1.2 (Inner product)

Hint 1. Compare

𝐵 (𝐯, 𝐯)

with

𝐵 (𝑐 𝐯, 𝑐 𝐯)

for a scalar

𝑐

Hint 2. Take

𝑐 = 𝑖

. What is

𝑖 \cdot 𝑖

Sketch.

𝐵 (𝑖 𝐯, 𝑖 𝐯) =

𝑖^{2} 𝐵 (𝐯, 𝐯) =

- 𝐵 (𝐯, 𝐯)

, and a real number whose negative is also nonnegative is

0

. A nonzero

𝐯

then violates definiteness.

Let

𝐯 \in 𝑉

. Linearity in each slot gives

𝐵 (𝑖 𝐯, 𝑖 𝐯) =

𝑖 \cdot

𝑖 𝐵 (𝐯, 𝐯) =

- 𝐵 (𝐯, 𝐯)

. Both

𝐵 (𝐯, 𝐯)

and

𝐵 (𝑖 𝐯, 𝑖 𝐯) =

- 𝐵 (𝐯, 𝐯)

are nonnegative real numbers, so

𝐵 (𝐯, 𝐯) = 0

. Since

𝑉 \neq {𝟎}

, there is some

𝐯 \neq 𝟎

, and

𝐵 (𝐯, 𝐯) = 0

violates definiteness. □

With the conjugate in the second slot the problem disappears: $⟨ 𝑖 𝐯, 𝑖 𝐯 ⟩ =$ $𝑖 𝑖 ⟨ 𝐯, 𝐯 ⟩ =$ $⟨ 𝐯, 𝐯 ⟩$ . This is what it should be, since multiplying by $𝑖$ rotates each coordinate of $𝐯 \in ℂ^{𝑛}$ without changing its modulus, so it should not change the length of $𝐯$ .

6.1.3Examples and non-examples

The dot product

The dot product on $ℂ^{𝑛}$ is most compactly written with the conjugate transpose, which Section 6.3 studies in detail.

Notation 6.1.7 (Conjugate transpose). For

𝐴 =

(𝑎_{𝑖 𝑗}) \in

ℂ^{𝑚 \times 𝑛}

, the conjugate transpose of

𝐴

is the matrix

𝐴^{*} \in ℂ^{𝑛 \times 𝑚}

whose

(𝑖, 𝑗)

entry is

𝑎_{𝑗 𝑖}

. That is,

𝐴^{*} = 𝐴^{𝖳}

, where

𝐴 = (𝑎_{𝑖 𝑗})

conjugates every entry (Definition 2.4.21). For a real matrix,

𝐴^{*} = 𝐴^{𝖳}

. For

𝐰 \in 𝔽^{𝑛}

𝐰^{*}

is the row

(𝑤_{1}, \dots, 𝑤_{𝑛})

, and for

𝐯 \in 𝔽^{𝑛}

the product

𝐰^{*} 𝐯

is a

1 \times 1

matrix, which we identify with its entry

\sum_{𝑘} 𝑤_{𝑘} 𝑣_{𝑘}

Definition 6.1.8 (Dot product). The dot product on $𝔽^{𝑛}$ is

𝐯 \cdot 𝐰 = \sum_{𝑘 = 1}^{𝑛} 𝑣_{𝑘} 𝑤_{𝑘} = 𝐰^{*} 𝐯

for $𝐯 = (𝑣_{1}, \dots, 𝑣_{𝑛})$ and $𝐰 = (𝑤_{1}, \dots, 𝑤_{𝑛})$ in $𝔽^{𝑛}$ . Over $ℝ$ it is $\sum_{𝑘} 𝑣_{𝑘} 𝑤_{𝑘} =$ $𝐰^{𝖳} 𝐯$ .

Checking the axioms for the dot product comes down to one elementary fact about real numbers, which we name because it is used constantly in this chapter.

Technique 6.1.9 (A sum of nonnegative terms). A sum of nonnegative real numbers is $0$ only if every term is $0$ : if $𝑡_{1}, \dots, 𝑡_{𝑛} \geq 0$ and some $𝑡_{𝑗} > 0$ , then $𝑡_{1} + \dots + 𝑡_{𝑛} \geq 𝑡_{𝑗} > 0$ . So to show that many quantities vanish, it suffices to exhibit them as squares, or squared moduli ${| 𝑧_{𝑘} |}^{2}$ , in a single sum, and to show that the sum is $0$ .

Reach for it when a positivity or definiteness axiom must be verified for a concrete formula, and whenever one equation should force many. The canonical example is definiteness of the dot product (Proposition 6.1.10), where $\sum_{𝑘} {| 𝑣_{𝑘} |}^{2} = 0$ forces every $𝑣_{𝑘} = 0$ . Integrals of nonnegative continuous functions behave the same way (Lemma 6.1.14), and Technique 6.1.18 applies the idea to the entries of a matrix.

Used in 27 places and 2 practice exercises

Exercise 1.5.48, Problem 3.15, Exercise 4.3.33, Proposition 6.1.10, Lemma 6.1.14, Proposition 6.1.13, Exercise 6.1.42, Exercise 6.1.46, Lemma 6.2.6, Exercise 6.2.43, Exercise 6.2.47, Proposition 6.3.13, Exercise 6.3.52, Theorem 6.4.33, Exercise 6.4.60, Exercise 6.7.12, Problem 6.5, Problem 6.12, Problem 6.15, Problem 6.16, Theorem 8.1.3, Lemma 8.2.4, Lemma 8.3.8, Notation 8.3.10, Exercise 8.3.40, Lemma 8.4.24, Exercise 8.4.35; 2 practice exercises.

Proposition 6.1.10 (The dot product is an inner product). The dot product is an inner product on

𝔽^{𝑛}

, and

𝐯 \cdot

𝐯 =

\sum_{𝑘 = 1}^{𝑛} {| 𝑣_{𝑘} |}^{2}

for all

𝐯 \in 𝔽^{𝑛}

Proof.routine

Uses. Definition 6.1.2 (Inner product), Definition 6.1.8 (Dot product), Notation 6.1.1 (Real and complex scalars)

Techniques. A sum of nonnegative terms (Technique 6.1.9)

Hint 1. Check the axioms one at a time with the coordinate formula

\sum_{𝑘} 𝑣_{𝑘} 𝑤_{𝑘}

Hint 2. Conjugate symmetry is

𝑤_{𝑘} 𝑣_{𝑘} = 𝑤_{𝑘} 𝑣_{𝑘}

for each term. For positivity,

𝑣_{𝑘} 𝑣_{𝑘} = {| 𝑣_{𝑘} |}^{2}

Hint 3. For definiteness, a sum of nonnegative real numbers is

0

only if each term is.

Sketch. Conjugating

𝐰 \cdot 𝐯

term by term gives

𝐯 \cdot 𝐰

, and linearity in

𝐯

is visible from the formula. Then

𝐯 \cdot

𝐯 =

\sum_{𝑘} {| 𝑣_{𝑘} |}^{2} \geq

0

, and it vanishes only if every

| 𝑣_{𝑘} | = 0

, that is,

𝐯 = 𝟎

Let $𝐯, 𝐰, 𝐱 \in 𝔽^{𝑛}$ and $𝑎, 𝑏 \in 𝔽$ .

Conjugate symmetry. $𝐰 \cdot 𝐯 =$ $\sum_{𝑘} 𝑤_{𝑘} 𝑣_{𝑘} =$ $\sum_{𝑘} 𝑤_{𝑘} 𝑣_{𝑘} =$ $𝐯 \cdot$ $𝐰$ .

Linearity in the first slot. The $𝑘$ th coordinate of $𝑎 𝐯 + 𝑏 𝐰$ is $𝑎 𝑣_{𝑘} + 𝑏 𝑤_{𝑘}$ , so

(𝑎 𝐯 + 𝑏 𝐰) \cdot 𝐱 = \sum_{𝑘} (𝑎 𝑣_{𝑘} + 𝑏 𝑤_{𝑘}) 𝑥_{𝑘} = 𝑎 \sum_{𝑘} 𝑣_{𝑘} 𝑥_{𝑘} + 𝑏 \sum_{𝑘} 𝑤_{𝑘} 𝑥_{𝑘} = 𝑎 (𝐯 \cdot 𝐱) + 𝑏 (𝐰 \cdot 𝐱) .

Positivity. $𝐯 \cdot$ $𝐯 =$ $\sum_{𝑘} 𝑣_{𝑘} 𝑣_{𝑘} =$ $\sum_{𝑘} {| 𝑣_{𝑘} |}^{2} \geq$ $0$ .

Definiteness. Suppose $𝐯 \cdot 𝐯 = 0$ . Then $\sum_{𝑘} {| 𝑣_{𝑘} |}^{2} = 0$ is a sum of nonnegative real numbers, so ${| 𝑣_{𝑘} |}^{2} = 0$ for every $𝑘$ . Writing $𝑣_{𝑘} = 𝑎_{𝑘} + 𝑏_{𝑘} 𝑖$ , this says $𝑎_{𝑘}^{2} + 𝑏_{𝑘}^{2} = 0$ , so $𝑎_{𝑘} = 𝑏_{𝑘} = 0$ and $𝑣_{𝑘} = 0$ . Hence $𝐯 = 𝟎$ . By Remark 6.1.3, this is all that definiteness requires. □

Unless another inner product is specified, $𝔽^{𝑛}$ carries the dot product. The lecture calls it “the dot product” precisely to distinguish it from the other inner products on $𝔽^{𝑛}$ below.

Example 6.1.11 (Computing in $ℂ^{2}$ ). Let $𝐯 = (1 + 𝑖, 2)$ and $𝐰 = (𝑖, 1 - 𝑖)$ . Then

\begin{aligned} 𝐯 \cdot 𝐰 & = (1 + 𝑖) 𝑖 + 2 (1 - 𝑖) = (1 + 𝑖) (- 𝑖) + 2 (1 + 𝑖) = (1 - 𝑖) + (2 + 2 𝑖) = 3 + 𝑖, \\ 𝐰 \cdot 𝐯 & = 𝑖 (1 + 𝑖) + (1 - 𝑖) 2 = 𝑖 (1 - 𝑖) + 2 - 2 𝑖 = 3 - 𝑖, \end{aligned}

as conjugate symmetry predicts. Also $𝐯 \cdot$ $𝐯 =$ ${| 1 + 𝑖 |}^{2} +$ ${| 2 |}^{2} =$ $2 +$ $4 =$ $6$ and $𝐰 \cdot 𝐰 = 1 + 2 = 3$ . Without the conjugate, $𝐯$ would have “squared length” ${(1 + 𝑖)}^{2} + 4 = 4 + 2 𝑖$ .

Exercise 6.1.12.routine In

ℂ^{3}

, let

𝐯 = (1, 𝑖, 2)

and

𝐰 = (𝑖, 1 + 𝑖, 1 - 𝑖)

. Compute

𝐯 \cdot 𝐰

𝐰 \cdot 𝐯

𝐯 \cdot 𝐯

, and

𝐰 \cdot 𝐰

, and find

𝑐 \in ℂ

with

(𝐯 - 𝑐 𝐰) \cdot 𝐰 = 0

Solution.

Uses. Definition 6.1.8 (Dot product), Proposition 6.1.10 (The dot product is an inner product)

Hint 1. Conjugate the coordinates of the second vector, multiply coordinatewise, and add.

Hint 2. For the last part, expand

(𝐯 - 𝑐 𝐰) \cdot 𝐰

by linearity in the first slot; it is a linear equation in

𝑐

Sketch.

𝐯 \cdot

𝐰 =

- 𝑖 +

(1 + 𝑖) +

(2 + 2 𝑖) =

3 +

2 𝑖

, and

𝐰 \cdot 𝐯 = 3 - 2 𝑖

. The squared lengths are

1 + 1 + 4 = 6

and

1 + 2 + 2 = 5

. Linearity gives

(𝐯 - 𝑐 𝐰) \cdot

𝐰 =

(3 + 2 𝑖) -

5 𝑐

, so

𝑐 = (3 + 2 𝑖) / 5

By Definition 6.1.8,

𝐯 \cdot 𝐰 = 1 \cdot 𝑖 + 𝑖 \cdot (1 + 𝑖) + 2 \cdot (1 - 𝑖) = - 𝑖 + 𝑖 (1 - 𝑖) + 2 (1 + 𝑖) = - 𝑖 + (𝑖 + 1) + (2 + 2 𝑖) = 3 + 2 𝑖,

and by conjugate symmetry $𝐰 \cdot 𝐯 = 3 + 2 𝑖 = 3 - 2 𝑖$ . Directly,

𝐰 \cdot 𝐯 = 𝑖 \cdot 1 + (1 + 𝑖) (- 𝑖) + (1 - 𝑖) 2 = 𝑖 + (1 - 𝑖) + (2 - 2 𝑖) = 3 - 2 𝑖 .

By Proposition 6.1.10, $𝐯 \cdot 𝐯 = 1 + 1 + 4 = 6$ and $𝐰 \cdot$ $𝐰 =$ $1 +$ ${| 1 + 𝑖 |}^{2} +$ ${| 1 - 𝑖 |}^{2} =$ $1 +$ $2 +$ $2 =$ $5$ .

By linearity in the first slot, $(𝐯 - 𝑐 𝐰) \cdot$ $𝐰 =$ $𝐯 \cdot$ $𝐰 -$ $𝑐 (𝐰 \cdot 𝐰) =$ $(3 + 2 𝑖) -$ $5 𝑐$ . This vanishes exactly for $𝑐 = (3 + 2 𝑖) / 5$ .

Continuous functions

Let $𝐶 ([0, 1], ℂ)$ be the set of continuous functions $𝑓 : [0, 1] \to ℂ$ , that is, of functions $𝑓 = 𝑢 + 𝑖 𝑣$ with $𝑢, 𝑣 \in 𝐶 [0, 1]$ (Example 1.2.12). It is a complex vector space under the pointwise operations, by the argument of Example 1.2.12. Such functions are integrated part by part, $\int_{0}^{1} 𝑓 =$ $\int_{0}^{1} 𝑢 +$ $𝑖 \int_{0}^{1} 𝑣$ . This integral is $ℂ$ -linear and commutes with conjugation:

\int_{0}^{1} (𝑎 𝑓 + 𝑏 𝑔) = 𝑎 \int_{0}^{1} 𝑓 + 𝑏 \int_{0}^{1} 𝑔 (𝑎, 𝑏 \in ℂ), \int_{0}^{1} 𝑓 = \int_{0}^{1} 𝑓 .

Both follow from the real case by splitting into real and imaginary parts. For instance, for $𝑎 = 𝑝 + 𝑞 𝑖$ we have $𝑎 𝑓 =$ $(𝑝 𝑢 - 𝑞 𝑣) +$ $𝑖 (𝑝 𝑣 + 𝑞 𝑢)$ , whose integral is $(𝑝 + 𝑞 𝑖) (\int 𝑢 + 𝑖 \int 𝑣)$ .

Proposition 6.1.13 (Integral inner products). The formula

⟨ 𝑓, 𝑔 ⟩ = \int_{0}^{1} 𝑓 (𝑡) 𝑔 (𝑡) d 𝑡

defines an inner product on $𝐶 ([0, 1], ℂ)$ . On the real vector space $𝐶 [0, 1]$ it reads $⟨ 𝑓, 𝑔 ⟩ =$ $\int_{0}^{1} 𝑓 (𝑡) 𝑔 (𝑡) d 𝑡$ , and it is an inner product there as well.

Intuition. Conjugate symmetry and linearity are properties of the integral, and positivity holds because

𝑓 𝑓 = {| 𝑓 |}^{2} \geq 0

. Definiteness is the one place where continuity matters. A function that is nonzero at only a single point has integral

0

, so

\int_{0}^{1} {| 𝑓 |}^{2} = 0

does not force

𝑓 = 0

for arbitrary integrable

𝑓

. A continuous function that is positive at one point, however, is positive on a whole interval around it, and so has positive integral. This is Technique 6.1.9 with an integral in place of a sum.

Lemma 6.1.14 (Nonnegative functions with zero integral). Let

ℎ \in 𝐶 [0, 1]

with

ℎ (𝑡) \geq 0

for all

𝑡 \in [0, 1]

. If

\int_{0}^{1} ℎ = 0

, then

ℎ = 0

Proof.standard

Uses. continuity of $ℎ$ , monotonicity of the integral: if $𝑔_{1} \leq 𝑔_{2}$ on an interval, then their integrals compare the same way

Techniques. A sum of nonnegative terms (Technique 6.1.9)

Hint 1. Prove the contrapositive: if

ℎ (𝑐) > 0

for some

𝑐

, then

\int_{0}^{1} ℎ > 0

Hint 2. By continuity,

ℎ > ℎ (𝑐) / 2

on all points of

[0, 1]

close enough to

𝑐

. Find an interval

𝐼 \subseteq [0, 1]

of positive length on which this holds.

Hint 3. Bound

\int_{0}^{1} ℎ

below by the integral over

𝐼

, using

ℎ \geq 0

off

𝐼

Sketch. If

ℎ (𝑐) > 0

, continuity gives an interval

𝐼 \subseteq [0, 1]

of positive length

ℓ

on which

ℎ > ℎ (𝑐) / 2

. Since

ℎ \geq 0

everywhere,

\int_{0}^{1} ℎ \geq

\int_{𝐼} ℎ \geq

ℓ ℎ (𝑐) / 2 >

0

We prove the contrapositive. Suppose $ℎ (𝑐) > 0$ for some $𝑐 \in [0, 1]$ , and let $𝜀 = ℎ (𝑐) / 2 > 0$ . By continuity of $ℎ$ at $𝑐$ , there is $𝛿 > 0$ such that $ℎ (𝑡) > 𝜀$ for all $𝑡 \in [0, 1]$ with $| 𝑡 - 𝑐 | < 𝛿$ . Let $ℓ = min (𝛿 / 2, 1 / 2)$ . Either $𝑐 + ℓ \leq 1$ or $𝑐 - ℓ \geq 0$ , since otherwise $1 < 𝑐 + ℓ$ and $𝑐 < ℓ$ , which give $1 < 2 ℓ \leq 1$ . So one of the intervals $𝐼 = [𝑐, 𝑐 + ℓ]$ or $𝐼 = [𝑐 - ℓ, 𝑐]$ lies in $[0, 1]$ . Every $𝑡 \in 𝐼$ satisfies $| 𝑡 - 𝑐 | \leq ℓ < 𝛿$ , so $ℎ > 𝜀$ on $𝐼$ . Since $ℎ \geq 0$ on all of $[0, 1]$ , monotonicity of the integral gives

\int_{0}^{1} ℎ \geq \int_{𝐼} ℎ \geq ℓ 𝜀 > 0 .

So if $\int_{0}^{1} ℎ = 0$ , then $ℎ (𝑐) > 0$ for no $𝑐$ , and as $ℎ \geq 0$ , $ℎ = 0$ . □

Proof of Proposition 6.1.13.standard

Uses. Definition 6.1.2 (Inner product), Lemma 6.1.14 (Nonnegative functions with zero integral), Notation 6.1.1 (Real and complex scalars), $ℂ$ -linearity of the integral, and $\int 𝑓 = \int 𝑓$

Techniques. A sum of nonnegative terms (Technique 6.1.9)

Hint 1. The products

𝑓 𝑔

are continuous, so all the integrals exist. Check conjugate symmetry and linearity with the two properties of the integral above.

Hint 2. For positivity,

𝑓 𝑓 = {| 𝑓 |}^{2}

is a nonnegative continuous real function.

Hint 3. For definiteness, apply Lemma 6.1.14 to

ℎ = {| 𝑓 |}^{2}

Sketch.

⟨ 𝑔, 𝑓 ⟩ =

\int 𝑔 𝑓 =

\int 𝑓 𝑔 =

⟨ 𝑓, 𝑔 ⟩

, and linearity in

𝑓

is linearity of the integral. Then

⟨ 𝑓, 𝑓 ⟩ = \int {| 𝑓 |}^{2} \geq 0

, and if it is

0

, Lemma 6.1.14 gives

{| 𝑓 |}^{2} = 0

, so

𝑓 = 0

For $𝑓, 𝑔 \in 𝐶 ([0, 1], ℂ)$ , the function $𝑓 𝑔$ is continuous, so $⟨ 𝑓, 𝑔 ⟩$ is defined. Let $𝑓,$ $𝑔,$ $ℎ \in$ $𝐶 ([0, 1], ℂ)$ and $𝑎, 𝑏 \in ℂ$ .

Conjugate symmetry. $⟨ 𝑔, 𝑓 ⟩ =$ $\int_{0}^{1} 𝑔 𝑓 =$ $\int_{0}^{1} 𝑓 𝑔 =$ $\int_{0}^{1} 𝑓 𝑔 =$ $⟨ 𝑓, 𝑔 ⟩$ .

Linearity in the first slot. $⟨ 𝑎 𝑓 + 𝑏 𝑔, ℎ ⟩ =$ $\int_{0}^{1} (𝑎 𝑓 ℎ + 𝑏 𝑔 ℎ) =$ $𝑎 ⟨ 𝑓, ℎ ⟩ +$ $𝑏 ⟨ 𝑔, ℎ ⟩$ .

Positivity. Writing $𝑓 = 𝑢 + 𝑖 𝑣$ , we have $𝑓 𝑓 =$ ${| 𝑓 |}^{2} =$ $𝑢^{2} +$ $𝑣^{2}$ , a continuous function with values in $[0, \infty)$ . So $⟨ 𝑓, 𝑓 ⟩ =$ $\int_{0}^{1} {| 𝑓 |}^{2} \geq$ $0$ .

Definiteness. If $⟨ 𝑓, 𝑓 ⟩ = 0$ , then Lemma 6.1.14 applied to $ℎ = {| 𝑓 |}^{2}$ gives ${| 𝑓 (𝑡) |}^{2} = 0$ , hence $𝑓 (𝑡) = 0$ , for every $𝑡$ . So $𝑓 = 𝟎$ .

For $𝐶 [0, 1]$ , the same argument applies with every bar erased. □

Nothing here depends on the interval being $[0, 1]$ . For any $𝑎 < 𝑏$ , the proofs of Lemma 6.1.14 and Proposition 6.1.13 go through on $[𝑎, 𝑏]$ , with $(𝑏 - 𝑎) / 2$ in place of $1 / 2$ in the choice of $ℓ$ , so $\int_{𝑎}^{𝑏} 𝑓 𝑔$ is an inner product on $𝐶 ([𝑎, 𝑏], ℂ)$ , and on $𝐶 [𝑎, 𝑏]$ with the bars erased.

Intuition (Infinitely many coordinates). Riemann sums explain why this integral behaves like a dot product. For large $𝑁$ ,

⟨ 𝑓, 𝑔 ⟩ \approx \frac{1}{𝑁} \sum_{𝑘 = 1}^{𝑁} 𝑓 (\frac{𝑘}{𝑁}) 𝑔 (\frac{𝑘}{𝑁}) = \frac{1}{𝑁} (𝐟_{𝑁} \cdot 𝐠_{𝑁}),

where $𝐟_{𝑁} =$ $(𝑓 (1 / 𝑁), 𝑓 (2 / 𝑁), \dots, 𝑓 (1)) \in$ $ℂ^{𝑁}$ records the values of $𝑓$ at $𝑁$ sample points. So $⟨ 𝑓, 𝑔 ⟩$ is a limit of rescaled dot products of longer and longer vectors, with one coordinate for each point of $[0, 1]$ , and $⟨ 𝑓, 𝑓 ⟩$ is the average of ${| 𝑓 |}^{2}$ .

Example 6.1.15 (Computing with functions). In $𝐶 [0, 1]$ , $⟨ 𝑡, 𝑡^{2} ⟩ =$ $\int_{0}^{1} 𝑡^{3} d 𝑡 =$ $1 / 4$ and $⟨ 1, 2 𝑡 - 1 ⟩ =$ $\int_{0}^{1} (2 𝑡 - 1) d 𝑡 =$ $0$ . In $𝐶 ([0, 1], ℂ)$ , let $𝑓 (𝑡) = 𝑡 + 𝑖$ and $𝑔 (𝑡) = 1 + 𝑖 𝑡$ . Then $𝑓 𝑔 =$ $(𝑡 + 𝑖) (1 - 𝑖 𝑡)$ , which expands to $2 𝑡 + 𝑖 (1 - 𝑡^{2})$ , so

⟨ 𝑓, 𝑔 ⟩ = \int_{0}^{1} 2 𝑡 d 𝑡 + 𝑖 \int_{0}^{1} (1 - 𝑡^{2}) d 𝑡 = 1 + \frac{2 𝑖}{3},

and $⟨ 𝑓, 𝑓 ⟩ =$ $\int_{0}^{1} (𝑡^{2} + 1) d 𝑡 =$ $4 / 3$ .

Matrices

The space $𝔽^{𝑚 \times 𝑛}$ is $𝔽^{𝑚 𝑛}$ with the coordinates arranged in a rectangle, so it inherits a dot product: multiply corresponding entries, conjugating those of the second matrix, and add. The trace expresses this sum compactly.

Proposition 6.1.16 (Frobenius inner product). For $𝐴 = (𝑎_{𝑖 𝑗})$ and $𝐵 = (𝑏_{𝑖 𝑗})$ in $𝔽^{𝑚 \times 𝑛}$ ,

tr (𝐵^{*} 𝐴) = \sum_{𝑖 = 1}^{𝑚} \sum_{𝑗 = 1}^{𝑛} 𝑎_{𝑖 𝑗} 𝑏_{𝑖 𝑗} .

Consequently $⟨ 𝐴, 𝐵 ⟩ = tr (𝐵^{*} 𝐴)$ is an inner product on $𝔽^{𝑚 \times 𝑛}$ , the Frobenius inner product, and $⟨ 𝐴, 𝐴 ⟩ =$ $\sum_{𝑖, 𝑗} {| 𝑎_{𝑖 𝑗} |}^{2}$ .

Proof.routine

Uses. Definition 2.4.26 (Trace), Definition 2.4.1 (Matrix product), Notation 6.1.7 (Conjugate transpose), Proposition 6.1.10 (The dot product is an inner product)

Hint 1. Compute the diagonal entries of

𝐵^{*} 𝐴

from the definition of the matrix product.

Hint 2. The

(𝑗, 𝑗)

entry of

𝐵^{*} 𝐴

\sum_{𝑖} 𝑏_{𝑖 𝑗} 𝑎_{𝑖 𝑗}

: the dot product of column

𝑗

𝐴

with column

𝑗

𝐵

Hint 3. For the axioms, notice that the double sum is the formula of Definition 6.1.8 applied to

𝑚 𝑛

coordinates.

Sketch. The

𝑗

th diagonal entry of

𝐵^{*} 𝐴

is the dot product of the

𝑗

th columns of

𝐴

and

𝐵

; summing over

𝑗

gives

\sum_{𝑖, 𝑗} 𝑎_{𝑖 𝑗} 𝑏_{𝑖 𝑗}

. That is the dot product of

𝐴

and

𝐵

regarded as vectors with

𝑚 𝑛

coordinates, so the axioms hold by Proposition 6.1.10.

The $(𝑗, 𝑖)$ entry of $𝐵^{*} \in 𝔽^{𝑛 \times 𝑚}$ is $𝑏_{𝑖 𝑗}$ . By Definition 2.4.1, the $(𝑗, 𝑗)$ entry of $𝐵^{*} 𝐴 \in 𝔽^{𝑛 \times 𝑛}$ is

{(𝐵^{*} 𝐴)}_{𝑗 𝑗} = \sum_{𝑖 = 1}^{𝑚} {(𝐵^{*})}_{𝑗 𝑖} 𝑎_{𝑖 𝑗} = \sum_{𝑖 = 1}^{𝑚} 𝑎_{𝑖 𝑗} 𝑏_{𝑖 𝑗},

and summing over $𝑗$ (Definition 2.4.26) gives the formula.

The right side of the formula is the dot product formula of Definition 6.1.8, with the $𝑚 𝑛$ pairs $(𝑖, 𝑗)$ in place of the indices $𝑘$ . The proof of Proposition 6.1.10 checks each axiom from that formula alone, term by term, so it applies verbatim: $⟨ 𝐴, 𝐵 ⟩ = tr (𝐵^{*} 𝐴)$ is an inner product on $𝔽^{𝑚 \times 𝑛}$ , and $⟨ 𝐴, 𝐴 ⟩ =$ $\sum_{𝑖, 𝑗} {| 𝑎_{𝑖 𝑗} |}^{2}$ . □

Example 6.1.17 (A Frobenius inner product). For $𝐴 = (\begin{matrix} 1 & 2 \\ 0 & - 1 \end{matrix})$ and $𝐵 = (\begin{matrix} 3 & 0 \\ 1 & 1 \end{matrix})$ in $ℝ^{2 \times 2}$ , the entrywise formula gives $⟨ 𝐴, 𝐵 ⟩ =$ $1 \cdot$ $3 +$ $2 \cdot$ $0 +$ $0 \cdot$ $1 +$ $(- 1) \cdot$ $1 =$ $2$ . With the trace,

𝐵^{𝖳} 𝐴 = (\begin{matrix} 3 & 1 \\ 0 & 1 \end{matrix}) (\begin{matrix} 1 & 2 \\ 0 & - 1 \end{matrix}) = (\begin{matrix} 3 & 5 \\ 0 & - 1 \end{matrix}), tr (𝐵^{𝖳} 𝐴) = 3 - 1 = 2 .

The identity $⟨ 𝐴, 𝐴 ⟩ =$ $tr (𝐴^{*} 𝐴) =$ $\sum_{𝑖, 𝑗} {| 𝑎_{𝑖 𝑗} |}^{2}$ is useful well beyond computing lengths.

Technique 6.1.18 (The trace of $𝐴^{*} 𝐴$ ). For every $𝐴 \in 𝔽^{𝑚 \times 𝑛}$ , $tr (𝐴^{*} 𝐴) =$ $\sum_{𝑖, 𝑗} {| 𝑎_{𝑖 𝑗} |}^{2}$ (Proposition 6.1.16). So $tr (𝐴^{*} 𝐴) \geq 0$ , with equality only for $𝐴 = 0$ , and one scalar equation controls every entry of $𝐴$ . To show that a matrix $𝐴$ is zero, it suffices to show $tr (𝐴^{*} 𝐴) = 0$ ; in particular, it suffices to show $𝐴^{*} 𝐴 = 0$ .

Reach for it when a matrix identity involves $𝐴^{*} 𝐴$ or $𝐴 𝐴^{*}$ (or $𝐴^{𝖳} 𝐴$ over $ℝ$ ) and the goal is a conclusion about $𝐴$ itself. It is Technique 6.1.9 applied to the entries of a matrix. The canonical example is Corollary 6.1.19.

Used in 9 places and 2 practice exercises

Corollary 6.1.19, Exercise 6.1.20, Exercise 6.1.54, Exercise 6.1.55, Exercise 6.6.50, Exercise 6.7.10, Exercise 7.5.41, Exercise 8.1.37, Exercise 8.1.38; 2 practice exercises.

Corollary 6.1.19 (

𝐴^{*} 𝐴 = 0

forces

𝐴 = 0

). Let

𝐴 \in 𝔽^{𝑚 \times 𝑛}

. If

𝐴^{*} 𝐴 = 0

𝐴 𝐴^{*} = 0

, then

𝐴 = 0

Proof.routine

Uses. Proposition 6.1.16 (Frobenius inner product), Notation 6.1.7 (Conjugate transpose)

Techniques. The trace of $𝐴^{*} 𝐴$ (Technique 6.1.18)

Hint 1. Take the trace of

𝐴^{*} 𝐴 = 0

Hint 2. For

𝐴 𝐴^{*} = 0

, apply the first case to

𝐴^{*}

, using

{(𝐴^{*})}^{*} = 𝐴

Sketch. If

𝐴^{*} 𝐴 = 0

, then

\sum_{𝑖, 𝑗} {| 𝑎_{𝑖 𝑗} |}^{2} =

tr (𝐴^{*} 𝐴) =

0

, so every

𝑎_{𝑖 𝑗} = 0

. Since

{(𝐴^{*})}^{*} = 𝐴

, the case

𝐴 𝐴^{*} = 0

is the first case for

𝐴^{*}

If $𝐴^{*} 𝐴 = 0$ , then by Proposition 6.1.16, $\sum_{𝑖, 𝑗} {| 𝑎_{𝑖 𝑗} |}^{2} =$ $tr (𝐴^{*} 𝐴) =$ $tr (0) =$ $0$ . This is a sum of nonnegative real numbers, so every ${| 𝑎_{𝑖 𝑗} |}^{2} = 0$ , and $𝐴 = 0$ .

The $(𝑖, 𝑗)$ entry of ${(𝐴^{*})}^{*}$ is the conjugate of the $(𝑗, 𝑖)$ entry of $𝐴^{*}$ , which is $𝑎_{𝑖 𝑗} = 𝑎_{𝑖 𝑗}$ , so ${(𝐴^{*})}^{*} = 𝐴$ . If $𝐴 𝐴^{*} = 0$ , then ${(𝐴^{*})}^{*} 𝐴^{*} = 0$ , and the first case applied to $𝐴^{*}$ gives $𝐴^{*} = 0$ . Hence $𝐴 = {(𝐴^{*})}^{*} = 0$ . □

The conjugate transpose is essential here. The nilpotent matrix $𝑁 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ is nonzero with $𝑁^{2} = 0$ , so $𝐴^{2} = 0$ does not force $𝐴 = 0$ . Over $ℂ$ , the transpose alone does not suffice either: the column $𝐚 = (1, 𝑖)$ is nonzero, but $𝐚^{𝖳} 𝐚 = 1 + 𝑖^{2} = 0$ . This is the failure of $(1, 𝑖) \cdot (1, 𝑖)$ without conjugation that we met at the start of the section.

Exercise 6.1.20.routine Let

𝐴 \in ℂ^{𝑛 \times 𝑛}

satisfy

𝐴^{*} = 𝐴

and

𝐴^{2} = 0

. Show that

𝐴 = 0

. Does the conclusion still hold if the hypothesis

𝐴^{*} = 𝐴

is dropped?

Solution.

Uses. Corollary 6.1.19 ( $𝐴^{*} 𝐴 = 0$ forces $𝐴 = 0$ )

Techniques. The trace of $𝐴^{*} 𝐴$ (Technique 6.1.18)

Hint 1. Rewrite

𝐴^{2}

using the hypothesis

𝐴^{*} = 𝐴

Hint 2.

𝐴^{*} 𝐴 = 𝐴 𝐴 = 0

. Now use Corollary 6.1.19.

Sketch.

𝐴^{*} 𝐴 = 𝐴^{2} = 0

, so

𝐴 = 0

by Corollary 6.1.19. Without

𝐴^{*} = 𝐴

the claim fails for

𝑁 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})

Since

𝐴^{*} = 𝐴

, we have

𝐴^{*} 𝐴 = 𝐴 𝐴 = 𝐴^{2} = 0

, so

𝐴 = 0

by Corollary 6.1.19. The hypothesis

𝐴^{*} = 𝐴

cannot be dropped:

𝑁 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})

satisfies

𝑁^{2} = 0

but

𝑁 \neq 0

. (Indeed

𝑁^{*} = (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}) \neq 𝑁

Inner products on $𝔽^{𝑛}$ from matrices

The dot product is not the only inner product on $𝔽^{𝑛}$ . Weighting the coordinates already gives others.

Example 6.1.21 (Weighted dot products). Let $𝑐_{1}, \dots, 𝑐_{𝑛}$ be positive real numbers. Then

{⟨ 𝐯, 𝐰 ⟩}_{𝑐} = \sum_{𝑘 = 1}^{𝑛} 𝑐_{𝑘} 𝑣_{𝑘} 𝑤_{𝑘}

is an inner product on $𝔽^{𝑛}$ . The proof of Proposition 6.1.10 goes through with $𝑐_{𝑘} {| 𝑣_{𝑘} |}^{2}$ in place of ${| 𝑣_{𝑘} |}^{2}$ , since $𝑐_{𝑘} {| 𝑣_{𝑘} |}^{2} \geq 0$ , with equality only if $𝑣_{𝑘} = 0$ . The weights must be positive: if $𝑐_{𝑘} \leq 0$ , then ${⟨ 𝐞_{𝑘}, 𝐞_{𝑘} ⟩}_{𝑐} =$ $𝑐_{𝑘} \leq$ $0$ while $𝐞_{𝑘} \neq 𝟎$ .

Different weights give different geometry on the same set. With weights $(1, 4)$ on $ℝ^{2}$ , the vector $𝐞_{2}$ has ${⟨ 𝐞_{2}, 𝐞_{2} ⟩}_{𝑐} = 4$ , so it is “twice as long” as $𝐞_{1}$ , and the vectors with ${⟨ 𝐱, 𝐱 ⟩}_{𝑐} = 1$ form the ellipse $𝑥_{1}^{2} + 4 𝑥_{2}^{2} = 1$ rather than the unit circle (Figure 6.1.2).

In fact every inner product on $𝔽^{𝑛}$ is given by a matrix. By sesquilinearity, $⟨ 𝐱, 𝐲 ⟩$ is determined by the $𝑛^{2}$ numbers $⟨ 𝐞_{𝑗}, 𝐞_{𝑖} ⟩$ , and the question is which arrays of $𝑛^{2}$ numbers occur.

Proposition 6.1.22 (Inner products on

𝔽^{𝑛}

(a)Let $⟨ \cdot, \cdot ⟩$ be an inner product on $𝔽^{𝑛}$ , and let $𝑀 \in 𝔽^{𝑛 \times 𝑛}$ have entries $𝑚_{𝑖 𝑗} =$ $⟨ 𝐞_{𝑗}, 𝐞_{𝑖} ⟩$ . Then $⟨ 𝐱, 𝐲 ⟩ = 𝐲^{*} 𝑀 𝐱$ for all $𝐱, 𝐲 \in 𝔽^{𝑛}$ .
(b)Conversely, let $𝑀 \in 𝔽^{𝑛 \times 𝑛}$ . The formula $⟨ 𝐱, 𝐲 ⟩ = 𝐲^{*} 𝑀 𝐱$ defines an inner product on $𝔽^{𝑛}$ if and only if $𝑀^{*} = 𝑀$ and $𝐱^{*} 𝑀 𝐱 > 0$ for every nonzero $𝐱 \in 𝔽^{𝑛}$ .

Proof.standard

Uses. Proposition 6.1.4 (Sesquilinearity), Definition 1.6.8 (Standard bases), Definition 2.3.8 (Matrix–vector product), Notation 6.1.7 (Conjugate transpose), Remark 6.1.3

Techniques. Expand by multilinearity (Technique 5.1.7), Check it on a basis (Technique 2.2.7)

Hint 1. For (a), write

𝐱 = \sum_{𝑗} 𝑥_{𝑗} 𝐞_{𝑗}

and

𝐲 = \sum_{𝑖} 𝑦_{𝑖} 𝐞_{𝑖}

, and expand using linearity in the first slot and Proposition 6.1.4 in the second.

Hint 2. For (b), the formula is linear in

𝐱

for every

𝑀

, and positivity together with definiteness says exactly that

𝐱^{*} 𝑀 𝐱 > 0

for

𝐱 \neq 𝟎

. So only conjugate symmetry needs thought.

Hint 3. Conjugate symmetry at the pair

(𝐞_{𝑗}, 𝐞_{𝑖})

reads

𝑚_{𝑖 𝑗} = 𝑚_{𝑗 𝑖}

. For the converse, write

𝐲^{*} 𝑀 𝐱 =

\sum_{𝑖, 𝑗} 𝑦_{𝑖} 𝑚_{𝑖 𝑗} 𝑥_{𝑗}

, conjugate, and rename the indices.

Sketch. Expanding by sesquilinearity,

⟨ 𝐱, 𝐲 ⟩ =

\sum_{𝑖, 𝑗} 𝑦_{𝑖} ⟨ 𝐞_{𝑗}, 𝐞_{𝑖} ⟩ 𝑥_{𝑗} =

𝐲^{*} 𝑀 𝐱

. For (b), linearity in

𝐱

is automatic, and positivity plus definiteness is the condition

𝐱^{*} 𝑀 𝐱 > 0

for

𝐱 \neq 𝟎

. Conjugate symmetry tested at

(𝐞_{𝑗}, 𝐞_{𝑖})

gives

𝑀 = 𝑀^{*}

; conversely, if

𝑀 = 𝑀^{*}

, conjugating the double sum for

⟨ 𝐲, 𝐱 ⟩

and swapping

𝑖

and

𝑗

gives

⟨ 𝐱, 𝐲 ⟩

(a) Write $𝐱 = \sum_{𝑗} 𝑥_{𝑗} 𝐞_{𝑗}$ and $𝐲 = \sum_{𝑖} 𝑦_{𝑖} 𝐞_{𝑖}$ . Linearity in the first slot and Proposition 6.1.4 give

⟨ 𝐱, 𝐲 ⟩ = \sum_{𝑗 = 1}^{𝑛} 𝑥_{𝑗} ⟨ 𝐞_{𝑗}, 𝐲 ⟩ = \sum_{𝑗 = 1}^{𝑛} \sum_{𝑖 = 1}^{𝑛} 𝑥_{𝑗} 𝑦_{𝑖} ⟨ 𝐞_{𝑗}, 𝐞_{𝑖} ⟩ = \sum_{𝑖 = 1}^{𝑛} 𝑦_{𝑖} (\sum_{𝑗 = 1}^{𝑛} 𝑚_{𝑖 𝑗} 𝑥_{𝑗}) .

The inner sum is the $𝑖$ th coordinate of $𝑀 𝐱$ (Definition 2.3.8), so the whole expression is $𝐲^{*} (𝑀 𝐱)$ .

(b) Write ${⟨ 𝐱, 𝐲 ⟩}_{𝑀} =$ $𝐲^{*} 𝑀 𝐱 =$ $\sum_{𝑖, 𝑗} 𝑦_{𝑖} 𝑚_{𝑖 𝑗} 𝑥_{𝑗}$ . For fixed $𝐲$ this is linear in $𝐱$ , for every $𝑀$ , so linearity in the first slot always holds. Positivity and definiteness together say that ${⟨ 𝐱, 𝐱 ⟩}_{𝑀} \geq 0$ for all $𝐱$ , with equality only for $𝐱 = 𝟎$ (Remark 6.1.3); since ${⟨ 𝟎, 𝟎 ⟩}_{𝑀} = 0$ in any case, this is the condition $𝐱^{*} 𝑀 𝐱 > 0$ for all $𝐱 \neq 𝟎$ . It remains to show that conjugate symmetry holds if and only if $𝑀^{*} = 𝑀$ .

Suppose conjugate symmetry holds. For all $𝑖, 𝑗$ ,

𝑚_{𝑖 𝑗} = 𝐞_{𝑖}^{*} 𝑀 𝐞_{𝑗} = {⟨ 𝐞_{𝑗}, 𝐞_{𝑖} ⟩}_{𝑀} = {⟨ 𝐞_{𝑖}, 𝐞_{𝑗} ⟩}_{𝑀} = 𝐞_{𝑗}^{*} 𝑀 𝐞_{𝑖} = 𝑚_{𝑗 𝑖},

which says $𝑀^{*} = 𝑀$ . Conversely, suppose $𝑚_{𝑖 𝑗} = 𝑚_{𝑗 𝑖}$ for all $𝑖, 𝑗$ . Then for all $𝐱, 𝐲$ ,

{⟨ 𝐲, 𝐱 ⟩}_{𝑀} = \sum_{𝑖, 𝑗} 𝑥_{𝑖} 𝑚_{𝑖 𝑗} 𝑦_{𝑗} = \sum_{𝑖, 𝑗} 𝑥_{𝑖} 𝑚_{𝑖 𝑗} 𝑦_{𝑗} = \sum_{𝑖, 𝑗} 𝑦_{𝑗} 𝑚_{𝑗 𝑖} 𝑥_{𝑖} = {⟨ 𝐱, 𝐲 ⟩}_{𝑀},

where the last step renames the summation indices $𝑖$ and $𝑗$ as $𝑗$ and $𝑖$ . So conjugate symmetry holds. □

A matrix $𝑀$ with $𝑀^{*} = 𝑀$ and $𝐱^{*} 𝑀 𝐱 > 0$ for all $𝐱 \neq 𝟎$ is called positive definite, and Section 12.3 gives practical tests for it. Parts (a) and (b) say that every inner product on $𝔽^{𝑛}$ is $𝐲^{*} 𝑀 𝐱$ for a positive definite $𝑀$ . The dot product is the case $𝑀 = 𝐼_{𝑛}$ , and the weighted dot products are the diagonal $𝑀$ with positive diagonal entries.

Example 6.1.23 (A tilted inner product on $ℝ^{2}$ ). Let $𝑀 = (\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix})$ . Then $⟨ 𝐱, 𝐲 ⟩ =$ $𝐲^{𝖳} 𝑀 𝐱 =$ $2 𝑥_{1} 𝑦_{1} +$ $𝑥_{1} 𝑦_{2} +$ $𝑥_{2} 𝑦_{1} +$ $2 𝑥_{2} 𝑦_{2}$ . Here $𝑀^{𝖳} = 𝑀$ , and for $𝐱 \neq 𝟎$ ,

𝐱^{𝖳} 𝑀 𝐱 = 2 𝑥_{1}^{2} + 2 𝑥_{1} 𝑥_{2} + 2 𝑥_{2}^{2} = 𝑥_{1}^{2} + 𝑥_{2}^{2} + {(𝑥_{1} + 𝑥_{2})}^{2} > 0,

a sum of squares that is not all zero. So Proposition 6.1.22 makes this an inner product. Measured with it, $(1, - 1)$ has $⟨ 𝐱, 𝐱 ⟩ = 2 - 2 + 2 = 2$ while $(1, 1)$ has $⟨ 𝐱, 𝐱 ⟩ = 6$ , and the vectors with $⟨ 𝐱, 𝐱 ⟩ = 1$ form a tilted ellipse (Figure 6.1.2).

Figure 6.1.2. The vectors

𝐱 \in ℝ^{2}

with

⟨ 𝐱, 𝐱 ⟩ = 1

(solid) for three inner products on

ℝ^{2}

: (a) the dot product; (b) the weighted dot product

𝑥_{1} 𝑦_{1} + 4 𝑥_{2} 𝑦_{2}

of Example 6.1.21, in which

𝐞_{2}

lies outside the curve because it has length

2

; (c) the inner product of Example 6.1.23. The dashed circle is the unit circle of the dot product.

Remark 6.1.24. Over

ℝ

, positivity of

𝐱^{𝖳} 𝑀 𝐱

does not force

𝑀

to be symmetric. For

𝑀 = (\begin{matrix} 1 & 1 \\ - 1 & 1 \end{matrix})

𝐱^{𝖳} 𝑀 𝐱 =

𝑥_{1}^{2} +

𝑥_{1} 𝑥_{2} -

𝑥_{2} 𝑥_{1} +

𝑥_{2}^{2} =

𝑥_{1}^{2} +

𝑥_{2}^{2} >

0

for

𝐱 \neq 𝟎

, but

𝐲^{𝖳} 𝑀 𝐱

is not an inner product, since

⟨ 𝐞_{2}, 𝐞_{1} ⟩ = 𝑚_{12} = 1

while

⟨ 𝐞_{1}, 𝐞_{2} ⟩ = 𝑚_{21} = - 1

. Over

ℂ

this cannot happen: if

𝐱^{*} 𝑀 𝐱

is real for every

𝐱 \in ℂ^{𝑛}

, then

𝑀^{*} = 𝑀

(Exercise 6.1.47).

Non-examples

Non-example 6.1.25 (Formulas that are not inner products).

(a)On $ℂ^{2}$ , $⟨ 𝐱, 𝐲 ⟩ =$ $𝑥_{1} 𝑦_{1} +$ $𝑥_{2} 𝑦_{2}$ is bilinear and symmetric, but $⟨ (𝑖, 0), (𝑖, 0) ⟩ =$ $𝑖^{2} =$ $- 1 <$ $0$ . This is Proposition 6.1.6 in action.
(b)On $ℂ^{𝑛}$ , $⟨ 𝐱, 𝐲 ⟩ =$ $\sum_{𝑘} 𝑥_{𝑘} 𝑦_{𝑘} =$ $𝐱^{*} 𝐲$ is conjugate symmetric, positive, and definite, and it is linear in the second slot. But $⟨ 𝑖 𝐱, 𝐲 ⟩ =$ $- 𝑖 ⟨ 𝐱, 𝐲 ⟩$ , so it is not linear in the first slot, and by our definition it is not an inner product. Many physics texts use this convention instead. It is the dot product with its arguments exchanged, $⟨ 𝐱, 𝐲 ⟩ = 𝐲 \cdot 𝐱$ , so results translate by swapping the slots.
(c)On $ℝ^{2 \times 2}$ , $⟨ 𝐴, 𝐵 ⟩ = tr (𝐴 𝐵)$ is bilinear and symmetric (as $tr (𝐴 𝐵) = tr (𝐵 𝐴)$ , Proposition 2.4.28). But $𝐽 = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ has $𝐽^{2} = - 𝐼_{2}$ , so $⟨ 𝐽, 𝐽 ⟩ = - 2 < 0$ . In the Frobenius inner product $tr (𝐵^{𝖳} 𝐴)$ , the transpose is what turns $⟨ 𝐴, 𝐴 ⟩$ into a sum of squares.
(d)On ${ℝ [𝑥]}_{\leq 2}$ , $⟨ 𝑝, 𝑞 ⟩ =$ $𝑝 (0) 𝑞 (0) +$ $𝑝 (1) 𝑞 (1)$ is symmetric, bilinear, and positive, but the nonzero polynomial $𝑝 = 𝑥^{2} - 𝑥$ has $⟨ 𝑝, 𝑝 ⟩ = 0$ . On ${ℝ [𝑥]}_{\leq 1}$ , the same formula is an inner product: if $𝑝 = 𝑎 + 𝑏 𝑥$ and ${𝑝 (0)}^{2} + {𝑝 (1)}^{2} = 0$ , then $𝑝 (0) = 𝑎 = 0$ and $𝑝 (1) = 𝑎 + 𝑏 = 0$ , so $𝑝 = 𝟎$ . Whether a formula defines an inner product depends on the space it is defined on.

6.1.4The norm

Definition 6.1.26 (Norm). Let $𝑉$ be an inner product space. The norm of $𝐯 \in 𝑉$ is

‖ 𝐯 ‖ = \sqrt{⟨ 𝐯, 𝐯 ⟩},

the nonnegative square root of the nonnegative real number $⟨ 𝐯, 𝐯 ⟩$ . A unit vector is a vector $𝐮$ with $‖ 𝐮 ‖ = 1$ .

The norm is the length of $𝐯$ as measured by the chosen inner product, and $‖ 𝐯 - 𝐰 ‖$ is the distance between $𝐯$ and $𝐰$ . For the dot product on $𝔽^{𝑛}$ , ${‖ 𝐯 ‖}^{2} = \sum_{𝑘} {| 𝑣_{𝑘} |}^{2}$ , so $‖ 𝐯 ‖$ is the Euclidean length. For the Frobenius inner product, ${‖ 𝐴 ‖}^{2} =$ $\sum_{𝑖, 𝑗} {| 𝑎_{𝑖 𝑗} |}^{2}$ , and $‖ 𝐴 ‖$ is often written ${‖ 𝐴 ‖}_{𝐹}$ . For the integral inner product, ${‖ 𝑓 ‖}^{2} = \int_{0}^{1} {| 𝑓 |}^{2}$ . A different inner product on the same space gives different lengths: with the weights $(1, 4)$ of Example 6.1.21, $‖ 𝐞_{2} ‖ = 2$ .

Proposition 6.1.27 (Basic properties of the norm). Let $𝑉$ be an inner product space, $𝐯 \in 𝑉$ , and $𝑎 \in 𝔽$ .

(a) $‖ 𝐯 ‖ \geq 0$ , and $‖ 𝐯 ‖ = 0$ if and only if $𝐯 = 𝟎$ .
(b) $‖ 𝑎 𝐯 ‖ = | 𝑎 | ‖ 𝐯 ‖$ .

Proof.routine

Uses. Definition 6.1.26 (Norm), Definition 6.1.2 (Inner product), Proposition 6.1.4 (Sesquilinearity), Notation 6.1.1 (Real and complex scalars)

Hint 1. Part (a) is a restatement of two axioms.

Hint 2. For (b), compare squares: compute

{‖ 𝑎 𝐯 ‖}^{2} =

⟨ 𝑎 𝐯, 𝑎 𝐯 ⟩

by taking

𝑎

out of each slot.

Hint 3. The second slot contributes

𝑎

, and

𝑎 𝑎 = {| 𝑎 |}^{2}

. Two nonnegative reals with the same square are equal.

Sketch. (a) is positivity and definiteness. For (b),

{‖ 𝑎 𝐯 ‖}^{2} =

𝑎 𝑎 ⟨ 𝐯, 𝐯 ⟩ =

{| 𝑎 |}^{2} {‖ 𝐯 ‖}^{2}

; take nonnegative square roots.

(a) The square root of a nonnegative real number is nonnegative. Also $‖ 𝐯 ‖ = 0$ if and only if $⟨ 𝐯, 𝐯 ⟩ = 0$ , which by definiteness happens if and only if $𝐯 = 𝟎$ .

(b) By linearity in the first slot and Proposition 6.1.4,

{‖ 𝑎 𝐯 ‖}^{2} = ⟨ 𝑎 𝐯, 𝑎 𝐯 ⟩ = 𝑎 𝑎 ⟨ 𝐯, 𝐯 ⟩ = {| 𝑎 |}^{2} {‖ 𝐯 ‖}^{2} .

Both $‖ 𝑎 𝐯 ‖$ and $| 𝑎 | ‖ 𝐯 ‖$ are nonnegative real numbers, and their squares are equal, so they are equal. □

In particular, if $𝐯 \neq 𝟎$ , then $𝐯 / ‖ 𝐯 ‖$ (that is, $(1 / ‖ 𝐯 ‖) 𝐯$ ) is a unit vector, since by (b) its norm is $‖ 𝐯 ‖ / ‖ 𝐯 ‖ = 1$ . Replacing $𝐯$ by $𝐯 / ‖ 𝐯 ‖$ is called normalizing $𝐯$ .

Exercise 6.1.28.routine Let

𝐯 \neq 𝟎

be a vector in an inner product space. Find all scalars

𝑎 \in 𝔽

with

‖ 𝑎 𝐯 ‖ = 1

, separately for

𝔽 = ℝ

and for

𝔽 = ℂ

Solution.

Uses. Proposition 6.1.27 (Basic properties of the norm)

Hint 1. By Proposition 6.1.27(b), the condition only involves

| 𝑎 |

Hint 2.

‖ 𝑎 𝐯 ‖ = 1

if and only if

| 𝑎 | = 1 / ‖ 𝐯 ‖

. How many real numbers, and which complex numbers, have a given modulus?

Sketch.

‖ 𝑎 𝐯 ‖ = | 𝑎 | ‖ 𝐯 ‖

, so the condition is

| 𝑎 | = 1 / ‖ 𝐯 ‖

. Over

ℝ

this gives

𝑎 = \pm 1 / ‖ 𝐯 ‖

; over

ℂ

it gives the circle of radius

1 / ‖ 𝐯 ‖

centred at

0

By Proposition 6.1.27,

‖ 𝑎 𝐯 ‖ = | 𝑎 | ‖ 𝐯 ‖

, and

‖ 𝐯 ‖ > 0

since

𝐯 \neq 𝟎

. So

‖ 𝑎 𝐯 ‖ = 1

if and only if

| 𝑎 | = 1 / ‖ 𝐯 ‖

. Over

ℝ

there are exactly two such scalars,

𝑎 = 1 / ‖ 𝐯 ‖

and

𝑎 = - 1 / ‖ 𝐯 ‖

. Over

ℂ

there are infinitely many: all

𝑎 = 𝑠 + 𝑡 𝑖

with

𝑠^{2} + 𝑡^{2} = 1 / {‖ 𝐯 ‖}^{2}

, a circle of radius

1 / ‖ 𝐯 ‖

about

0

in the complex plane. So a nonzero vector in a complex space has a whole circle of normalizations, which differ by factors of modulus

1

Expanding norms

The computation that produced (6.1.1) expanded ${‖ 𝐯 - 𝐰 ‖}^{2}$ . In a general inner product space the same expansion works, with one change over $ℂ$ .

Proposition 6.1.29 (Expanding a norm). Let $𝑉$ be an inner product space, $𝐱, 𝐲 \in 𝑉$ , and $𝑎 \in 𝔽$ . Then

{‖ 𝐱 + 𝑎 𝐲 ‖}^{2} = {‖ 𝐱 ‖}^{2} + 2 Re (𝑎 ⟨ 𝐱, 𝐲 ⟩) + {| 𝑎 |}^{2} {‖ 𝐲 ‖}^{2} .

In particular,

{‖ 𝐱 \pm 𝐲 ‖}^{2} = {‖ 𝐱 ‖}^{2} \pm 2 Re ⟨ 𝐱, 𝐲 ⟩ + {‖ 𝐲 ‖}^{2} .

Proof.routine

Uses. Definition 6.1.26 (Norm), Definition 6.1.2 (Inner product), Proposition 6.1.4 (Sesquilinearity), Notation 6.1.1 (Real and complex scalars)

Techniques. Expand by multilinearity (Technique 5.1.7)

Hint 1. Write

{‖ 𝐱 + 𝑎 𝐲 ‖}^{2} =

⟨ 𝐱 + 𝑎 𝐲, 𝐱 + 𝑎 𝐲 ⟩

and expand both slots, the second with Proposition 6.1.4.

Hint 2. Of the four terms, the two cross terms are complex conjugates of each other. Use

𝑧 + 𝑧 = 2 Re 𝑧

Sketch. Expanding gives

{‖ 𝐱 ‖}^{2} +

𝑎 ⟨ 𝐱, 𝐲 ⟩ +

𝑎 ⟨ 𝐲, 𝐱 ⟩ +

{| 𝑎 |}^{2} {‖ 𝐲 ‖}^{2}

, and by conjugate symmetry the cross terms are

𝑧

and

𝑧

for

𝑧 = 𝑎 ⟨ 𝐱, 𝐲 ⟩

. They add to

2 Re 𝑧

. Take

𝑎 = \pm 1

for the second formula.

By linearity in the first slot and Proposition 6.1.4,

\begin{aligned} {‖ 𝐱 + 𝑎 𝐲 ‖}^{2} & = ⟨ 𝐱 + 𝑎 𝐲, 𝐱 + 𝑎 𝐲 ⟩ \\ = ⟨ 𝐱, 𝐱 ⟩ + 𝑎 ⟨ 𝐱, 𝐲 ⟩ + 𝑎 ⟨ 𝐲, 𝐱 ⟩ + 𝑎 𝑎 ⟨ 𝐲, 𝐲 ⟩ . \end{aligned}

Let $𝑧 = 𝑎 ⟨ 𝐱, 𝐲 ⟩$ . By conjugate symmetry, $𝑎 ⟨ 𝐲, 𝐱 ⟩ =$ $𝑎 ⟨ 𝐱, 𝐲 ⟩ =$ $𝑎 ⟨ 𝐱, 𝐲 ⟩ =$ $𝑧$ , so the two middle terms add to $𝑧 + 𝑧 = 2 Re 𝑧$ . Since $𝑎 𝑎 = {| 𝑎 |}^{2}$ , this is the first formula. For $𝑎 = 1$ it reads ${‖ 𝐱 + 𝐲 ‖}^{2} =$ ${‖ 𝐱 ‖}^{2} +$ $2 Re ⟨ 𝐱, 𝐲 ⟩ +$ ${‖ 𝐲 ‖}^{2}$ , and for $𝑎 = - 1$ the middle term is $2 Re (- ⟨ 𝐱, 𝐲 ⟩) =$ $- 2 Re ⟨ 𝐱, 𝐲 ⟩$ . □

Intuition (The law of cosines, everywhere). Over $ℝ$ the cross term is $2 ⟨ 𝐱, 𝐲 ⟩$ , and in $ℝ^{2}$ with the dot product, (6.1.1) turns ${‖ 𝐱 - 𝐲 ‖}^{2} =$ ${‖ 𝐱 ‖}^{2} -$ $2 𝐱 \cdot$ $𝐲 +$ ${‖ 𝐲 ‖}^{2}$ back into the law of cosines. So Proposition 6.1.29 is the law of cosines of an arbitrary inner product space; the computation at the start of the section, run backwards.

Over $ℂ$ , only the real part of $⟨ 𝐱, 𝐲 ⟩$ affects $‖ 𝐱 \pm 𝐲 ‖$ . The imaginary part appears once $𝐲$ is multiplied by $𝑖$ : with $𝑎 = 𝑖$ , $Re (𝑖 ⟨ 𝐱, 𝐲 ⟩) =$ $Re (- 𝑖 ⟨ 𝐱, 𝐲 ⟩) =$ $Im ⟨ 𝐱, 𝐲 ⟩$ , so

{‖ 𝐱 + 𝑖 𝐲 ‖}^{2} = {‖ 𝐱 ‖}^{2} + 2 Im ⟨ 𝐱, 𝐲 ⟩ + {‖ 𝐲 ‖}^{2} .

When $⟨ 𝐱, 𝐲 ⟩ = 0$ , the expansion gives ${‖ 𝐱 + 𝐲 ‖}^{2} =$ ${‖ 𝐱 ‖}^{2} +$ ${‖ 𝐲 ‖}^{2}$ , the Pythagorean theorem of Section 6.3.

The parallelogram law

Vectors $𝐱$ and $𝐲$ span a parallelogram whose diagonals are $𝐱 + 𝐲$ and $𝐱 - 𝐲$ (Figure 6.1.3). In a rectangle the two diagonals are equally long, and Pythagoras says that each has squared length ${‖ 𝐱 ‖}^{2} + {‖ 𝐲 ‖}^{2}$ . Skewing the rectangle lengthens one diagonal and shortens the other. Adding the two expansions in Proposition 6.1.29 shows that the changes in their squares cancel exactly.

Proposition 6.1.30 (Parallelogram law). For all $𝐱, 𝐲$ in an inner product space,

{‖ 𝐱 + 𝐲 ‖}^{2} + {‖ 𝐱 - 𝐲 ‖}^{2} = 2 {‖ 𝐱 ‖}^{2} + 2 {‖ 𝐲 ‖}^{2} .

Proof.routine

Uses. Proposition 6.1.29 (Expanding a norm)

Hint 1. Expand each term on the left with Proposition 6.1.29.

Hint 2. The cross terms

\pm 2 Re ⟨ 𝐱, 𝐲 ⟩

have opposite signs.

Sketch. Add the expansions of

{‖ 𝐱 + 𝐲 ‖}^{2}

and

{‖ 𝐱 - 𝐲 ‖}^{2}

; the cross terms cancel.

By Proposition 6.1.29,

\begin{aligned} {‖ 𝐱 + 𝐲 ‖}^{2} + {‖ 𝐱 - 𝐲 ‖}^{2} & = ({‖ 𝐱 ‖}^{2} + 2 Re ⟨ 𝐱, 𝐲 ⟩ + {‖ 𝐲 ‖}^{2}) + ({‖ 𝐱 ‖}^{2} - 2 Re ⟨ 𝐱, 𝐲 ⟩ + {‖ 𝐲 ‖}^{2}) \\ = 2 {‖ 𝐱 ‖}^{2} + 2 {‖ 𝐲 ‖}^{2}, \end{aligned}

as desired. □

Figure 6.1.3. The parallelogram law: the squares of the diagonals

𝐱 + 𝐲

and

𝐱 - 𝐲

add up to the squares of the four sides,

2 {‖ 𝐱 ‖}^{2} + 2 {‖ 𝐲 ‖}^{2}

Unlike Proposition 6.1.29, the parallelogram law mentions only norms, not the inner product. That makes it a test: a notion of length that violates it cannot be the norm of any inner product (Aside (Norms that do not come from inner products)).

Exercise 6.1.31.routine In an inner product space,

‖ 𝐱 ‖ = 3

‖ 𝐲 ‖ = 4

, and

‖ 𝐱 + 𝐲 ‖ = 6

. Find

‖ 𝐱 - 𝐲 ‖

and

Re ⟨ 𝐱, 𝐲 ⟩

Solution.

Uses. Proposition 6.1.30 (Parallelogram law), Proposition 6.1.29 (Expanding a norm)

Hint 1. The parallelogram law relates the four given or unknown norms.

Hint 2. For the real part, solve the expansion of

{‖ 𝐱 + 𝐲 ‖}^{2}

for

Re ⟨ 𝐱, 𝐲 ⟩

Sketch.

{‖ 𝐱 - 𝐲 ‖}^{2} =

2 \cdot

9 +

2 \cdot

16 -

36 =

14

, and

36 =

9 +

2 Re ⟨ 𝐱, 𝐲 ⟩ +

16

gives

Re ⟨ 𝐱, 𝐲 ⟩ = 11 / 2

By Proposition 6.1.30,

{‖ 𝐱 - 𝐲 ‖}^{2} = 2 {‖ 𝐱 ‖}^{2} + 2 {‖ 𝐲 ‖}^{2} - {‖ 𝐱 + 𝐲 ‖}^{2} = 18 + 32 - 36 = 14,

so $‖ 𝐱 - 𝐲 ‖ = \sqrt{14}$ . By Proposition 6.1.29, $36 =$ $9 +$ $2 Re ⟨ 𝐱, 𝐲 ⟩ +$ $16$ , so $Re ⟨ 𝐱, 𝐲 ⟩ = 11 / 2$ . (As a check, Proposition 6.1.29 with the minus sign gives $9 - 11 + 16 = 14$ .)

Polarization

The norm is defined from the inner product. Polarization goes the other way. Over $ℝ$ , subtracting the two expansions in Proposition 6.1.29 isolates the cross term: ${‖ 𝐱 + 𝐲 ‖}^{2} -$ ${‖ 𝐱 - 𝐲 ‖}^{2} =$ $4 ⟨ 𝐱, 𝐲 ⟩$ . Over $ℂ$ the same subtraction yields only $4 Re ⟨ 𝐱, 𝐲 ⟩$ , but the imaginary part is also a real part, since $Im ⟨ 𝐱, 𝐲 ⟩ =$ $Re (- 𝑖 ⟨ 𝐱, 𝐲 ⟩) =$ $Re ⟨ 𝐱, 𝑖 𝐲 ⟩$ .

Proposition 6.1.32 (Polarization identities). Let $𝑉$ be an inner product space and $𝐱, 𝐲 \in 𝑉$ .

(a)If $𝔽 = ℝ$ , then
$⟨ 𝐱, 𝐲 ⟩ = \frac{1}{4} ({‖ 𝐱 + 𝐲 ‖}^{2} - {‖ 𝐱 - 𝐲 ‖}^{2}) .$
(b)If $𝔽 = ℂ$ , then
$\begin{aligned} ⟨ 𝐱, 𝐲 ⟩ & = \frac{1}{4} \sum_{𝑘 = 0}^{3} 𝑖^{𝑘} {‖ 𝐱 + 𝑖^{𝑘} 𝐲 ‖}^{2} \\ = \frac{1}{4} ({‖ 𝐱 + 𝐲 ‖}^{2} - {‖ 𝐱 - 𝐲 ‖}^{2} + 𝑖 {‖ 𝐱 + 𝑖 𝐲 ‖}^{2} - 𝑖 {‖ 𝐱 - 𝑖 𝐲 ‖}^{2}) . \end{aligned}$

Proof (via real and imaginary parts).standard

Uses. Proposition 6.1.29 (Expanding a norm), Proposition 6.1.4 (Sesquilinearity), Notation 6.1.1 (Real and complex scalars)

Hint 1. Subtract the two identities of Proposition 6.1.29.

Hint 2. Over

ℂ

this gives only

Re ⟨ 𝐱, 𝐲 ⟩

. Write

Im ⟨ 𝐱, 𝐲 ⟩

as the real part of another inner product.

Hint 3.

⟨ 𝐱, 𝑖 𝐲 ⟩ =

- 𝑖 ⟨ 𝐱, 𝐲 ⟩

and

Re (- 𝑖 𝑧) = Im 𝑧

. Apply the real-part formula with

𝑖 𝐲

in place of

𝐲

Sketch. Subtracting the expansions of

{‖ 𝐱 \pm 𝐲 ‖}^{2}

gives

4 Re ⟨ 𝐱, 𝐲 ⟩

, which is

4 ⟨ 𝐱, 𝐲 ⟩

over

ℝ

. Over

ℂ

, applying this to the pair

𝐱, 𝑖 𝐲

gives

4 Re ⟨ 𝐱, 𝑖 𝐲 ⟩ =

4 Im ⟨ 𝐱, 𝐲 ⟩ =

{‖ 𝐱 + 𝑖 𝐲 ‖}^{2} -

{‖ 𝐱 - 𝑖 𝐲 ‖}^{2}

. Then combine

⟨ 𝐱, 𝐲 ⟩ = Re + 𝑖 Im

Subtracting the two identities of Proposition 6.1.29 gives, for all $𝐱, 𝐲$ in any inner product space,

Re ⟨ 𝐱, 𝐲 ⟩ = \frac{1}{4} ({‖ 𝐱 + 𝐲 ‖}^{2} - {‖ 𝐱 - 𝐲 ‖}^{2}) .

(6.1.2)

(a) If $𝔽 = ℝ$ , then $⟨ 𝐱, 𝐲 ⟩$ is real, so it equals its real part, and (6.1.2) is (a).

(b) Let $𝔽 = ℂ$ . By Proposition 6.1.4, $⟨ 𝐱, 𝑖 𝐲 ⟩ =$ $𝑖 ⟨ 𝐱, 𝐲 ⟩ =$ $- 𝑖 ⟨ 𝐱, 𝐲 ⟩$ , and $Re (- 𝑖 𝑧) = Im 𝑧$ for every $𝑧 \in ℂ$ . So (6.1.2) applied to $𝐱$ and $𝑖 𝐲$ gives

Im ⟨ 𝐱, 𝐲 ⟩ = Re ⟨ 𝐱, 𝑖 𝐲 ⟩ = \frac{1}{4} ({‖ 𝐱 + 𝑖 𝐲 ‖}^{2} - {‖ 𝐱 - 𝑖 𝐲 ‖}^{2}) .

Since $⟨ 𝐱, 𝐲 ⟩ =$ $Re ⟨ 𝐱, 𝐲 ⟩ +$ $𝑖 Im ⟨ 𝐱, 𝐲 ⟩$ , adding (6.1.2) to $𝑖$ times this identity gives the second line of (b). The terms of $\sum_{𝑘 = 0}^{3} 𝑖^{𝑘} {‖ 𝐱 + 𝑖^{𝑘} 𝐲 ‖}^{2}$ for $𝑘 = 0, 1, 2, 3$ are ${‖ 𝐱 + 𝐲 ‖}^{2}$ , $𝑖 {‖ 𝐱 + 𝑖 𝐲 ‖}^{2}$ , $- {‖ 𝐱 - 𝐲 ‖}^{2}$ , and $- 𝑖 {‖ 𝐱 - 𝑖 𝐲 ‖}^{2}$ , so the two lines agree. □

Proof (by averaging over powers of $𝑖$ ).standard

Uses. Proposition 6.1.29 (Expanding a norm), Notation 6.1.1 (Real and complex scalars)

Hint 1. Expand each

{‖ 𝐱 + 𝑖^{𝑘} 𝐲 ‖}^{2}

with Proposition 6.1.29, writing

2 Re 𝑤

𝑤 + 𝑤

Hint 2. After multiplying by

𝑖^{𝑘}

, the terms carry the factors

𝑖^{𝑘}

1

, and

𝑖^{2 𝑘}

. Which of these survive summing over

𝑘 = 0, 1, 2, 3

Sketch. With

𝑧 = ⟨ 𝐱, 𝐲 ⟩

𝑖^{𝑘} {‖ 𝐱 + 𝑖^{𝑘} 𝐲 ‖}^{2} =

𝑖^{𝑘} ({‖ 𝐱 ‖}^{2} + {‖ 𝐲 ‖}^{2}) +

𝑧 +

𝑖^{2 𝑘} 𝑧

. Summing over

𝑘

kills the first and last terms, because

1 + 𝑖 + 𝑖^{2} + 𝑖^{3} = 0

and

1 - 1 + 1 - 1 = 0

, and leaves

4 𝑧

Let $𝔽 = ℂ$ , let $𝑧 = ⟨ 𝐱, 𝐲 ⟩$ , and let $𝑐 = 𝑖^{𝑘}$ for some $𝑘 \in {0, 1, 2, 3}$ , so that $| 𝑐 | = 1$ and $𝑐 = 𝑐^{- 1}$ . By Proposition 6.1.29, and since $2 Re 𝑤 = 𝑤 + 𝑤$ ,

{‖ 𝐱 + 𝑐 𝐲 ‖}^{2} = {‖ 𝐱 ‖}^{2} + {‖ 𝐲 ‖}^{2} + 𝑐 𝑧 + 𝑐 𝑧 .

Multiplying by $𝑐$ and using $𝑐 𝑐 = 1$ ,

𝑐 {‖ 𝐱 + 𝑐 𝐲 ‖}^{2} = 𝑐 ({‖ 𝐱 ‖}^{2} + {‖ 𝐲 ‖}^{2}) + 𝑧 + 𝑐^{2} 𝑧 .

Now sum over $𝑐 = 1, 𝑖, - 1, - 𝑖$ . The coefficients $𝑐$ add to $1 + 𝑖 - 1 - 𝑖 = 0$ , and the coefficients $𝑐^{2}$ add to $1 - 1 + 1 - 1 = 0$ , so

\sum_{𝑘 = 0}^{3} 𝑖^{𝑘} {‖ 𝐱 + 𝑖^{𝑘} 𝐲 ‖}^{2} = 4 𝑧 = 4 ⟨ 𝐱, 𝐲 ⟩,

which is (b). Part (a) is the same computation with $𝑐 = \pm 1$ and $𝑧$ real: then $𝑐 {‖ 𝐱 + 𝑐 𝐲 ‖}^{2} =$ $𝑐 ({‖ 𝐱 ‖}^{2} + {‖ 𝐲 ‖}^{2}) +$ $2 𝑧$ , and summing over $𝑐 = \pm 1$ gives ${‖ 𝐱 + 𝐲 ‖}^{2} -$ ${‖ 𝐱 - 𝐲 ‖}^{2} =$ $4 𝑧$ . □

The first proof explains why there are four terms: two give the real part, and two more give the imaginary part. The second explains the coefficients. Multiplying by $𝑖^{𝑘}$ and summing over $𝑘$ is a filter: it keeps the term that carries no power of $𝑖$ and discards the terms that carry $𝑖^{𝑘}$ and $𝑖^{2 𝑘}$ , because the powers of $𝑖$ add to $0$ . The same filter works with any $𝑚 \geq 3$ equally spaced points on the unit circle in place of $1, 𝑖, - 1, - 𝑖$ , and it is the idea behind the discrete Fourier transform.

Technique 6.1.33 (Polarization). An inner product is determined by its values $⟨ 𝐯, 𝐯 ⟩$ on single vectors: by Proposition 6.1.32, $⟨ 𝐱, 𝐲 ⟩$ is a combination of the numbers ${‖ 𝐱 + 𝑐 𝐲 ‖}^{2}$ with $𝑐 \in {\pm 1, \pm 𝑖}$ . So a statement about all pairs $(𝐱, 𝐲)$ can often be reduced to a statement about single vectors, applied to $𝐱 + 𝑐 𝐲$ for a few scalars $𝑐$ .

Reach for it when the hypothesis or the conclusion involves only lengths, or only “quadratic” quantities such as $⟨ 𝑇 𝐯, 𝐯 ⟩$ , and you need information about two different vectors at once. The canonical example is Proposition 6.1.32 itself. Over $ℝ$ there is a caveat: a symmetric function $𝐵 (𝐱, 𝐲)$ that is linear in each slot is determined by its diagonal values $𝐵 (𝐯, 𝐯)$ , but without symmetry this fails. For instance, $𝐵 (𝐱, 𝐲) =$ $𝑥_{1} 𝑦_{2} -$ $𝑥_{2} 𝑦_{1}$ on $ℝ^{2}$ satisfies $𝐵 (𝐯, 𝐯) = 0$ for all $𝐯$ without being $0$ . Over $ℂ$ no symmetry is needed (Proposition 6.1.41).

Used in 12 places and 3 practice exercises

Corollary 6.1.34, Exercise 6.1.35, Proposition 6.1.41, Exercise 6.1.45, Exercise 6.1.51, Exercise 6.1.53, Exercise 6.1.59, Exercise 6.7.8, Problem 6.8, Proposition 8.1.4, Proposition 8.1.13, Theorem 8.4.4; 3 practice exercises.

Corollary 6.1.34 (The norm determines the inner product). Let

{⟨ \cdot, \cdot ⟩}_{1}

and

{⟨ \cdot, \cdot ⟩}_{2}

be inner products on

𝑉

with

{⟨ 𝐯, 𝐯 ⟩}_{1} = {⟨ 𝐯, 𝐯 ⟩}_{2}

for all

𝐯 \in 𝑉

. Then

{⟨ 𝐱, 𝐲 ⟩}_{1} = {⟨ 𝐱, 𝐲 ⟩}_{2}

for all

𝐱, 𝐲 \in 𝑉

Proof.routine

Uses. Proposition 6.1.32 (Polarization identities)

Techniques. Polarization (Technique 6.1.33)

Hint 1. Write both

{⟨ 𝐱, 𝐲 ⟩}_{1}

and

{⟨ 𝐱, 𝐲 ⟩}_{2}

with Proposition 6.1.32.

Hint 2. The right sides of the polarization identities only involve norms, and the two norms agree.

Sketch. Both inner products equal the same combination of the numbers

{‖ 𝐱 + 𝑐 𝐲 ‖}^{2}

𝑐 \in {\pm 1, \pm 𝑖}

, and these do not depend on which inner product defines the norm.

The two inner products define the same norm,

‖ 𝐯 ‖ =

\sqrt{{⟨ 𝐯, 𝐯 ⟩}_{1}} =

\sqrt{{⟨ 𝐯, 𝐯 ⟩}_{2}}

. By Proposition 6.1.32, applied once to each inner product, both

{⟨ 𝐱, 𝐲 ⟩}_{1}

and

{⟨ 𝐱, 𝐲 ⟩}_{2}

equal

\frac{1}{4} ({‖ 𝐱 + 𝐲 ‖}^{2} - {‖ 𝐱 - 𝐲 ‖}^{2})

𝔽 = ℝ

, and

\frac{1}{4} \sum_{𝑘 = 0}^{3} 𝑖^{𝑘} {‖ 𝐱 + 𝑖^{𝑘} 𝐲 ‖}^{2}

𝔽 = ℂ

. So they are equal. □

Combined with Proposition 6.1.27(b), the inner product is even determined by its unit sphere ${𝐯 \in 𝑉 | ‖ 𝐯 ‖ = 1}$ : for $𝐯 \neq 𝟎$ , $‖ 𝐯 ‖$ is the unique $𝑡 > 0$ for which $𝐯 / 𝑡$ lies on the sphere. So each curve in Figure 6.1.2 determines its inner product completely.

Exercise 6.1.35.routine Vectors $𝐱, 𝐲$ in a complex inner product space satisfy

{‖ 𝐱 + 𝐲 ‖}^{2} = 10, {‖ 𝐱 - 𝐲 ‖}^{2} = 2, {‖ 𝐱 + 𝑖 𝐲 ‖}^{2} = 4, {‖ 𝐱 - 𝑖 𝐲 ‖}^{2} = 8 .

Find $⟨ 𝐱, 𝐲 ⟩$ and ${‖ 𝐱 ‖}^{2} + {‖ 𝐲 ‖}^{2}$ .

Solution.

Uses. Proposition 6.1.32 (Polarization identities), Proposition 6.1.30 (Parallelogram law)

Techniques. Polarization (Technique 6.1.33)

Hint 1. Substitute into Proposition 6.1.32(b).

Hint 2. For

{‖ 𝐱 ‖}^{2} + {‖ 𝐲 ‖}^{2}

, use the parallelogram law, for the pair

𝐱, 𝐲

or for the pair

𝐱, 𝑖 𝐲

Sketch.

⟨ 𝐱, 𝐲 ⟩ =

(10 - 2 + 4 𝑖 - 8 𝑖) / 4 =

2 -

𝑖

. The parallelogram law gives

2 ({‖ 𝐱 ‖}^{2} + {‖ 𝐲 ‖}^{2}) =

10 +

2 =

12

, so the sum is

6

By Proposition 6.1.32(b),

⟨ 𝐱, 𝐲 ⟩ = \frac{1}{4} (10 - 2 + 4 𝑖 - 8 𝑖) = \frac{1}{4} (8 - 4 𝑖) = 2 - 𝑖 .

By Proposition 6.1.30, $2 {‖ 𝐱 ‖}^{2} +$ $2 {‖ 𝐲 ‖}^{2} =$ $10 +$ $2 =$ $12$ , so ${‖ 𝐱 ‖}^{2} + {‖ 𝐲 ‖}^{2} = 6$ . The data are consistent: the parallelogram law for $𝐱$ and $𝑖 𝐲$ , where $‖ 𝑖 𝐲 ‖ = ‖ 𝐲 ‖$ , gives $4 + 8 = 12$ as well.

6.1.5Inner products distinguish vectors

The lecture’s next observation is that a vector is determined by its inner products with all vectors. The proof is a one-line instance of a move used throughout this chapter, so we name it first.

Technique 6.1.36 (Show a vector is zero via its norm). By definiteness, $𝐯 = 𝟎$ as soon as $⟨ 𝐯, 𝐯 ⟩ = 0$ . So to prove that a vector is zero, compute $⟨ 𝐯, 𝐯 ⟩$ , or ${‖ 𝐯 ‖}^{2}$ , and show that it vanishes. Often one knows that $⟨ 𝐯, 𝐱 ⟩ = 0$ for many $𝐱$ , and the move is to take $𝐱 = 𝐯$ . To prove $𝐯 = 𝐰$ , apply this to $𝐯 - 𝐰$ , expanding ${‖ 𝐯 - 𝐰 ‖}^{2}$ with Proposition 6.1.29 if necessary.

Reach for it when the hypotheses are statements about inner products and the conclusion is an equation between vectors. The canonical example is Proposition 6.1.37. In Section 6.5 the same move shows that $𝐴^{*} 𝐴 𝐱 = 𝟎$ implies $𝐴 𝐱 = 𝟎$ .

Used in 19 places

Proposition 6.1.37, Corollary 6.1.38, Exercise 6.1.50, Proposition 6.3.30, Theorem 6.5.5, Proposition 6.5.9, Theorem 6.6.21, Proposition 6.6.23, Exercise 6.6.51, Exercise 6.7.9, Problem 6.3, Problem 6.8, Problem 6.9, Problem 6.11, Proposition 8.1.4, Exercise 8.1.6, Proposition 8.1.14, Theorem 8.2.5, Lemma 8.3.8.

Proposition 6.1.37 (Inner products distinguish vectors). Let

𝑉

be an inner product space and

𝐯, 𝐰 \in 𝑉

. If

⟨ 𝐯, 𝐱 ⟩ = ⟨ 𝐰, 𝐱 ⟩

for all

𝐱 \in 𝑉

, then

𝐯 = 𝐰

Proof.routine

Uses. Definition 6.1.2 (Inner product), Proposition 6.1.4 (Sesquilinearity)

Techniques. Show a vector is zero via its norm (Technique 6.1.36)

Hint 1. Reduce to the case

𝐰 = 𝟎

by looking at

𝐯 - 𝐰

Hint 2. If

⟨ 𝐮, 𝐱 ⟩ = 0

for every

𝐱

, which choice of

𝐱

tells you the most about

𝐮

Hint 3. Take

𝐱 = 𝐮

and use definiteness.

Sketch. By linearity,

⟨ 𝐯 - 𝐰, 𝐱 ⟩ = 0

for every

𝐱

. Choosing

𝐱 = 𝐯 - 𝐰

gives

{‖ 𝐯 - 𝐰 ‖}^{2} = 0

, so

𝐯 = 𝐰

First suppose $𝐰 = 𝟎$ . By Proposition 6.1.4, $⟨ 𝐯, 𝐱 ⟩ = ⟨ 𝟎, 𝐱 ⟩ = 0$ for all $𝐱 \in 𝑉$ . In particular, for $𝐱 = 𝐯$ we get $⟨ 𝐯, 𝐯 ⟩ = 0$ , and $𝐯 = 𝟎$ by definiteness.

For the general case, linearity in the first slot gives $⟨ 𝐯 - 𝐰, 𝐱 ⟩ =$ $⟨ 𝐯, 𝐱 ⟩ -$ $⟨ 𝐰, 𝐱 ⟩ =$ $0 =$ $⟨ 𝟎, 𝐱 ⟩$ for all $𝐱 \in 𝑉$ . By the first case, $𝐯 - 𝐰 = 𝟎$ , that is, $𝐯 = 𝐰$ , as desired. □

The same holds with the vectors in the second slot: if $⟨ 𝐱, 𝐯 ⟩ = ⟨ 𝐱, 𝐰 ⟩$ for all $𝐱$ , conjugating gives $⟨ 𝐯, 𝐱 ⟩ = ⟨ 𝐰, 𝐱 ⟩$ for all $𝐱$ . Nor do we need every $𝐱$ : the vectors of a spanning list suffice (Exercise 6.1.50).

Corollary 6.1.38 (Linear maps are determined by inner products). Let

𝑉

be a vector space over

𝔽

, let

𝑊

be an inner product space, and let

𝑆, 𝑇 : 𝑉 \to 𝑊

be linear. If

⟨ 𝑆 𝐯, 𝐰 ⟩ =

⟨ 𝑇 𝐯, 𝐰 ⟩

for all

𝐯 \in 𝑉

and all

𝐰 \in 𝑊

, then

𝑆 = 𝑇

Proof.routine

Uses. Proposition 6.1.37 (Inner products distinguish vectors)

Techniques. Show a vector is zero via its norm (Technique 6.1.36)

Hint 1. Fix

𝐯 \in 𝑉

. What does the hypothesis say about the two vectors

𝑆 𝐯

and

𝑇 𝐯

𝑊

Hint 2. Apply Proposition 6.1.37 in

𝑊

Sketch. For each fixed

𝐯

⟨ 𝑆 𝐯, 𝐰 ⟩ =

⟨ 𝑇 𝐯, 𝐰 ⟩

for all

𝐰 \in 𝑊

, so

𝑆 𝐯 = 𝑇 𝐯

by Proposition 6.1.37.

Here and below we write

𝑇 𝐯

for

𝑇 (𝐯)

. Fix

𝐯 \in 𝑉

. By hypothesis, the vectors

𝑆 𝐯

and

𝑇 𝐯

𝑊

satisfy

⟨ 𝑆 𝐯, 𝐰 ⟩ =

⟨ 𝑇 𝐯, 𝐰 ⟩

for all

𝐰 \in 𝑊

, so

𝑆 𝐯 = 𝑇 𝐯

by Proposition 6.1.37. Since

𝐯

was arbitrary,

𝑆 = 𝑇

. □

The proof never used that $𝑆$ and $𝑇$ are linear, so the corollary holds for arbitrary functions $𝑉 \to 𝑊$ . It is how adjoints will be shown to be unique (Section 6.6).

Testing on the diagonal

Corollary 6.1.38 tests $⟨ 𝑇 𝐯, 𝐰 ⟩$ on all pairs of vectors. For an operator $𝑇 \in L (𝑉)$ , it is natural to ask whether the values $⟨ 𝑇 𝐯, 𝐯 ⟩$ , with the same vector in both slots, are already enough. Over $ℝ$ they are not.

Example 6.1.39 (A rotation). Let $𝑇 : ℝ^{2} \to ℝ^{2}$ be $𝑇 𝐱 =$ $𝐽 𝐱 =$ $(- 𝑥_{2}, 𝑥_{1})$ , where $𝐽 = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ ; this is the rotation by $𝜋 / 2$ . With the dot product,

⟨ 𝑇 𝐱, 𝐱 ⟩ = (- 𝑥_{2}) 𝑥_{1} + 𝑥_{1} 𝑥_{2} = 0 for every 𝐱 \in ℝ^{2},

as a rotation by a right angle moves every vector to a perpendicular one. But $𝑇 \neq 0$ .

Over $ℂ$ , however, the values $⟨ 𝑇 𝐯, 𝐯 ⟩$ do determine $𝑇$ . The proof feeds the hypothesis well-chosen vectors, a move that polarization has already used.

Technique 6.1.40 (Plug in clever test vectors). A hypothesis that holds for all vectors holds in particular for sums and multiples of the vectors you care about. Applying it to $𝐯 = 𝐱 + 𝐲$ , $𝐱 - 𝐲$ , $𝐱 + 𝑖 𝐲$ , or to the very vector you want to control, turns one identity into a system of equations, which can then be solved for the quantity you want.

Reach for it when a hypothesis is quantified over all vectors but involves each vector only quadratically, like $‖ 𝐯 ‖$ or $⟨ 𝑇 𝐯, 𝐯 ⟩$ , and you need information about pairs of vectors. Over $ℂ$ , the pair $𝐱 + 𝐲$ and $𝐱 + 𝑖 𝐲$ is usually the right choice, because together they separate real and imaginary parts. The canonical example is Proposition 6.1.41. The polarization identities (Proposition 6.1.32) and the choice $𝐱 = 𝐯 - 𝐰$ in Proposition 6.1.37 are instances of the same move.

Used in 14 places and 1 practice exercise

Proposition 6.1.41, Exercise 6.1.46, Exercise 6.1.47, Exercise 6.1.52, Proposition 6.3.5, Exercise 6.3.52, Exercise 6.7.7, Exercise 6.7.8, Exercise 6.7.15, Problem 6.6, Problem 7.8, Proposition 8.1.4, Theorem 8.3.23, Exercise 8.4.34; 1 practice exercise.

Proposition 6.1.41 (Over

ℂ

, the values

⟨ 𝑇 𝐯, 𝐯 ⟩

determine

𝑇

). Let

𝑉

be a complex inner product space and

𝑇 \in L (𝑉)

. If

⟨ 𝑇 𝐯, 𝐯 ⟩ = 0

for all

𝐯 \in 𝑉

, then

𝑇 = 0

. Consequently, if

𝑆, 𝑇 \in L (𝑉)

satisfy

⟨ 𝑆 𝐯, 𝐯 ⟩ =

⟨ 𝑇 𝐯, 𝐯 ⟩

for all

𝐯 \in 𝑉

, then

𝑆 = 𝑇

Intuition. By Corollary 6.1.38 it suffices to show

⟨ 𝑇 𝐱, 𝐲 ⟩ = 0

for all

𝐱, 𝐲

, but the hypothesis only speaks about one vector at a time. Plugging

𝐯 = 𝐱 + 𝐲

into it isolates the sum

⟨ 𝑇 𝐱, 𝐲 ⟩ +

⟨ 𝑇 𝐲, 𝐱 ⟩

. Over

ℝ

that is all one can get, and for the rotation of Example 6.1.39 this sum is

0

for all

𝐱, 𝐲

although its terms are not. Over

ℂ

, plugging in

𝐯 = 𝐱 + 𝑖 𝐲

produces a second combination,

- 𝑖

times the difference

⟨ 𝑇 𝐱, 𝐲 ⟩ -

⟨ 𝑇 𝐲, 𝐱 ⟩

, and two independent equations in the two unknowns can be solved.

Proof (with test vectors).standard

Uses. Corollary 6.1.38 (Linear maps are determined by inner products), Proposition 6.1.4 (Sesquilinearity), Proposition 2.1.23 ( $L (𝑉, 𝑊)$ is a vector space)

Techniques. Plug in clever test vectors (Technique 6.1.40)

Hint 1. It suffices to show

⟨ 𝑇 𝐱, 𝐲 ⟩ = 0

for all

𝐱, 𝐲 \in 𝑉

. Apply the hypothesis to a vector built from

𝐱

and

𝐲

Hint 2.

𝐯 = 𝐱 + 𝐲

gives

⟨ 𝑇 𝐱, 𝐲 ⟩ +

⟨ 𝑇 𝐲, 𝐱 ⟩ =

0

. That alone cannot suffice, since the rotation satisfies it. Find a second equation.

Hint 3. Try

𝐯 = 𝐱 + 𝑖 𝐲

, and remember that

𝑖

leaves the second slot as

- 𝑖

Sketch. Expanding

⟨ 𝑇 𝐯, 𝐯 ⟩ = 0

𝐯 = 𝐱 + 𝐲

and at

𝐯 = 𝐱 + 𝑖 𝐲

gives

⟨ 𝑇 𝐱, 𝐲 ⟩ +

⟨ 𝑇 𝐲, 𝐱 ⟩ =

0

and

- 𝑖 ⟨ 𝑇 𝐱, 𝐲 ⟩ +

𝑖 ⟨ 𝑇 𝐲, 𝐱 ⟩ =

0

. Dividing the second by

- 𝑖

and adding,

⟨ 𝑇 𝐱, 𝐲 ⟩ = 0

for all

𝐱, 𝐲

, so

𝑇 = 0

. Apply this to

𝑆 - 𝑇

for the last claim.

Let $𝐱, 𝐲 \in 𝑉$ . Since $𝑇$ is linear and the inner product is sesquilinear,

⟨ 𝑇 (𝐱 + 𝐲), 𝐱 + 𝐲 ⟩ = ⟨ 𝑇 𝐱, 𝐱 ⟩ + ⟨ 𝑇 𝐱, 𝐲 ⟩ + ⟨ 𝑇 𝐲, 𝐱 ⟩ + ⟨ 𝑇 𝐲, 𝐲 ⟩ .

The left side and the first and last terms on the right are $0$ by hypothesis, so

⟨ 𝑇 𝐱, 𝐲 ⟩ + ⟨ 𝑇 𝐲, 𝐱 ⟩ = 0 .

(6.1.3)

Similarly, $𝑇 (𝐱 + 𝑖 𝐲) =$ $𝑇 𝐱 +$ $𝑖 𝑇 𝐲$ , and Proposition 6.1.4 gives

⟨ 𝑇 𝐱 + 𝑖 𝑇 𝐲, 𝐱 + 𝑖 𝐲 ⟩ = ⟨ 𝑇 𝐱, 𝐱 ⟩ + 𝑖 ⟨ 𝑇 𝐱, 𝐲 ⟩ + 𝑖 ⟨ 𝑇 𝐲, 𝐱 ⟩ + 𝑖 𝑖 ⟨ 𝑇 𝐲, 𝐲 ⟩ .

Again the left side and the first and last terms vanish, so $- 𝑖 ⟨ 𝑇 𝐱, 𝐲 ⟩ +$ $𝑖 ⟨ 𝑇 𝐲, 𝐱 ⟩ =$ $0$ . Dividing by $- 𝑖$ ,

⟨ 𝑇 𝐱, 𝐲 ⟩ - ⟨ 𝑇 𝐲, 𝐱 ⟩ = 0 .

Adding this to (6.1.3) gives $2 ⟨ 𝑇 𝐱, 𝐲 ⟩ = 0$ , so $⟨ 𝑇 𝐱, 𝐲 ⟩ =$ $0 =$ $⟨ 0 𝐱, 𝐲 ⟩$ for all $𝐱, 𝐲 \in 𝑉$ . By Corollary 6.1.38, $𝑇 = 0$ .

For the last statement, $𝑆 - 𝑇 \in L (𝑉)$ (Proposition 2.1.23), and $⟨ (𝑆 - 𝑇) 𝐯, 𝐯 ⟩ =$ $⟨ 𝑆 𝐯, 𝐯 ⟩ -$ $⟨ 𝑇 𝐯, 𝐯 ⟩ =$ $0$ for all $𝐯$ . So $𝑆 - 𝑇 = 0$ . □

Proof (by polarizing a sesquilinear function).standard

Uses. Corollary 6.1.38 (Linear maps are determined by inner products), Proposition 6.1.4 (Sesquilinearity), Proposition 6.1.32 (Polarization identities)

Techniques. Polarization (Technique 6.1.33)

Hint 1. The function

𝐵 (𝐱, 𝐲) =

⟨ 𝑇 𝐱, 𝐲 ⟩

is linear in

𝐱

and conjugate-linear in

𝐲

, like an inner product, but it need not be conjugate symmetric or positive. Try to redo the second proof of Proposition 6.1.32 for

𝐵

; where that proof wrote

𝑧

for

⟨ 𝐲, 𝐱 ⟩

, keep

𝐵 (𝐲, 𝐱)

instead.

Hint 2. Expand

𝐵 (𝐱 + 𝑖^{𝑘} 𝐲, 𝐱 + 𝑖^{𝑘} 𝐲)

, multiply by

𝑖^{𝑘}

, and sum over

𝑘 = 0, \dots, 3

Sketch. Redo the roots-of-unity computation of Proposition 6.1.32 with

𝐵 (𝐲, 𝐱)

in place of

𝑧

; that term still carries the factor

𝑐^{2}

, which sums to

0

. So any

𝐵

that is linear in the first slot and conjugate-linear in the second satisfies

𝐵 (𝐱, 𝐲) =

\frac{1}{4} \sum_{𝑘} 𝑖^{𝑘} 𝐵 (𝐱 + 𝑖^{𝑘} 𝐲, 𝐱 + 𝑖^{𝑘} 𝐲)

. For

𝐵 (𝐱, 𝐲) =

⟨ 𝑇 𝐱, 𝐲 ⟩

, every term on the right is

0

Let $𝐵 (𝐱, 𝐲) =$ $⟨ 𝑇 𝐱, 𝐲 ⟩$ . It is linear in $𝐱$ , as $𝑇$ is linear and the inner product is linear in the first slot, and it satisfies $𝐵 (𝐱, 𝑎 𝐲 + 𝑏 𝐳) =$ $𝑎 𝐵 (𝐱, 𝐲) +$ $𝑏 𝐵 (𝐱, 𝐳)$ by Proposition 6.1.4. For $𝑐 \in ℂ$ with $| 𝑐 | = 1$ , expanding gives

𝐵 (𝐱 + 𝑐 𝐲, 𝐱 + 𝑐 𝐲) = 𝐵 (𝐱, 𝐱) + 𝑐 𝐵 (𝐱, 𝐲) + 𝑐 𝐵 (𝐲, 𝐱) + 𝐵 (𝐲, 𝐲),

using $𝑐 𝑐 = 1$ . Multiply by $𝑐$ and sum over $𝑐 = 𝑖^{𝑘}$ , $𝑘 = 0, 1, 2, 3$ . Since $\sum_{𝑘} 𝑖^{𝑘} = 0$ , $\sum_{𝑘} 𝑖^{𝑘} 𝑖^{𝑘} = 4$ , and $\sum_{𝑘} 𝑖^{2 𝑘} = 0$ ,

𝐵 (𝐱, 𝐲) = \frac{1}{4} \sum_{𝑘 = 0}^{3} 𝑖^{𝑘} 𝐵 (𝐱 + 𝑖^{𝑘} 𝐲, 𝐱 + 𝑖^{𝑘} 𝐲) .

(6.1.4)

We never used conjugate symmetry or positivity, only how $𝐵$ treats scalars in each slot. For our $𝐵$ , every term on the right is of the form $⟨ 𝑇 𝐯, 𝐯 ⟩ = 0$ . So $⟨ 𝑇 𝐱, 𝐲 ⟩ = 0$ for all $𝐱, 𝐲$ , and $𝑇 = 0$ by Corollary 6.1.38. The last statement follows as in the first proof. □

The second proof shows more. Over $ℂ$ , every function $𝐵 : 𝑉 \times 𝑉 \to ℂ$ that is linear in the first slot and conjugate-linear in the second is determined by its diagonal values $𝐵 (𝐯, 𝐯)$ , through (6.1.4), with no symmetry assumption. Over $ℝ$ the analogous computation gives

\frac{1}{4} (𝐵 (𝐱 + 𝐲, 𝐱 + 𝐲) - 𝐵 (𝐱 - 𝐲, 𝐱 - 𝐲)) = \frac{1}{2} (𝐵 (𝐱, 𝐲) + 𝐵 (𝐲, 𝐱)),

which recovers only the symmetric part of $𝐵$ . For the rotation of Example 6.1.39, $𝐵 (𝐱, 𝐲) =$ $(𝐽 𝐱) \cdot$ $𝐲 =$ $𝑥_{1} 𝑦_{2} -$ $𝑥_{2} 𝑦_{1}$ is antisymmetric, so its symmetric part is $0$ although $𝐵$ is not.

Aside (Norms that do not come from inner products).Beyond 21-242 The properties of $‖ \cdot ‖$ that matter in analysis are (a) and (b) of Proposition 6.1.27, together with the triangle inequality $‖ 𝐯 + 𝐰 ‖ \leq ‖ 𝐯 ‖ + ‖ 𝐰 ‖$ , which Section 6.2 proves for the norm of an inner product. Any function $𝑉 \to ℝ$ with these three properties is called a norm on $𝑉$ , and a vector space with a norm is a normed space. Every inner product space is a normed space, but $𝔽^{𝑛}$ carries many norms that do not come from inner products, for instance

{‖ 𝐱 ‖}_{1} = \sum_{𝑘 = 1}^{𝑛} | 𝑥_{𝑘} |, {‖ 𝐱 ‖}_{\infty} = max_{𝑘} | 𝑥_{𝑘} | .

The first is the taxicab norm, the distance driven along a grid of streets; the second is the maximum norm. Checking the three properties for each is a short exercise. Their unit spheres in $ℝ^{2}$ are a diamond and a square (Figure 6.1.4), while those of inner products look like the ellipses of Figure 6.1.2.

Figure 6.1.4. The unit spheres of

{‖ \cdot ‖}_{1}

(diamond), of the Euclidean norm (circle), and of

{‖ \cdot ‖}_{\infty}

(square) in

ℝ^{2}

. The marked points

𝐞_{1} \pm 𝐞_{2}

test the parallelogram law.

Corners suggest that no inner product is involved, and the parallelogram law (Proposition 6.1.30) makes this precise. In $ℝ^{2}$ ,

\begin{aligned} {‖ 𝐞_{1} + 𝐞_{2} ‖}_{1}^{2} + {‖ 𝐞_{1} - 𝐞_{2} ‖}_{1}^{2} & = 4 + 4 = 8, \\ {‖ 𝐞_{1} + 𝐞_{2} ‖}_{\infty}^{2} + {‖ 𝐞_{1} - 𝐞_{2} ‖}_{\infty}^{2} & = 1 + 1 = 2, \end{aligned}

while $2 {‖ 𝐞_{1} ‖}^{2} + 2 {‖ 𝐞_{2} ‖}^{2} = 4$ for both norms. So neither satisfies the parallelogram law, and neither is the norm of any inner product on $ℝ^{2}$ .

The parallelogram law is the only obstruction. The Jordan–von Neumann theorem says that a norm on a real or complex vector space is the norm of an inner product if and only if it satisfies the parallelogram law. The inner product is then given by the polarization identities, as it must be by Corollary 6.1.34. The real case is Exercise 6.1.61.

Exercises

Exercise 6.1.42.standard Decide which of the following define inner products.

(a)On $ℝ^{3}$ : $⟨ 𝐱, 𝐲 ⟩ =$ $𝑥_{1} 𝑦_{1} +$ $2 𝑥_{2} 𝑦_{2} +$ $3 𝑥_{3} 𝑦_{3} +$ $𝑥_{1} 𝑦_{3} +$ $𝑥_{3} 𝑦_{1}$ .
(b)On $ℝ^{2 \times 2}$ : $⟨ 𝐴, 𝐵 ⟩ =$ $tr (𝐴) tr (𝐵)$ .
(c)On $ℂ^{2}$ : $⟨ 𝐱, 𝐲 ⟩ =$ $𝑥_{1} 𝑦_{1} +$ $𝑖 𝑥_{2} 𝑦_{2}$ .
(d)On ${ℝ [𝑥]}_{\leq 2}$ : $⟨ 𝑝, 𝑞 ⟩ =$ $𝑝 (- 1) 𝑞 (- 1) +$ $𝑝 (0) 𝑞 (0) +$ $𝑝 (1) 𝑞 (1)$ .
(e)On $ℂ^{2}$ : $⟨ 𝐱, 𝐲 ⟩ =$ $2 𝑥_{1} 𝑦_{1} -$ $𝑖 𝑥_{2} 𝑦_{1} +$ $𝑖 𝑥_{1} 𝑦_{2} +$ $𝑥_{2} 𝑦_{2}$ .

Solution.

Uses. Definition 6.1.2 (Inner product), Proposition 6.1.22 (Inner products on $𝔽^{𝑛}$ ), Proposition 6.1.4 (Sesquilinearity), Definition 1.2.17 (Evaluation)

Techniques. A sum of nonnegative terms (Technique 6.1.9)

Hint 1. (a) and (e) have the form

𝐲^{*} 𝑀 𝐱

; write down

𝑀

and use Proposition 6.1.22. To show

𝐱^{*} 𝑀 𝐱 > 0

, complete squares.

Hint 2. (b) Look for a nonzero matrix with

⟨ 𝐴, 𝐴 ⟩ = 0

. (c) Compute

⟨ 𝐞_{2}, 𝐞_{2} ⟩

Hint 3. (d) A polynomial

𝑎 + 𝑏 𝑥 + 𝑐 𝑥^{2}

that vanishes at

- 1

0

, and

1

: solve for

𝑎, 𝑏, 𝑐

. (e) Try to write

𝐱^{*} 𝑀 𝐱

{| 𝑥_{1} |}^{2} +

{| 𝑥_{2} + 𝑖 𝑥_{1} |}^{2}

Sketch. (a) Yes:

𝑀

is symmetric and

𝐱^{𝖳} 𝑀 𝐱 =

{(𝑥_{1} + 𝑥_{3})}^{2} +

2 𝑥_{2}^{2} +

2 𝑥_{3}^{2}

. (b) No:

𝐴 = (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}) \neq 0

has

⟨ 𝐴, 𝐴 ⟩ = 0

. (c) No:

⟨ 𝐞_{2}, 𝐞_{2} ⟩ = 𝑖

is not real. (d) Yes: it is symmetric, bilinear, a sum of squares on the diagonal, and a polynomial of degree at most

2

vanishing at three points is

0

. (e) Yes:

𝑀 = (\begin{matrix} 2 & - 𝑖 \\ 𝑖 & 1 \end{matrix})

is self-conjugate-transpose and

𝐱^{*} 𝑀 𝐱 =

{| 𝑥_{1} |}^{2} +

{| 𝑥_{2} + 𝑖 𝑥_{1} |}^{2}

(a) Yes. The formula is $𝐲^{𝖳} 𝑀 𝐱$ with $𝑀 = (\begin{matrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & 0 & 3 \end{matrix})$ , which is symmetric, and

𝐱^{𝖳} 𝑀 𝐱 = 𝑥_{1}^{2} + 2 𝑥_{2}^{2} + 3 𝑥_{3}^{2} + 2 𝑥_{1} 𝑥_{3} = {(𝑥_{1} + 𝑥_{3})}^{2} + 2 𝑥_{2}^{2} + 2 𝑥_{3}^{2} .

This is a sum of nonnegative terms, and it is $0$ only if $𝑥_{2} = 𝑥_{3} = 0$ and $𝑥_{1} + 𝑥_{3} = 0$ , that is, $𝐱 = 𝟎$ . By Proposition 6.1.22(b), it is an inner product.

(b) No. Definiteness fails: $𝐴 = (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix})$ is nonzero, but $⟨ 𝐴, 𝐴 ⟩ = {tr (𝐴)}^{2} = 0$ . (The formula is symmetric, bilinear, and positive.)

(c) No. $⟨ 𝐞_{2}, 𝐞_{2} ⟩ = 𝑖$ is not a nonnegative real number, so positivity fails. (Conjugate symmetry fails as well, since $𝑖 \neq 𝑖$ .)

(d) Yes. Evaluation at a point is linear (Definition 1.2.17), so the formula is linear in $𝑝$ , and it is visibly symmetric. Also $⟨ 𝑝, 𝑝 ⟩ =$ ${𝑝 (- 1)}^{2} +$ ${𝑝 (0)}^{2} +$ ${𝑝 (1)}^{2} \geq$ $0$ . If $⟨ 𝑝, 𝑝 ⟩ = 0$ , each term vanishes, so $𝑝 = 𝑎 + 𝑏 𝑥 + 𝑐 𝑥^{2}$ satisfies $𝑝 (0) = 𝑎 = 0$ , $𝑝 (1) = 𝑎 + 𝑏 + 𝑐 = 0$ , and $𝑝 (- 1) = 𝑎 - 𝑏 + 𝑐 = 0$ . Then $𝑏 + 𝑐 = 0$ and $- 𝑏 + 𝑐 = 0$ , so $𝑏 = 𝑐 = 0$ and $𝑝 = 𝟎$ .

(e) Yes. Reading off the coefficient of $𝑦_{𝑖} 𝑥_{𝑗}$ as $𝑚_{𝑖 𝑗}$ , the formula is $𝐲^{*} 𝑀 𝐱$ with $𝑀 = (\begin{matrix} 2 & - 𝑖 \\ 𝑖 & 1 \end{matrix})$ . The conjugate transpose of $𝑀$ is $(\begin{matrix} 2 & 𝑖 \\ - 𝑖 & 1 \end{matrix}) =$ $(\begin{matrix} 2 & - 𝑖 \\ 𝑖 & 1 \end{matrix}) =$ $𝑀$ . Also,

𝐱^{*} 𝑀 𝐱 = 2 {| 𝑥_{1} |}^{2} - 𝑖 𝑥_{1} 𝑥_{2} + 𝑖 𝑥_{1} 𝑥_{2} + {| 𝑥_{2} |}^{2} .

On the other hand, by Proposition 6.1.29 in $ℂ^{1}$ (or by direct expansion), ${| 𝑥_{2} + 𝑖 𝑥_{1} |}^{2} =$ ${| 𝑥_{2} |}^{2} +$ $𝑖 𝑥_{1} 𝑥_{2} -$ $𝑖 𝑥_{1} 𝑥_{2} +$ ${| 𝑥_{1} |}^{2}$ . Hence $𝐱^{*} 𝑀 𝐱 =$ ${| 𝑥_{1} |}^{2} +$ ${| 𝑥_{2} + 𝑖 𝑥_{1} |}^{2}$ , which is $0$ only if $𝑥_{1} = 0$ and then $𝑥_{2} = 0$ . By Proposition 6.1.22(b), this is an inner product.

Exercise 6.1.43.standard The proof of Proposition 6.1.13 works on any closed interval $[𝑎, 𝑏]$ in place of $[0, 1]$ . Decide which of the following are inner products.

(a) $⟨ 𝑓, 𝑔 ⟩ =$ $\int_{- 1}^{1} 𝑡^{2} 𝑓 (𝑡) 𝑔 (𝑡) d 𝑡$ on $𝐶 [- 1, 1]$ .
(b) $⟨ 𝑓, 𝑔 ⟩ =$ $\int_{- 1}^{1} 𝑡 𝑓 (𝑡) 𝑔 (𝑡) d 𝑡$ on $𝐶 [- 1, 1]$ .
(c) $⟨ 𝑓, 𝑔 ⟩ =$ $\int_{0}^{1 / 2} 𝑓 (𝑡) 𝑔 (𝑡) d 𝑡$ on $𝐶 [0, 1]$ .
(d) $⟨ 𝑓, 𝑔 ⟩ =$ $𝑓 (0) 𝑔 (0) +$ $\int_{0}^{1} 𝑓^{'} (𝑡) 𝑔^{'} (𝑡) d 𝑡$ on the space $𝐶^{1} [0, 1]$ of continuously differentiable functions $[0, 1] \to ℝ$ .

Solution.

Uses. Proposition 6.1.13 (Integral inner products), Lemma 6.1.14 (Nonnegative functions with zero integral)

Hint 1. In each case symmetry and bilinearity come from the integral. Only positivity and definiteness need thought.

Hint 2. (b) Find

𝑓

that is large where

𝑡 < 0

and small where

𝑡 > 0

. (c) Find a nonzero continuous function that vanishes on

[0, 1 / 2]

Hint 3. For (a) and (d), use Lemma 6.1.14 (on the relevant interval), then continuity or the mean value theorem.

Sketch. (a) Yes:

𝑡^{2} {𝑓 (𝑡)}^{2} = 0

for all

𝑡

forces

𝑓 = 0

off

0

, and then at

0

by continuity. (b) No:

𝑓 (𝑡) = 1 - 𝑡

gives

⟨ 𝑓, 𝑓 ⟩ = - 4 / 3

. (c) No:

𝑓 (𝑡) =

max (0, 𝑡 - 1 / 2)

is nonzero with

⟨ 𝑓, 𝑓 ⟩ = 0

. (d) Yes:

⟨ 𝑓, 𝑓 ⟩ = 0

gives

𝑓 (0) = 0

and

𝑓^{'} = 0

, so

𝑓 = 0

In all four cases the formula is symmetric and bilinear, by linearity of the integral and of $𝑓 \mapsto 𝑓^{'}$ . So only positivity and definiteness are at stake.

(a) Yes. $⟨ 𝑓, 𝑓 ⟩ =$ $\int_{- 1}^{1} 𝑡^{2} {𝑓 (𝑡)}^{2} d 𝑡 \geq$ $0$ , as the integrand is continuous and nonnegative. If $⟨ 𝑓, 𝑓 ⟩ = 0$ , then Lemma 6.1.14 (on $[- 1, 1]$ ) gives $𝑡^{2} {𝑓 (𝑡)}^{2} = 0$ for all $𝑡$ , so $𝑓 (𝑡) = 0$ for all $𝑡 \neq 0$ . By continuity, $𝑓 (0) =$ ${lim}_{𝑡 \to 0} 𝑓 (𝑡) =$ $0$ as well. So $𝑓 = 0$ .

(b) No. Positivity fails for $𝑓 (𝑡) = 1 - 𝑡$ :

⟨ 𝑓, 𝑓 ⟩ = \int_{- 1}^{1} 𝑡 {(1 - 𝑡)}^{2} d 𝑡 = \int_{- 1}^{1} (𝑡 - 2 𝑡^{2} + 𝑡^{3}) d 𝑡 = 0 - \frac{4}{3} + 0 = - \frac{4}{3} < 0 .

(c) No. Definiteness fails: $𝑓 (𝑡) =$ $max (0, 𝑡 - 1 / 2)$ is continuous and nonzero, but $𝑓 = 0$ on $[0, 1 / 2]$ , so $⟨ 𝑓, 𝑓 ⟩ = 0$ . (Positivity does hold.)

(d) Yes. $⟨ 𝑓, 𝑓 ⟩ =$ ${𝑓 (0)}^{2} +$ $\int_{0}^{1} 𝑓^{'} {(𝑡)}^{2} d 𝑡 \geq$ $0$ . If $⟨ 𝑓, 𝑓 ⟩ = 0$ , both nonnegative terms vanish. So $𝑓 (0) = 0$ , and Lemma 6.1.14 applied to the continuous function ${(𝑓^{'})}^{2}$ gives $𝑓^{'} = 0$ on $[0, 1]$ . By the mean value theorem $𝑓$ is constant, so $𝑓 = 𝑓 (0) = 0$ . Without the term $𝑓 (0) 𝑔 (0)$ , definiteness would fail for the constant function $1$ .

Exercise 6.1.44.routine In a complex inner product space, $‖ 𝐱 ‖ = 1$ , $‖ 𝐲 ‖ = 2$ , and $⟨ 𝐱, 𝐲 ⟩ = 1 - 𝑖$ . Compute

(a) $⟨ 2 𝐱 + 𝑖 𝐲, 𝐱 - 𝐲 ⟩$ ;
(b) ${‖ 𝐱 - 𝑖 𝐲 ‖}^{2}$ ;
(c) ${‖ (1 + 𝑖) 𝐱 + 𝐲 ‖}^{2}$ .
(d)Find $𝑐 \in ℂ$ with $⟨ 𝐱 - 𝑐 𝐲, 𝐲 ⟩ = 0$ , and compute ${‖ 𝐱 - 𝑐 𝐲 ‖}^{2}$ .

Solution.

Uses. Proposition 6.1.4 (Sesquilinearity), Proposition 6.1.29 (Expanding a norm)

Techniques. Expand by multilinearity (Technique 5.1.7)

Hint 1. Expand each inner product into the four values

⟨ 𝐱, 𝐱 ⟩ = 1

⟨ 𝐱, 𝐲 ⟩ = 1 - 𝑖

⟨ 𝐲, 𝐱 ⟩ = 1 + 𝑖

⟨ 𝐲, 𝐲 ⟩ = 4

. Scalars leave the second slot conjugated.

Hint 2. For (b) and (c), Proposition 6.1.29 saves work:

{‖ 𝐱 + 𝑎 𝐲 ‖}^{2} =

{‖ 𝐱 ‖}^{2} +

2 Re (𝑎 ⟨ 𝐱, 𝐲 ⟩) +

{| 𝑎 |}^{2} {‖ 𝐲 ‖}^{2}

Sketch. (a)

2 -

2 (1 - 𝑖) +

𝑖 (1 + 𝑖) -

4 𝑖 =

- 1 -

𝑖

. (b)

1 +

2 Re (𝑖 (1 - 𝑖)) +

4 =

7

. (c)

{| 1 + 𝑖 |}^{2} +

2 Re ((1 + 𝑖) (1 - 𝑖)) +

4 =

10

. (d)

𝑐 = (1 - 𝑖) / 4

and

{‖ 𝐱 - 𝑐 𝐲 ‖}^{2} =

1 -

1 +

1 / 2 =

1 / 2

We use $⟨ 𝐱, 𝐱 ⟩ = 1$ , $⟨ 𝐲, 𝐲 ⟩ = 4$ , $⟨ 𝐱, 𝐲 ⟩ = 1 - 𝑖$ , and $⟨ 𝐲, 𝐱 ⟩ = 1 - 𝑖 = 1 + 𝑖$ .

(a) Expanding,

\begin{aligned} ⟨ 2 𝐱 + 𝑖 𝐲, 𝐱 - 𝐲 ⟩ & = 2 ⟨ 𝐱, 𝐱 ⟩ - 2 ⟨ 𝐱, 𝐲 ⟩ + 𝑖 ⟨ 𝐲, 𝐱 ⟩ - 𝑖 ⟨ 𝐲, 𝐲 ⟩ \\ = 2 - (2 - 2 𝑖) + (𝑖 - 1) - 4 𝑖 = - 1 - 𝑖 . \end{aligned}

(b) By Proposition 6.1.29 with $𝑎 = - 𝑖$ , $𝑎 = 𝑖$ , and $| 𝑎 | = 1$ ,

{‖ 𝐱 - 𝑖 𝐲 ‖}^{2} = 1 + 2 Re (𝑖 (1 - 𝑖)) + 4 = 1 + 2 Re (1 + 𝑖) + 4 = 7 .

(c) Here the scalar sits on $𝐱$ , so we expand directly:

{‖ (1 + 𝑖) 𝐱 + 𝐲 ‖}^{2} = {| 1 + 𝑖 |}^{2} ⟨ 𝐱, 𝐱 ⟩ + (1 + 𝑖) ⟨ 𝐱, 𝐲 ⟩ + (1 + 𝑖) ⟨ 𝐲, 𝐱 ⟩ + ⟨ 𝐲, 𝐲 ⟩,

which is

2 + (1 + 𝑖) (1 - 𝑖) + (1 - 𝑖) (1 + 𝑖) + 4 = 2 + 2 + 2 + 4 = 10 .

(d) By linearity in the first slot, $⟨ 𝐱 - 𝑐 𝐲, 𝐲 ⟩ =$ $(1 - 𝑖) -$ $4 𝑐$ , which vanishes exactly for $𝑐 = (1 - 𝑖) / 4$ . Then $𝑐 ⟨ 𝐱, 𝐲 ⟩ =$ $(1 + 𝑖) (1 - 𝑖) / 4 =$ $1 / 2$ and ${| 𝑐 |}^{2} = 2 / 16 = 1 / 8$ , so by Proposition 6.1.29 with $𝑎 = - 𝑐$ ,

{‖ 𝐱 - 𝑐 𝐲 ‖}^{2} = 1 - 2 \cdot \frac{1}{2} + \frac{1}{8} \cdot 4 = \frac{1}{2} .

Exercise 6.1.45.standard

(a)An inner product on $ℝ^{2}$ satisfies $‖ 𝐞_{1} ‖ = 1$ , $‖ 𝐞_{2} ‖ = 2$ , and $‖ 𝐞_{1} + 𝐞_{2} ‖ = 2$ . Show that there is exactly one such inner product, and find a formula for $⟨ 𝐱, 𝐲 ⟩$ .
(b)Compute $‖ 𝐞_{1} - 𝐞_{2} ‖$ for the inner product of (a) in two ways: from your formula, and from the parallelogram law.
(c)Show that no inner product on $ℝ^{2}$ satisfies $‖ 𝐞_{1} ‖ = 1$ , $‖ 𝐞_{2} ‖ = 2$ , and $‖ 𝐞_{1} + 𝐞_{2} ‖ = 4$ .

Solution.

Uses. Proposition 6.1.22 (Inner products on $𝔽^{𝑛}$ ), Proposition 6.1.29 (Expanding a norm), Proposition 6.1.30 (Parallelogram law)

Techniques. Polarization (Technique 6.1.33)

Hint 1. By Proposition 6.1.22(a), an inner product on

ℝ^{2}

is determined by

⟨ 𝐞_{1}, 𝐞_{1} ⟩

⟨ 𝐞_{2}, 𝐞_{2} ⟩

, and

𝑐 = ⟨ 𝐞_{1}, 𝐞_{2} ⟩

. Find

𝑐

from

{‖ 𝐞_{1} + 𝐞_{2} ‖}^{2}

Hint 2. In (a), you must also check that the matrix you find gives an inner product: complete the square.

Hint 3. In (c), the same computation forces a value of

𝑐

. Find a vector

𝐱

with

⟨ 𝐱, 𝐱 ⟩ < 0

for that value.

Sketch. (a)

4 = 1 + 2 𝑐 + 4

forces

𝑐 = - 1 / 2

, so

⟨ 𝐱, 𝐲 ⟩ =

𝑥_{1} 𝑦_{1} -

(𝑥_{1} 𝑦_{2} + 𝑥_{2} 𝑦_{1}) / 2 +

4 𝑥_{2} 𝑦_{2}

, which is an inner product because

𝑥_{1}^{2} -

𝑥_{1} 𝑥_{2} +

4 𝑥_{2}^{2} =

{(𝑥_{1} - 𝑥_{2} / 2)}^{2} +

15 𝑥_{2}^{2} / 4

. (b) Both give

{‖ 𝐞_{1} - 𝐞_{2} ‖}^{2} = 6

. (c)

16 = 5 + 2 𝑐

forces

𝑐 = 11 / 2

, and then

(2, - 1)

would have

⟨ 𝐱, 𝐱 ⟩ =

4 -

22 +

4 <

0

(a) Suppose $⟨ \cdot, \cdot ⟩$ is such an inner product, and let $𝑐 =$ $⟨ 𝐞_{1}, 𝐞_{2} ⟩ =$ $⟨ 𝐞_{2}, 𝐞_{1} ⟩$ (it is real and symmetric). By Proposition 6.1.29, ${‖ 𝐞_{1} + 𝐞_{2} ‖}^{2} =$ $1 +$ $2 𝑐 +$ $4$ , so $4 = 5 + 2 𝑐$ and $𝑐 = - 1 / 2$ . By Proposition 6.1.22(a), $⟨ 𝐱, 𝐲 ⟩ = 𝐲^{𝖳} 𝑀 𝐱$ with $𝑀 = (\begin{matrix} 1 & - 1 / 2 \\ - 1 / 2 & 4 \end{matrix})$ , that is,

⟨ 𝐱, 𝐲 ⟩ = 𝑥_{1} 𝑦_{1} - \frac{1}{2} (𝑥_{1} 𝑦_{2} + 𝑥_{2} 𝑦_{1}) + 4 𝑥_{2} 𝑦_{2} .

So there is at most one such inner product. Conversely, $𝑀$ is symmetric and

𝐱^{𝖳} 𝑀 𝐱 = 𝑥_{1}^{2} - 𝑥_{1} 𝑥_{2} + 4 𝑥_{2}^{2} = {(𝑥_{1} - \frac{𝑥_{2}}{2})}^{2} + \frac{15}{4} 𝑥_{2}^{2},

which is positive unless $𝑥_{2} = 0$ and $𝑥_{1} = 𝑥_{2} / 2 = 0$ . By Proposition 6.1.22(b), this is an inner product, and it has ${‖ 𝐞_{1} ‖}^{2} = 1$ , ${‖ 𝐞_{2} ‖}^{2} = 4$ , and ${‖ 𝐞_{1} + 𝐞_{2} ‖}^{2} =$ $1 -$ $1 +$ $4 =$ $4$ , as required.

(b) From the formula, ${‖ 𝐞_{1} - 𝐞_{2} ‖}^{2} =$ $1 -$ $2 𝑐 +$ $4 =$ $1 +$ $1 +$ $4 =$ $6$ . From Proposition 6.1.30, ${‖ 𝐞_{1} - 𝐞_{2} ‖}^{2} =$ $2 \cdot$ $1 +$ $2 \cdot$ $4 -$ $4 =$ $6$ . So $‖ 𝐞_{1} - 𝐞_{2} ‖ = \sqrt{6}$ .

(c) Suppose such an inner product exists. The argument of (a) gives $16 = 5 + 2 𝑐$ , so $𝑐 = 11 / 2$ , and $⟨ 𝐱, 𝐱 ⟩ =$ $𝑥_{1}^{2} +$ $11 𝑥_{1} 𝑥_{2} +$ $4 𝑥_{2}^{2}$ for all $𝐱$ . For $𝐱 = (2, - 1)$ this is $4 - 22 + 4 = - 14 < 0$ , contradicting positivity. (In Section 6.2 this also follows from the triangle inequality, as $4 > 1 + 2$ .)

Exercise 6.1.46.standard For which $𝑎, 𝑏, 𝑐 \in ℝ$ is

⟨ 𝐱, 𝐲 ⟩ = 𝑎 𝑥_{1} 𝑦_{1} + 𝑏 (𝑥_{1} 𝑦_{2} + 𝑥_{2} 𝑦_{1}) + 𝑐 𝑥_{2} 𝑦_{2}

an inner product on $ℝ^{2}$ ?

Solution.

Uses. Proposition 6.1.22 (Inner products on $𝔽^{𝑛}$ )

Techniques. Plug in clever test vectors (Technique 6.1.40), A sum of nonnegative terms (Technique 6.1.9)

Hint 1. The formula is

𝐲^{𝖳} 𝑀 𝐱

for the symmetric matrix

𝑀 = (\begin{matrix} 𝑎 & 𝑏 \\ 𝑏 & 𝑐 \end{matrix})

, so by Proposition 6.1.22 the question is when

𝑞 (𝐱) =

𝑎 𝑥_{1}^{2} +

2 𝑏 𝑥_{1} 𝑥_{2} +

𝑐 𝑥_{2}^{2} >

0

for all

𝐱 \neq 𝟎

Hint 2. Necessary conditions come from test vectors: try

𝐞_{1}

, and then a vector that makes the cross term as harmful as possible, such as

(- 𝑏, 𝑎)

Hint 3. For sufficiency, when

𝑎 > 0

, complete the square in

𝑎 𝑞 (𝐱)

Sketch. Exactly when

𝑎 > 0

and

𝑎 𝑐 - 𝑏^{2} > 0

. Necessity:

𝑞 (𝐞_{1}) = 𝑎

and

𝑞 (- 𝑏, 𝑎) =

𝑎 (𝑎 𝑐 - 𝑏^{2})

. Sufficiency:

𝑎 𝑞 (𝐱) =

{(𝑎 𝑥_{1} + 𝑏 𝑥_{2})}^{2} +

(𝑎 𝑐 - 𝑏^{2}) 𝑥_{2}^{2}

The formula is $𝐲^{𝖳} 𝑀 𝐱$ with $𝑀 = (\begin{matrix} 𝑎 & 𝑏 \\ 𝑏 & 𝑐 \end{matrix})$ , which is symmetric. By Proposition 6.1.22(b), it is an inner product if and only if $𝑞 (𝐱) =$ $𝐱^{𝖳} 𝑀 𝐱 =$ $𝑎 𝑥_{1}^{2} +$ $2 𝑏 𝑥_{1} 𝑥_{2} +$ $𝑐 𝑥_{2}^{2}$ is positive for all $𝐱 \neq 𝟎$ . We show that this happens exactly when $𝑎 > 0$ and $𝑎 𝑐 - 𝑏^{2} > 0$ .

Suppose $𝑞 (𝐱) > 0$ for all $𝐱 \neq 𝟎$ . Then $𝑎 = 𝑞 (𝐞_{1}) > 0$ . The vector $(- 𝑏, 𝑎)$ is nonzero since $𝑎 \neq 0$ , and

𝑞 (- 𝑏, 𝑎) = 𝑎 𝑏^{2} - 2 𝑏 \cdot 𝑏 𝑎 + 𝑐 𝑎^{2} = 𝑎 (𝑎 𝑐 - 𝑏^{2}),

which is positive; dividing by $𝑎 > 0$ gives $𝑎 𝑐 - 𝑏^{2} > 0$ .

Conversely, suppose $𝑎 > 0$ and $𝑎 𝑐 - 𝑏^{2} > 0$ . Completing the square,

𝑎 𝑞 (𝐱) = 𝑎^{2} 𝑥_{1}^{2} + 2 𝑎 𝑏 𝑥_{1} 𝑥_{2} + 𝑎 𝑐 𝑥_{2}^{2} = {(𝑎 𝑥_{1} + 𝑏 𝑥_{2})}^{2} + (𝑎 𝑐 - 𝑏^{2}) 𝑥_{2}^{2} .

Both terms are nonnegative, so $𝑎 𝑞 (𝐱) \geq 0$ , and it is $0$ only if $𝑥_{2} = 0$ and $𝑎 𝑥_{1} + 𝑏 𝑥_{2} = 0$ , that is, $𝐱 = 𝟎$ . Since $𝑎 > 0$ , $𝑞 (𝐱) > 0$ for $𝐱 \neq 𝟎$ .

The number $𝑎 𝑐 - 𝑏^{2}$ is $det 𝑀$ ; Section 12.3 extends this test to $𝑛 \times 𝑛$ matrices.

Exercise 6.1.47.standard Let

𝑀 \in ℂ^{𝑛 \times 𝑛}

, and suppose

𝐱^{*} 𝑀 𝐱 \in ℝ

for every

𝐱 \in ℂ^{𝑛}

. Show that

𝑀^{*} = 𝑀

. Conclude that over

ℂ

, the condition

𝑀^{*} = 𝑀

in Proposition 6.1.22(b) is implied by the condition

𝐱^{*} 𝑀 𝐱 > 0

for

𝐱 \neq 𝟎

. (By Remark 6.1.24, this fails over

ℝ

Solution.

Uses. Notation 6.1.7 (Conjugate transpose), Notation 6.1.1 (Real and complex scalars)

Techniques. Plug in clever test vectors (Technique 6.1.40)

Hint 1. Compute

𝐱^{*} 𝑀 𝐱

for

𝐱 = 𝐞_{𝑗}

𝐞_{𝑗} + 𝐞_{𝑘}

, and

𝐞_{𝑗} + 𝑖 𝐞_{𝑘}

Hint 2. With

𝐱 = 𝐞_{𝑗} + 𝐞_{𝑘}

you learn that

𝑚_{𝑗 𝑘} + 𝑚_{𝑘 𝑗}

is real; with

𝐞_{𝑗} + 𝑖 𝐞_{𝑘}

, that

𝑖 (𝑚_{𝑗 𝑘} - 𝑚_{𝑘 𝑗})

is real. What do these say about real and imaginary parts?

Sketch.

𝐞_{𝑗}^{*} 𝑀 𝐞_{𝑗} =

𝑚_{𝑗 𝑗}

is real. For

𝑗 \neq 𝑘

𝐱 = 𝐞_{𝑗} + 𝐞_{𝑘}

gives

𝑚_{𝑗 𝑘} + 𝑚_{𝑘 𝑗} \in ℝ

, so

Im 𝑚_{𝑗 𝑘} =

- Im 𝑚_{𝑘 𝑗}

; and

𝐱 = 𝐞_{𝑗} + 𝑖 𝐞_{𝑘}

gives

𝑖 (𝑚_{𝑗 𝑘} - 𝑚_{𝑘 𝑗}) \in

ℝ

, so

Re 𝑚_{𝑗 𝑘} = Re 𝑚_{𝑘 𝑗}

. Together,

𝑚_{𝑗 𝑘} = 𝑚_{𝑘 𝑗}

For $𝐱 \in ℂ^{𝑛}$ , $𝐱^{*} 𝑀 𝐱 =$ $\sum_{𝑝, 𝑞} 𝑥_{𝑝} 𝑚_{𝑝 𝑞} 𝑥_{𝑞}$ . Fix $𝑗$ . For $𝐱 = 𝐞_{𝑗}$ this is $𝑚_{𝑗 𝑗}$ , so $𝑚_{𝑗 𝑗} \in ℝ$ , that is, $𝑚_{𝑗 𝑗} = 𝑚_{𝑗 𝑗}$ .

Now fix $𝑗 \neq 𝑘$ . For $𝐱 = 𝐞_{𝑗} + 𝐞_{𝑘}$ , only the terms with $𝑝, 𝑞 \in {𝑗, 𝑘}$ survive, and

𝐱^{*} 𝑀 𝐱 = 𝑚_{𝑗 𝑗} + 𝑚_{𝑗 𝑘} + 𝑚_{𝑘 𝑗} + 𝑚_{𝑘 𝑘} .

Since this and $𝑚_{𝑗 𝑗}, 𝑚_{𝑘 𝑘}$ are real, $𝑚_{𝑗 𝑘} + 𝑚_{𝑘 𝑗} \in ℝ$ , so $Im 𝑚_{𝑗 𝑘} =$ $- Im 𝑚_{𝑘 𝑗}$ . For $𝐱 = 𝐞_{𝑗} + 𝑖 𝐞_{𝑘}$ we have $𝑥_{𝑗} = 1$ and $𝑥_{𝑘} = 𝑖$ , so

𝐱^{*} 𝑀 𝐱 = 𝑚_{𝑗 𝑗} + 𝑚_{𝑗 𝑘} 𝑖 + 𝑖 𝑚_{𝑘 𝑗} + 𝑖 𝑚_{𝑘 𝑘} 𝑖 = 𝑚_{𝑗 𝑗} + 𝑖 (𝑚_{𝑗 𝑘} - 𝑚_{𝑘 𝑗}) + 𝑚_{𝑘 𝑘} .

So $𝑖 (𝑚_{𝑗 𝑘} - 𝑚_{𝑘 𝑗})$ is real, which means $𝑚_{𝑗 𝑘} - 𝑚_{𝑘 𝑗}$ has real part $0$ : $Re 𝑚_{𝑗 𝑘} = Re 𝑚_{𝑘 𝑗}$ . Together, $𝑚_{𝑗 𝑘} =$ $Re 𝑚_{𝑘 𝑗} -$ $𝑖 Im 𝑚_{𝑘 𝑗} =$ $𝑚_{𝑘 𝑗}$ . Hence the $(𝑘, 𝑗)$ entry of $𝑀^{*}$ , which is $𝑚_{𝑗 𝑘}$ , equals $𝑚_{𝑘 𝑗}$ , and $𝑀^{*} = 𝑀$ .

If $𝐱^{*} 𝑀 𝐱 > 0$ for every $𝐱 \neq 𝟎$ , then $𝐱^{*} 𝑀 𝐱$ is real for every $𝐱$ (it is $0$ at $𝟎$ ), so $𝑀^{*} = 𝑀$ by what we just proved.

Exercise 6.1.48.standard Let $𝐱, 𝐲, 𝐳$ be vectors in an inner product space, and let $𝐦 = (𝐱 + 𝐲) / 2$ . Prove Apollonius’s identity

{‖ 𝐳 - 𝐱 ‖}^{2} + {‖ 𝐳 - 𝐲 ‖}^{2} = \frac{1}{2} {‖ 𝐱 - 𝐲 ‖}^{2} + 2 {‖ 𝐳 - 𝐦 ‖}^{2} .

In $ℝ^{2}$ , what does it say about the triangle with vertices $𝐱, 𝐲, 𝐳$ ?

Solution.

Uses. Proposition 6.1.30 (Parallelogram law), Proposition 6.1.27 (Basic properties of the norm)

Hint 1. Write

𝐳 - 𝐱

and

𝐳 - 𝐲

𝐮 + 𝐰

and

𝐮 - 𝐰

for suitable

𝐮, 𝐰

Hint 2. Take

𝐮 = 𝐳 - 𝐦

and

𝐰 = (𝐲 - 𝐱) / 2

, then apply the parallelogram law.

Sketch. With

𝐮 = 𝐳 - 𝐦

and

𝐰 = (𝐲 - 𝐱) / 2

, we have

𝐳 - 𝐱 = 𝐮 + 𝐰

and

𝐳 - 𝐲 = 𝐮 - 𝐰

. The parallelogram law gives

2 {‖ 𝐮 ‖}^{2} + 2 {‖ 𝐰 ‖}^{2}

, and

2 {‖ 𝐰 ‖}^{2} = {‖ 𝐱 - 𝐲 ‖}^{2} / 2

Let $𝐮 = 𝐳 - 𝐦$ and $𝐰 = (𝐲 - 𝐱) / 2$ . Since $𝐦 -$ $𝐱 =$ $(𝐲 - 𝐱) / 2 =$ $𝐰$ and $𝐦 -$ $𝐲 =$ $(𝐱 - 𝐲) / 2 =$ $- 𝐰$ ,

𝐳 - 𝐱 = (𝐳 - 𝐦) + (𝐦 - 𝐱) = 𝐮 + 𝐰, 𝐳 - 𝐲 = 𝐮 - 𝐰 .

By Proposition 6.1.30 and Proposition 6.1.27(b),

{‖ 𝐳 - 𝐱 ‖}^{2} + {‖ 𝐳 - 𝐲 ‖}^{2} = 2 {‖ 𝐮 ‖}^{2} + 2 {‖ 𝐰 ‖}^{2} = 2 {‖ 𝐳 - 𝐦 ‖}^{2} + 2 \cdot \frac{1}{4} {‖ 𝐲 - 𝐱 ‖}^{2},

which is the identity, since $‖ 𝐲 - 𝐱 ‖ = ‖ 𝐱 - 𝐲 ‖$ .

In $ℝ^{2}$ , $𝐦$ is the midpoint of the side from $𝐱$ to $𝐲$ , and $‖ 𝐳 - 𝐦 ‖$ is the length of the median from $𝐳$ . The identity expresses the length of a median through the three side lengths:

{‖ 𝐳 - 𝐦 ‖}^{2} = \frac{1}{4} (2 {‖ 𝐳 - 𝐱 ‖}^{2} + 2 {‖ 𝐳 - 𝐲 ‖}^{2} - {‖ 𝐱 - 𝐲 ‖}^{2}) .

Exercise 6.1.49.standard Let $𝐱, 𝐲$ be vectors in an inner product space.

(a)Show that $‖ 𝐱 ‖ = ‖ 𝐲 ‖$ if and only if $Re ⟨ 𝐱 + 𝐲, 𝐱 - 𝐲 ⟩ = 0$ . Interpret the case $𝔽 = ℝ$ , $𝑉 = ℝ^{2}$ geometrically.
(b)Over $ℂ$ , show that $⟨ 𝐱 + 𝐲, 𝐱 - 𝐲 ⟩ = 0$ if and only if $‖ 𝐱 ‖ = ‖ 𝐲 ‖$ and $⟨ 𝐱, 𝐲 ⟩ \in ℝ$ .

Solution.

Uses. Proposition 6.1.4 (Sesquilinearity), Notation 6.1.1 (Real and complex scalars)

Hint 1. Expand

⟨ 𝐱 + 𝐲, 𝐱 - 𝐲 ⟩

into four terms.

Hint 2. The two cross terms are

- ⟨ 𝐱, 𝐲 ⟩

and

⟨ 𝐲, 𝐱 ⟩ = ⟨ 𝐱, 𝐲 ⟩

, whose sum is

- 2 𝑖 Im ⟨ 𝐱, 𝐲 ⟩

Sketch.

⟨ 𝐱 + 𝐲, 𝐱 - 𝐲 ⟩ =

({‖ 𝐱 ‖}^{2} - {‖ 𝐲 ‖}^{2}) -

2 𝑖 Im ⟨ 𝐱, 𝐲 ⟩

. Its real part is

{‖ 𝐱 ‖}^{2} - {‖ 𝐲 ‖}^{2}

, and it vanishes exactly when both the real part and

Im ⟨ 𝐱, 𝐲 ⟩

do. In

ℝ^{2}

: a parallelogram is a rhombus if and only if its diagonals are perpendicular.

Expanding with linearity in the first slot and Proposition 6.1.4,

⟨ 𝐱 + 𝐲, 𝐱 - 𝐲 ⟩ = {‖ 𝐱 ‖}^{2} - ⟨ 𝐱, 𝐲 ⟩ + ⟨ 𝐲, 𝐱 ⟩ - {‖ 𝐲 ‖}^{2} .

With $𝑧 = ⟨ 𝐱, 𝐲 ⟩$ , conjugate symmetry gives $- 𝑧 + 𝑧 = - 2 𝑖 Im 𝑧$ , so

⟨ 𝐱 + 𝐲, 𝐱 - 𝐲 ⟩ = ({‖ 𝐱 ‖}^{2} - {‖ 𝐲 ‖}^{2}) - 2 𝑖 Im ⟨ 𝐱, 𝐲 ⟩ .

(a) The real part of the right side is ${‖ 𝐱 ‖}^{2} - {‖ 𝐲 ‖}^{2}$ , which is $0$ if and only if $‖ 𝐱 ‖ = ‖ 𝐲 ‖$ (norms are nonnegative). In $ℝ^{2}$ with the dot product, $𝐱 + 𝐲$ and $𝐱 - 𝐲$ are the diagonals of the parallelogram on $𝐱$ and $𝐲$ , so (a) says that the parallelogram is a rhombus (all sides equal) if and only if its diagonals are perpendicular, provided $𝐱, 𝐲$ are linearly independent so that the parallelogram is not degenerate.

(b) A complex number is $0$ if and only if its real and imaginary parts are. So $⟨ 𝐱 + 𝐲, 𝐱 - 𝐲 ⟩ = 0$ if and only if $‖ 𝐱 ‖ = ‖ 𝐲 ‖$ and $Im ⟨ 𝐱, 𝐲 ⟩ = 0$ , that is, $⟨ 𝐱, 𝐲 ⟩ \in ℝ$ . For instance, $𝐱 = (1, 0)$ and $𝐲 = (0, 1)$ in $ℂ^{2}$ qualify, but $𝐱 = (1)$ and $𝐲 = (𝑖)$ in $ℂ^{1}$ do not: they have equal norms, while $⟨ 𝐱 + 𝐲, 𝐱 - 𝐲 ⟩ =$ $(1 + 𝑖) (1 - 𝑖) =$ ${(1 + 𝑖)}^{2} =$ $2 𝑖$ .

Exercise 6.1.50.routine Let

𝑉

be an inner product space, and let

𝐯_{1}, \dots, 𝐯_{𝑘}

span

𝑉

. Show that if

𝐱, 𝐲 \in 𝑉

satisfy

⟨ 𝐱, 𝐯_{𝑗} ⟩ =

⟨ 𝐲, 𝐯_{𝑗} ⟩

for

𝑗 = 1, \dots, 𝑘

, then

𝐱 = 𝐲

. Give an example showing that the spanning hypothesis cannot be dropped.

Solution.

Uses. Proposition 6.1.37 (Inner products distinguish vectors), Proposition 6.1.4 (Sesquilinearity), Definition 1.3.10 (Span)

Techniques. Show a vector is zero via its norm (Technique 6.1.36)

Hint 1. By Proposition 6.1.37, it suffices to show

⟨ 𝐱 - 𝐲, 𝐰 ⟩ = 0

for every

𝐰 \in 𝑉

Hint 2. Write

𝐰 = \sum_{𝑗} 𝑐_{𝑗} 𝐯_{𝑗}

and expand the second slot with Proposition 6.1.4.

Sketch. Every

𝐰 = \sum_{𝑗} 𝑐_{𝑗} 𝐯_{𝑗}

, so

⟨ 𝐱 - 𝐲, 𝐰 ⟩ =

\sum_{𝑗} 𝑐_{𝑗} ⟨ 𝐱 - 𝐲, 𝐯_{𝑗} ⟩ =

0

, and Proposition 6.1.37 gives

𝐱 = 𝐲

. In

ℝ^{2}

with the list

𝐞_{1}

, the vectors

𝐞_{2}

and

𝟎

have the same inner product with

𝐞_{1}

Let $𝐮 = 𝐱 - 𝐲$ , so that $⟨ 𝐮, 𝐯_{𝑗} ⟩ = 0$ for every $𝑗$ . Let $𝐰 \in 𝑉$ . Since the list spans $𝑉$ , $𝐰 = \sum_{𝑗} 𝑐_{𝑗} 𝐯_{𝑗}$ for some scalars $𝑐_{𝑗}$ , and by Proposition 6.1.4,

⟨ 𝐮, 𝐰 ⟩ = \sum_{𝑗 = 1}^{𝑘} 𝑐_{𝑗} ⟨ 𝐮, 𝐯_{𝑗} ⟩ = 0 = ⟨ 𝟎, 𝐰 ⟩ .

By Proposition 6.1.37, $𝐮 = 𝟎$ , that is, $𝐱 = 𝐲$ .

Without the spanning hypothesis the conclusion fails: in $ℝ^{2}$ with the list consisting of $𝐞_{1}$ alone, $⟨ 𝐞_{2}, 𝐞_{1} ⟩ =$ $0 =$ $⟨ 𝟎, 𝐞_{1} ⟩$ , but $𝐞_{2} \neq 𝟎$ .

Exercise 6.1.51.standard Let

𝑉, 𝑊

be inner product spaces over the same

𝔽

, and let

𝑇 : 𝑉 \to 𝑊

be linear with

‖ 𝑇 𝐯 ‖ = ‖ 𝐯 ‖

for all

𝐯 \in 𝑉

. Show that

⟨ 𝑇 𝐯, 𝑇 𝐰 ⟩ =

⟨ 𝐯, 𝐰 ⟩

for all

𝐯, 𝐰 \in 𝑉

, and that

𝑇

is injective. (Maps of this kind are studied in Section 8.4.)

Solution.

Uses. Proposition 6.1.32 (Polarization identities), Proposition 6.1.27 (Basic properties of the norm), Theorem 2.6.11 (Injectivity is decided at $𝟎$ )

Techniques. Polarization (Technique 6.1.33)

Hint 1. Express

⟨ 𝑇 𝐯, 𝑇 𝐰 ⟩

through norms with Proposition 6.1.32.

Hint 2. Each norm that appears is

‖ 𝑇 𝐯 +

𝑐 𝑇 𝐰 ‖ =

‖ 𝑇 (𝐯 + 𝑐 𝐰) ‖

. For injectivity, look at the kernel.

Sketch. By polarization in

𝑊

and linearity,

⟨ 𝑇 𝐯, 𝑇 𝐰 ⟩

is a combination of the numbers

{‖ 𝑇 (𝐯 + 𝑐 𝐰) ‖}^{2} =

{‖ 𝐯 + 𝑐 𝐰 ‖}^{2}

, which is the same combination that polarization in

𝑉

gives for

⟨ 𝐯, 𝐰 ⟩

. If

𝑇 𝐯 = 𝟎

, then

‖ 𝐯 ‖ = ‖ 𝑇 𝐯 ‖ = 0

Suppose $𝔽 = ℂ$ ; the real case is the same with $𝑐$ ranging over $\pm 1$ and the coefficients of Proposition 6.1.32(a). For $𝑐 \in ℂ$ , linearity gives $𝑇 𝐯 +$ $𝑐 𝑇 𝐰 =$ $𝑇 (𝐯 + 𝑐 𝐰)$ , so by hypothesis $‖ 𝑇 𝐯 +$ $𝑐 𝑇 𝐰 ‖ =$ $‖ 𝐯 +$ $𝑐 𝐰 ‖$ . By Proposition 6.1.32 in $𝑊$ and then in $𝑉$ ,

⟨ 𝑇 𝐯, 𝑇 𝐰 ⟩ = \frac{1}{4} \sum_{𝑘 = 0}^{3} 𝑖^{𝑘} {‖ 𝑇 𝐯 + 𝑖^{𝑘} 𝑇 𝐰 ‖}^{2} = \frac{1}{4} \sum_{𝑘 = 0}^{3} 𝑖^{𝑘} {‖ 𝐯 + 𝑖^{𝑘} 𝐰 ‖}^{2} = ⟨ 𝐯, 𝐰 ⟩ .

For injectivity, if $𝑇 𝐯 = 𝟎$ , then $‖ 𝐯 ‖ = ‖ 𝑇 𝐯 ‖ = 0$ , so $𝐯 = 𝟎$ by Proposition 6.1.27. So $ker 𝑇 = {𝟎}$ , and $𝑇$ is injective by Theorem 2.6.11.

Exercise 6.1.52.standard Let $𝑉$ be a real inner product space and $𝑇 \in L (𝑉)$ .

(a)Prove or disprove: if $⟨ 𝑇 𝐯, 𝐯 ⟩ = 0$ for all $𝐯 \in 𝑉$ , then $𝑇 = 0$ .
(b)Show that $⟨ 𝑇 𝐯, 𝐯 ⟩ = 0$ for all $𝐯 \in 𝑉$ if and only if $⟨ 𝑇 𝐯, 𝐰 ⟩ =$ $- ⟨ 𝐯, 𝑇 𝐰 ⟩$ for all $𝐯, 𝐰 \in 𝑉$ .
(c)Suppose $⟨ 𝑇 𝐯, 𝐰 ⟩ =$ $⟨ 𝐯, 𝑇 𝐰 ⟩$ for all $𝐯, 𝐰 \in 𝑉$ . Show that if $⟨ 𝑇 𝐯, 𝐯 ⟩ = 0$ for all $𝐯$ , then $𝑇 = 0$ .

Solution.

Uses. Example 6.1.39 (A rotation), Corollary 6.1.38 (Linear maps are determined by inner products), Proposition 6.1.4 (Sesquilinearity)

Techniques. Plug in clever test vectors (Technique 6.1.40)

Hint 1. (a) Look at Example 6.1.39.

Hint 2. (b) Apply the hypothesis to

𝐯 + 𝐰

, and use that a real inner product is symmetric.

Hint 3. (c) Combine (b) with the hypothesis to get

⟨ 𝑇 𝐯, 𝐰 ⟩ =

- ⟨ 𝑇 𝐯, 𝐰 ⟩

Sketch. (a) False: the rotation by

𝜋 / 2

ℝ^{2}

. (b) Expanding

⟨ 𝑇 (𝐯 + 𝐰), 𝐯 + 𝐰 ⟩ =

0

gives

⟨ 𝑇 𝐯, 𝐰 ⟩ +

⟨ 𝑇 𝐰, 𝐯 ⟩ =

0

, and

⟨ 𝑇 𝐰, 𝐯 ⟩ =

⟨ 𝐯, 𝑇 𝐰 ⟩

by symmetry; conversely put

𝐰 = 𝐯

. (c) Then

⟨ 𝑇 𝐯, 𝐰 ⟩ =

- ⟨ 𝑇 𝐯, 𝐰 ⟩

, so it is

0

and

𝑇 = 0

(a) False. The rotation $𝑇 𝐱 = (- 𝑥_{2}, 𝑥_{1})$ of $ℝ^{2}$ satisfies $⟨ 𝑇 𝐱, 𝐱 ⟩ = 0$ for every $𝐱$ but is not $0$ (Example 6.1.39).

(b) Suppose $⟨ 𝑇 𝐯, 𝐯 ⟩ = 0$ for all $𝐯$ . For $𝐯, 𝐰 \in 𝑉$ , expanding as in the proof of Proposition 6.1.41,

0 = ⟨ 𝑇 (𝐯 + 𝐰), 𝐯 + 𝐰 ⟩ = ⟨ 𝑇 𝐯, 𝐰 ⟩ + ⟨ 𝑇 𝐰, 𝐯 ⟩ .

A real inner product is symmetric, so $⟨ 𝑇 𝐰, 𝐯 ⟩ =$ $⟨ 𝐯, 𝑇 𝐰 ⟩$ , and hence $⟨ 𝑇 𝐯, 𝐰 ⟩ =$ $- ⟨ 𝐯, 𝑇 𝐰 ⟩$ . Conversely, if this holds for all $𝐯, 𝐰$ , then taking $𝐰 = 𝐯$ gives $⟨ 𝑇 𝐯, 𝐯 ⟩ =$ $- ⟨ 𝐯, 𝑇 𝐯 ⟩ =$ $- ⟨ 𝑇 𝐯, 𝐯 ⟩$ , so $⟨ 𝑇 𝐯, 𝐯 ⟩ = 0$ .

(c) By (b) and the hypothesis on $𝑇$ , $⟨ 𝑇 𝐯, 𝐰 ⟩ =$ $- ⟨ 𝐯, 𝑇 𝐰 ⟩ =$ $- ⟨ 𝑇 𝐯, 𝐰 ⟩$ for all $𝐯, 𝐰$ . So $⟨ 𝑇 𝐯, 𝐰 ⟩ =$ $0 =$ $⟨ 0 𝐯, 𝐰 ⟩$ , and $𝑇 = 0$ by Corollary 6.1.38.

Exercise 6.1.53.hard Let

𝑉

be a complex inner product space and

𝑇 \in L (𝑉)

. Show that

⟨ 𝑇 𝐯, 𝐯 ⟩ \in ℝ

for all

𝐯 \in 𝑉

if and only if

⟨ 𝑇 𝐯, 𝐰 ⟩ =

⟨ 𝐯, 𝑇 𝐰 ⟩

for all

𝐯, 𝐰 \in 𝑉

. (Such operators are called self-adjoint; see Section 6.6.)

Solution.

Uses. Proposition 6.1.41 (Over $ℂ$ , the values $⟨ 𝑇 𝐯, 𝐯 ⟩$ determine $𝑇$ ), Proposition 6.1.4 (Sesquilinearity), Notation 6.1.1 (Real and complex scalars)

Techniques. Polarization (Technique 6.1.33)

Hint 1. One direction is a one-line computation with conjugate symmetry.

Hint 2. For the other, consider

𝐵 (𝐯, 𝐰) =

⟨ 𝑇 𝐯, 𝐰 ⟩ -

⟨ 𝐯, 𝑇 𝐰 ⟩

. How does

𝐵

treat scalars in each slot, and what is

𝐵 (𝐯, 𝐯)

Hint 3.

𝐵

is linear in the first slot, conjugate-linear in the second, and

𝐵 (𝐯, 𝐯) =

2 𝑖 Im ⟨ 𝑇 𝐯, 𝐯 ⟩

. Use (6.1.4).

Sketch. If

𝑇

satisfies the identity, then

⟨ 𝑇 𝐯, 𝐯 ⟩ =

⟨ 𝐯, 𝑇 𝐯 ⟩ =

⟨ 𝑇 𝐯, 𝐯 ⟩

is real. Conversely,

𝐵 (𝐯, 𝐰) =

⟨ 𝑇 𝐯, 𝐰 ⟩ -

⟨ 𝐯, 𝑇 𝐰 ⟩

is linear in

𝐯

and conjugate-linear in

𝐰

, and

𝐵 (𝐯, 𝐯) =

2 𝑖 Im ⟨ 𝑇 𝐯, 𝐯 ⟩ =

0

. By (6.1.4),

𝐵 = 0

Suppose $⟨ 𝑇 𝐯, 𝐰 ⟩ =$ $⟨ 𝐯, 𝑇 𝐰 ⟩$ for all $𝐯, 𝐰$ . Then $⟨ 𝑇 𝐯, 𝐯 ⟩ =$ $⟨ 𝐯, 𝑇 𝐯 ⟩ =$ $⟨ 𝑇 𝐯, 𝐯 ⟩$ by conjugate symmetry, so $⟨ 𝑇 𝐯, 𝐯 ⟩$ is real.

Conversely, suppose $⟨ 𝑇 𝐯, 𝐯 ⟩ \in ℝ$ for all $𝐯$ , and define

𝐵 (𝐯, 𝐰) = ⟨ 𝑇 𝐯, 𝐰 ⟩ - ⟨ 𝐯, 𝑇 𝐰 ⟩ .

The first term is linear in $𝐯$ and conjugate-linear in $𝐰$ , as in the second proof of Proposition 6.1.41. So is the second term: it is linear in $𝐯$ by linearity in the first slot, and by Proposition 6.1.4,

⟨ 𝐯, 𝑇 (𝑎 𝐰 + 𝑏 𝐳) ⟩ = ⟨ 𝐯, 𝑎 𝑇 𝐰 + 𝑏 𝑇 𝐳 ⟩ = 𝑎 ⟨ 𝐯, 𝑇 𝐰 ⟩ + 𝑏 ⟨ 𝐯, 𝑇 𝐳 ⟩ .

Hence $𝐵$ is linear in the first slot and conjugate-linear in the second. Its diagonal values are

𝐵 (𝐯, 𝐯) = ⟨ 𝑇 𝐯, 𝐯 ⟩ - ⟨ 𝑇 𝐯, 𝐯 ⟩ = 2 𝑖 Im ⟨ 𝑇 𝐯, 𝐯 ⟩ = 0 .

The identity (6.1.4) holds for every such $𝐵$ , as its derivation used only how $𝐵$ treats scalars in each slot. So $𝐵 (𝐯, 𝐰) =$ $\frac{1}{4} \sum_{𝑘} 𝑖^{𝑘} 𝐵 (𝐯 + 𝑖^{𝑘} 𝐰, 𝐯 + 𝑖^{𝑘} 𝐰) =$ $0$ for all $𝐯, 𝐰$ , which is the claim.

Exercise 6.1.54.hard Let

𝐴 \in ℂ^{𝑛 \times 𝑛}

satisfy

𝐴^{*} = 𝐴

and

𝐴^{𝑘} = 0

for some

𝑘 \geq 1

. Show that

𝐴 = 0

. (Start by proving

{(𝐵 𝐶)}^{*} = 𝐶^{*} 𝐵^{*}

for square matrices

𝐵, 𝐶

Solution.

Uses. Corollary 6.1.19 ( $𝐴^{*} 𝐴 = 0$ forces $𝐴 = 0$ ), Proposition 2.4.23 (Transpose of a product), Definition 2.4.1 (Matrix product)

Techniques. The trace of $𝐴^{*} 𝐴$ (Technique 6.1.18)

Hint 1. Show

{(𝐵 𝐶)}^{*} = 𝐶^{*} 𝐵^{*}

entrywise, and deduce

{(𝐴^{𝑚})}^{*} = 𝐴^{𝑚}

for every

𝑚 \geq 1

Hint 2. If

𝐴^{2 𝑚} = 0

, then

{(𝐴^{𝑚})}^{*} 𝐴^{𝑚} =

𝐴^{2 𝑚} =

0

. What does Corollary 6.1.19 say?

Hint 3. Choose

𝑗

with

2^{𝑗} \geq 𝑘

and halve the exponent

𝑗

times.

Sketch.

{(𝐴^{𝑚})}^{*} =

{(𝐴^{*})}^{𝑚} =

𝐴^{𝑚}

, so

𝐴^{2 𝑚} = 0

gives

{(𝐴^{𝑚})}^{*} 𝐴^{𝑚} = 0

and

𝐴^{𝑚} = 0

. Starting from

𝐴^{2^{𝑗}} = 0

with

2^{𝑗} \geq 𝑘

, halving

𝑗

times gives

𝐴 = 0

First, for $𝐵, 𝐶 \in ℂ^{𝑛 \times 𝑛}$ , the $(𝑖, 𝑗)$ entry of ${(𝐵 𝐶)}^{*}$ is

{(𝐵 𝐶)}_{𝑗 𝑖} = \sum_{𝑙} 𝑏_{𝑗 𝑙} 𝑐_{𝑙 𝑖} = \sum_{𝑙} 𝑐_{𝑙 𝑖} 𝑏_{𝑗 𝑙} = \sum_{𝑙} {(𝐶^{*})}_{𝑖 𝑙} {(𝐵^{*})}_{𝑙 𝑗},

which is the $(𝑖, 𝑗)$ entry of $𝐶^{*} 𝐵^{*}$ (Definition 2.4.1). So ${(𝐵 𝐶)}^{*} = 𝐶^{*} 𝐵^{*}$ . By induction on $𝑚$ , using $𝐴^{*} = 𝐴$ ,

{(𝐴^{𝑚})}^{*} = {(𝐴^{𝑚 - 1} 𝐴)}^{*} = 𝐴^{*} {(𝐴^{𝑚 - 1})}^{*} = 𝐴 𝐴^{𝑚 - 1} = 𝐴^{𝑚} for all 𝑚 \geq 1 .

Next, if $𝑚 \geq 1$ and $𝐴^{2 𝑚} = 0$ , then ${(𝐴^{𝑚})}^{*} 𝐴^{𝑚} =$ $𝐴^{𝑚} 𝐴^{𝑚} =$ $𝐴^{2 𝑚} =$ $0$ , so $𝐴^{𝑚} = 0$ by Corollary 6.1.19.

Now choose $𝑗 \geq 0$ with $2^{𝑗} \geq 𝑘$ . Then $𝐴^{2^{𝑗}} =$ $𝐴^{2^{𝑗} - 𝑘} 𝐴^{𝑘} =$ $0$ (with $𝐴^{0} = 𝐼$ ). If $𝑗 = 0$ , this says $𝐴 = 0$ . Otherwise, applying the previous paragraph with $𝑚 =$ $2^{𝑗 - 1},$ $2^{𝑗 - 2},$ $\dots,$ $1$ in turn gives $𝐴^{2^{𝑗 - 1}} = 0$ , then $𝐴^{2^{𝑗 - 2}} = 0$ , and finally $𝐴^{1} = 0$ .

Exercise 6.1.55.routine Let $𝐴 \in ℝ^{𝑛 \times 𝑛}$ be skew-symmetric, $𝐴^{𝖳} = - 𝐴$ .

(a)Show that $tr (𝐴^{2}) \leq 0$ , with equality only if $𝐴 = 0$ .
(b)Deduce that no real skew-symmetric matrix satisfies $𝐴^{2} = 𝐼_{𝑛}$ . Show that $𝐴^{2} = - 𝐼_{2}$ is possible.

Solution.

Uses. Proposition 6.1.16 (Frobenius inner product)

Techniques. The trace of $𝐴^{*} 𝐴$ (Technique 6.1.18)

Hint 1. Rewrite

𝐴^{2}

using

𝐴 = - 𝐴^{𝖳}

Hint 2.

𝐴^{2} = - 𝐴^{𝖳} 𝐴

, and

tr (𝐴^{𝖳} 𝐴)

is a sum of squares.

Sketch.

tr (𝐴^{2}) =

- tr (𝐴^{𝖳} 𝐴) =

- \sum_{𝑖, 𝑗} 𝑎_{𝑖 𝑗}^{2} \leq

0

, with equality only for

𝐴 = 0

. If

𝐴^{2} = 𝐼_{𝑛}

, then

tr (𝐴^{2}) = 𝑛 > 0

. The rotation

𝐽 = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})

has

𝐽^{2} = - 𝐼_{2}

(a) Since $𝐴 = - 𝐴^{𝖳}$ , $𝐴^{2} =$ $(- 𝐴^{𝖳}) 𝐴 =$ $- 𝐴^{𝖳} 𝐴$ . By Proposition 6.1.16 (over $ℝ$ , $𝐴^{*} = 𝐴^{𝖳}$ ),

tr (𝐴^{2}) = - tr (𝐴^{𝖳} 𝐴) = - \sum_{𝑖, 𝑗} 𝑎_{𝑖 𝑗}^{2} \leq 0,

with equality only if every $𝑎_{𝑖 𝑗} = 0$ .

(b) If $𝐴^{2} = 𝐼_{𝑛}$ , then $tr (𝐴^{2}) = 𝑛 > 0$ , contradicting (a). On the other hand, $𝐽 = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ is skew-symmetric with $𝐽^{2} = - 𝐼_{2}$ , and indeed $tr (𝐽^{2}) =$ $- 2 =$ $- \sum_{𝑖, 𝑗} 𝑗_{𝑖 𝑗}^{2}$ .

Exercise 6.1.56.standard For $𝐮, 𝐱 \in 𝔽^{𝑚}$ and $𝐯, 𝐲 \in 𝔽^{𝑛}$ , the products $𝐮 𝐯^{*}$ and $𝐱 𝐲^{*}$ are $𝑚 \times 𝑛$ matrices. Show that, with the Frobenius inner product on $𝔽^{𝑚 \times 𝑛}$ and dot products on $𝔽^{𝑚}$ and $𝔽^{𝑛}$ ,

⟨ 𝐮 𝐯^{*}, 𝐱 𝐲^{*} ⟩ = (𝐮 \cdot 𝐱) (𝐲 \cdot 𝐯),

and deduce that $‖ 𝐮 𝐯^{*} ‖ = ‖ 𝐮 ‖ ‖ 𝐯 ‖$ .

Solution.

Uses. Proposition 6.1.16 (Frobenius inner product), Definition 6.1.8 (Dot product), Definition 2.4.1 (Matrix product)

Hint 1. The

(𝑖, 𝑗)

entry of

𝐮 𝐯^{*}

𝑢_{𝑖} 𝑣_{𝑗}

Hint 2. Use the entrywise formula of Proposition 6.1.16; the double sum factors into a product of two single sums.

Sketch. By Proposition 6.1.16 the inner product is

\sum_{𝑖, 𝑗} 𝑢_{𝑖} 𝑣_{𝑗} 𝑥_{𝑖} 𝑦_{𝑗}

, and this double sum factors as

(\sum_{𝑖} 𝑢_{𝑖} 𝑥_{𝑖}) (\sum_{𝑗} 𝑦_{𝑗} 𝑣_{𝑗})

. With

𝐱 = 𝐮

and

𝐲 = 𝐯

this is

{‖ 𝐮 ‖}^{2} {‖ 𝐯 ‖}^{2}

By Definition 2.4.1, the $(𝑖, 𝑗)$ entry of $𝐮 𝐯^{*}$ is $𝑢_{𝑖} 𝑣_{𝑗}$ , and that of $𝐱 𝐲^{*}$ is $𝑥_{𝑖} 𝑦_{𝑗}$ . By Proposition 6.1.16,

⟨ 𝐮 𝐯^{*}, 𝐱 𝐲^{*} ⟩ = \sum_{𝑖 = 1}^{𝑚} \sum_{𝑗 = 1}^{𝑛} 𝑢_{𝑖} 𝑣_{𝑗} 𝑥_{𝑖} 𝑦_{𝑗} = \sum_{𝑖, 𝑗} 𝑢_{𝑖} 𝑥_{𝑖} 𝑦_{𝑗} 𝑣_{𝑗} = (\sum_{𝑖} 𝑢_{𝑖} 𝑥_{𝑖}) (\sum_{𝑗} 𝑦_{𝑗} 𝑣_{𝑗}),

which is $(𝐮 \cdot 𝐱) (𝐲 \cdot 𝐯)$ . Taking $𝐱 = 𝐮$ and $𝐲 = 𝐯$ gives ${‖ 𝐮 𝐯^{*} ‖}^{2} =$ $(𝐮 \cdot 𝐮) (𝐯 \cdot 𝐯) =$ ${‖ 𝐮 ‖}^{2} {‖ 𝐯 ‖}^{2}$ , and taking nonnegative square roots, $‖ 𝐮 𝐯^{*} ‖ = ‖ 𝐮 ‖ ‖ 𝐯 ‖$ .

Exercise 6.1.57.hard Let $𝐱_{1}, \dots, 𝐱_{𝑛}$ be unit vectors in an inner product space.

(a)Show that
$\sum_{1 \leq 𝑖 < 𝑗 \leq 𝑛} {‖ 𝐱_{𝑖} - 𝐱_{𝑗} ‖}^{2} = 𝑛^{2} - {‖ 𝐱_{1} + \dots + 𝐱_{𝑛} ‖}^{2} .$
(b)Deduce that $\sum_{𝑖 < 𝑗} {‖ 𝐱_{𝑖} - 𝐱_{𝑗} ‖}^{2} \leq$ $𝑛^{2}$ , with equality if and only if $𝐱_{1} + \dots + 𝐱_{𝑛} = 𝟎$ .
(c)For three unit vectors in $ℝ^{2}$ , find the largest possible value of $\sum_{𝑖 < 𝑗} {‖ 𝐱_{𝑖} - 𝐱_{𝑗} ‖}^{2}$ and a configuration attaining it.

Solution.

Uses. Proposition 6.1.29 (Expanding a norm), Proposition 6.1.4 (Sesquilinearity)

Techniques. Expand by multilinearity (Technique 5.1.7)

Hint 1. Sum over all ordered pairs

(𝑖, 𝑗)

instead; the terms with

𝑖 = 𝑗

are

0

, and each unordered pair is counted twice.

Hint 2. Expand

{‖ 𝐱_{𝑖} - 𝐱_{𝑗} ‖}^{2} =

2 -

2 Re ⟨ 𝐱_{𝑖}, 𝐱_{𝑗} ⟩

, and recognize

\sum_{𝑖, 𝑗} ⟨ 𝐱_{𝑖}, 𝐱_{𝑗} ⟩

⟨ 𝐬, 𝐬 ⟩

for

𝐬 = \sum_{𝑖} 𝐱_{𝑖}

Sketch. Each term is

2 - 2 Re ⟨ 𝐱_{𝑖}, 𝐱_{𝑗} ⟩

, and

\sum_{𝑖, 𝑗} ⟨ 𝐱_{𝑖}, 𝐱_{𝑗} ⟩ =

{‖ 𝐬 ‖}^{2}

, so the sum over all ordered pairs is

2 𝑛^{2} - 2 {‖ 𝐬 ‖}^{2}

; the sum over

𝑖 < 𝑗

is half of this. (b) is immediate, and in (c) the maximum

9

is attained by three unit vectors at angles

2 𝜋 / 3

, which sum to

𝟎

(a) Let $𝐬 = 𝐱_{1} + \dots + 𝐱_{𝑛}$ . By sesquilinearity, $\sum_{𝑖, 𝑗} ⟨ 𝐱_{𝑖}, 𝐱_{𝑗} ⟩ =$ $⟨ 𝐬, 𝐬 ⟩ =$ ${‖ 𝐬 ‖}^{2}$ , the sum running over all ordered pairs. By Proposition 6.1.29 and $‖ 𝐱_{𝑖} ‖ = 1$ , ${‖ 𝐱_{𝑖} - 𝐱_{𝑗} ‖}^{2} =$ $2 -$ $2 Re ⟨ 𝐱_{𝑖}, 𝐱_{𝑗} ⟩$ . Summing over all ordered pairs,

\sum_{𝑖, 𝑗} {‖ 𝐱_{𝑖} - 𝐱_{𝑗} ‖}^{2} = 2 𝑛^{2} - 2 Re \sum_{𝑖, 𝑗} ⟨ 𝐱_{𝑖}, 𝐱_{𝑗} ⟩ = 2 𝑛^{2} - 2 {‖ 𝐬 ‖}^{2} .

The terms with $𝑖 = 𝑗$ vanish, and each pair $𝑖 < 𝑗$ occurs twice, as $(𝑖, 𝑗)$ and $(𝑗, 𝑖)$ , with the same value. So the sum over $𝑖 < 𝑗$ is half of this, $𝑛^{2} - {‖ 𝐬 ‖}^{2}$ .

(b) Since ${‖ 𝐬 ‖}^{2} \geq 0$ , the sum is at most $𝑛^{2}$ , with equality if and only if $‖ 𝐬 ‖ = 0$ , that is, $𝐬 = 𝟎$ by Proposition 6.1.27.

(c) By (b) the sum is at most $9$ , with equality exactly when $𝐱_{1} + 𝐱_{2} + 𝐱_{3} = 𝟎$ . This happens for $𝐱_{1} = (1, 0)$ , $𝐱_{2} = (- 1 / 2, \sqrt{3} / 2)$ , and $𝐱_{3} = (- 1 / 2, - \sqrt{3} / 2)$ , the vertices of an equilateral triangle inscribed in the unit circle. Each side has squared length ${(3 / 2)}^{2} + {(\sqrt{3} / 2)}^{2} = 3$ , and $3 + 3 + 3 = 9$ .

Exercise 6.1.58.hard This exercise shows that for $𝑛 \geq 3$ , the formula $\sum_{𝑘} 𝑥_{𝑘} 𝑦_{𝑘}$ on $𝔽_{𝑝}^{𝑛}$ is never definite. (For $𝑛 = 1$ it is, since $𝑥^{2} = 0$ forces $𝑥 = 0$ .)

(a)Find a nonzero $𝐱 \in 𝔽_{5}^{2}$ with $𝑥_{1}^{2} + 𝑥_{2}^{2} = 0$ .
(b)Show that for every prime $𝑝$ there is a nonzero $𝐱 \in 𝔽_{𝑝}^{3}$ with $𝑥_{1}^{2} + 𝑥_{2}^{2} + 𝑥_{3}^{2} = 0$ , and deduce the same in $𝔽_{𝑝}^{𝑛}$ for every $𝑛 \geq 3$ .

Solution.

Uses. Proposition 1.1.24 ( $ℤ / 𝑛 ℤ$ is a field if and only if $𝑛$ is prime), Proposition 1.1.17 (No zero divisors)

Hint 1. (a) List the squares in

𝔽_{5}

Hint 2. (b) For odd

𝑝

, it suffices to find

𝑎, 𝑏 \in 𝔽_{𝑝}

with

𝑎^{2} + 𝑏^{2} = - 1

, and then take

𝐱 = (𝑎, 𝑏, 1)

. Compare the sets

{𝑎^{2}}

and

{- 1 - 𝑏^{2}}

Hint 3. Count:

𝑎^{2} = 𝑐^{2}

if and only if

𝑎 = \pm 𝑐

, so there are exactly

(𝑝 + 1) / 2

squares in

𝔽_{𝑝}

. Two subsets of

𝔽_{𝑝}

of that size must meet.

Sketch. (a)

(1, 2)

, since

1 + 4 = 5 = 0

. (b) For

𝑝 = 2

take

(1, 1, 0)

. For odd

𝑝

, the set

𝑆

of squares has

(𝑝 + 1) / 2

elements, and so does

- 1 - 𝑆

; since

(𝑝 + 1) / 2 +

(𝑝 + 1) / 2 >

𝑝

, they share an element

𝑎^{2} = - 1 - 𝑏^{2}

, and

(𝑎, 𝑏, 1)

works.

(a) The squares in $𝔽_{5}$ are $0, 1, 4, 4, 1$ (of $0, 1, 2, 3, 4$ ). Since $1 + 4 = 5 = 0$ , the vector $𝐱 = (1, 2)$ works.

(b) For $𝑝 = 2$ , $𝐱 = (1, 1, 0)$ works, since $1 + 1 = 0$ in $𝔽_{2}$ . Let $𝑝$ be odd. It suffices to find $𝑎, 𝑏 \in 𝔽_{𝑝}$ with $𝑎^{2} + 𝑏^{2} + 1 = 0$ , because then $𝐱 = (𝑎, 𝑏, 1) \neq 𝟎$ works.

Let $𝑆 = {𝑎^{2} | 𝑎 \in 𝔽_{𝑝}}$ . For $𝑎, 𝑐 \in 𝔽_{𝑝}$ , $𝑎^{2} = 𝑐^{2}$ if and only if $(𝑎 - 𝑐) (𝑎 + 𝑐) = 0$ , which by Proposition 1.1.17 means $𝑎 = 𝑐$ or $𝑎 = - 𝑐$ . For $𝑎 \neq 0$ we have $𝑎 \neq - 𝑎$ , since $𝑎 = - 𝑎$ would give $2 𝑎 = 0$ , and $2 \neq 0$ in $𝔽_{𝑝}$ for odd $𝑝$ . So the $𝑝 - 1$ nonzero elements fall into $(𝑝 - 1) / 2$ pairs ${𝑎, - 𝑎}$ , each pair having one square, and distinct pairs having distinct squares. Together with $0^{2} = 0$ , $𝑆$ has exactly $(𝑝 + 1) / 2$ elements.

The map $𝑠 \mapsto - 1 - 𝑠$ is a bijection of $𝔽_{𝑝}$ , so $𝑇 =$ ${- 1 - 𝑏^{2} | 𝑏 \in 𝔽_{𝑝}}$ also has $(𝑝 + 1) / 2$ elements. Since $| 𝑆 | +$ $| 𝑇 | =$ $𝑝 +$ $1 >$ $𝑝 =$ $| 𝔽_{𝑝} |$ , the sets $𝑆$ and $𝑇$ intersect: $𝑎^{2} = - 1 - 𝑏^{2}$ for some $𝑎, 𝑏 \in 𝔽_{𝑝}$ . Then $𝑎^{2} + 𝑏^{2} + 1 = 0$ , as desired. For $𝑛 > 3$ , pad with zeros: $(𝑥_{1}, 𝑥_{2}, 𝑥_{3}, 0, \dots, 0)$ is nonzero with $\sum_{𝑘} 𝑥_{𝑘}^{2} = 0$ .

Exercise 6.1.59.standard Find the error in the following argument, and decide whether the claim is true.

Claim. If $𝐵 : ℝ^{𝑛} \times ℝ^{𝑛} \to ℝ$ is linear in each slot and $𝐵 (𝐯, 𝐯) = 0$ for all $𝐯 \in ℝ^{𝑛}$ , then $𝐵 = 0$ .

Proof. By the polarization identity, for all $𝐱, 𝐲$ ,

𝐵 (𝐱, 𝐲) = \frac{1}{4} (𝐵 (𝐱 + 𝐲, 𝐱 + 𝐲) - 𝐵 (𝐱 - 𝐲, 𝐱 - 𝐲)) = \frac{1}{4} (0 - 0) = 0 .

Solution.

Uses. Proposition 6.1.32 (Polarization identities), Proposition 5.1.10 (Alternating implies antisymmetric)

Techniques. Polarization (Technique 6.1.33)

Hint 1. Which properties of an inner product did the proof of Proposition 6.1.32(a) use? Does

𝐵

have them?

Hint 2. Expand

𝐵 (𝐱 + 𝐲, 𝐱 + 𝐲) -

𝐵 (𝐱 - 𝐲, 𝐱 - 𝐲)

for a

𝐵

that is only bilinear.

Hint 3. Look for a counterexample among

2 \times 2

determinants.

Sketch. The real polarization identity needs symmetry: for a bilinear

𝐵

, the right side equals

(𝐵 (𝐱, 𝐲) + 𝐵 (𝐲, 𝐱)) / 2

. The claim is false:

𝐵 (𝐱, 𝐲) =

𝑥_{1} 𝑦_{2} -

𝑥_{2} 𝑦_{1}

vanishes on the diagonal but

𝐵 (𝐞_{1}, 𝐞_{2}) = 1

. What is true is

𝐵 (𝐲, 𝐱) =

- 𝐵 (𝐱, 𝐲)

The error is the first equality. The real polarization identity was derived for inner products, and its proof used symmetry: the two cross terms of ${‖ 𝐱 + 𝐲 ‖}^{2}$ were combined into $2 ⟨ 𝐱, 𝐲 ⟩$ . For a bilinear $𝐵$ without symmetry, expanding gives

𝐵 (𝐱 + 𝐲, 𝐱 + 𝐲) - 𝐵 (𝐱 - 𝐲, 𝐱 - 𝐲) = 2 𝐵 (𝐱, 𝐲) + 2 𝐵 (𝐲, 𝐱),

so the right side of the displayed formula is $(𝐵 (𝐱, 𝐲) + 𝐵 (𝐲, 𝐱)) / 2$ , not $𝐵 (𝐱, 𝐲)$ .

The claim is false. On $ℝ^{2}$ , $𝐵 (𝐱, 𝐲) =$ $𝑥_{1} 𝑦_{2} -$ $𝑥_{2} 𝑦_{1} =$ $det (\begin{matrix} 𝑥_{1} & 𝑦_{1} \\ 𝑥_{2} & 𝑦_{2} \end{matrix})$ is linear in each slot and $𝐵 (𝐯, 𝐯) = 0$ for every $𝐯$ , but $𝐵 (𝐞_{1}, 𝐞_{2}) = 1$ . What the argument does prove, correctly read, is that $𝐵 (𝐱, 𝐲) +$ $𝐵 (𝐲, 𝐱) =$ $0$ : a bilinear function that vanishes on the diagonal is antisymmetric. This is the argument of Proposition 5.1.10, which applies to any bilinear function of two vectors (Remark 5.1.6).

Exercise 6.1.60.standard Let $𝑅 [0, 1]$ be the real vector space of Riemann-integrable functions $[0, 1] \to ℝ$ . Find the error in the following argument, and decide whether the claim is true.

Claim. $⟨ 𝑓, 𝑔 ⟩ =$ $\int_{0}^{1} 𝑓 (𝑡) 𝑔 (𝑡) d 𝑡$ is an inner product on $𝑅 [0, 1]$ .

Proof. Products of integrable functions are integrable, so $⟨ 𝑓, 𝑔 ⟩$ is defined. Symmetry and linearity follow from the properties of the integral, and $⟨ 𝑓, 𝑓 ⟩ = \int_{0}^{1} 𝑓^{2} \geq 0$ since $𝑓^{2} \geq 0$ . Finally, if $\int_{0}^{1} 𝑓^{2} = 0$ , then since $𝑓^{2} \geq 0$ we must have $𝑓^{2} = 0$ , so $𝑓 = 0$ .

Solution.

Uses. Lemma 6.1.14 (Nonnegative functions with zero integral), Proposition 6.1.13 (Integral inner products)

Hint 1. Which hypothesis of Lemma 6.1.14 is missing for an integrable

𝑓

Hint 2. Find a nonzero integrable function whose square has integral

0

. Change a continuous function at one point.

Sketch. The step “

\int 𝑓^{2} = 0

and

𝑓^{2} \geq 0

imply

𝑓^{2} = 0

” needs continuity (Lemma 6.1.14). The function equal to

1

𝑡 = 1 / 2

and

0

elsewhere is integrable, nonzero, and has

\int 𝑓^{2} = 0

. So the claim is false: definiteness fails on

𝑅 [0, 1]

The error is in the last step. A nonnegative function with integral $0$ must vanish if it is continuous (Lemma 6.1.14), but not in general. Let $𝑓 (1 / 2) = 1$ and $𝑓 (𝑡) = 0$ for $𝑡 \neq 1 / 2$ . Then $𝑓$ is Riemann-integrable and $𝑓^{2} = 𝑓$ , and $\int_{0}^{1} 𝑓^{2} = 0$ , since every Riemann sum over a partition of mesh $𝛿$ is at most $2 𝛿$ (the point $1 / 2$ lies in at most two subintervals). But $𝑓 \neq 0$ .

So the claim is false: the formula is symmetric, bilinear, and positive on $𝑅 [0, 1]$ , but not definite. This is why Proposition 6.1.13 is stated for continuous functions.

Exercise 6.1.61.challengeBeyond 21-242 (Jordan–von Neumann, real case.) Let $𝑉$ be a real vector space with a function $‖ \cdot ‖ : 𝑉 \to ℝ$ satisfying, for all $𝐯, 𝐰 \in 𝑉$ and $𝑎 \in ℝ$ : $‖ 𝐯 ‖ \geq 0$ with equality only for $𝐯 = 𝟎$ ; $‖ 𝑎 𝐯 ‖ = | 𝑎 | ‖ 𝐯 ‖$ ; $‖ 𝐯 + 𝐰 ‖ \leq ‖ 𝐯 ‖ + ‖ 𝐰 ‖$ ; and the parallelogram law ${‖ 𝐯 + 𝐰 ‖}^{2} +$ ${‖ 𝐯 - 𝐰 ‖}^{2} =$ $2 {‖ 𝐯 ‖}^{2} +$ $2 {‖ 𝐰 ‖}^{2}$ . Define

⟨ 𝐱, 𝐲 ⟩ = \frac{1}{4} ({‖ 𝐱 + 𝐲 ‖}^{2} - {‖ 𝐱 - 𝐲 ‖}^{2}) .

(a)Show that $⟨ 𝐱, 𝐲 ⟩ = ⟨ 𝐲, 𝐱 ⟩$ , $⟨ 𝐱, 𝐱 ⟩ = {‖ 𝐱 ‖}^{2}$ , and $⟨ - 𝐱, 𝐲 ⟩ = - ⟨ 𝐱, 𝐲 ⟩$ .
(b)Show that $⟨ 𝐱 + 𝐳, 𝐲 ⟩ +$ $⟨ 𝐱 - 𝐳, 𝐲 ⟩ =$ $2 ⟨ 𝐱, 𝐲 ⟩$ for all $𝐱, 𝐲, 𝐳$ .
(c)Deduce that $⟨ 𝐮 + 𝐰, 𝐲 ⟩ =$ $⟨ 𝐮, 𝐲 ⟩ +$ $⟨ 𝐰, 𝐲 ⟩$ for all $𝐮, 𝐰, 𝐲$ .
(d)Show that $⟨ 𝑡 𝐱, 𝐲 ⟩ =$ $𝑡 ⟨ 𝐱, 𝐲 ⟩$ for all $𝑡 \in ℝ$ , and conclude that $⟨ \cdot, \cdot ⟩$ is an inner product on $𝑉$ whose norm is $‖ \cdot ‖$ .

Solution.Beyond 21-242

Uses. Proposition 6.1.32 (Polarization identities), Proposition 6.1.30 (Parallelogram law), Exercise 2.1.51

Hint 1. (a) uses only

‖ - 𝐯 ‖ = ‖ 𝐯 ‖

and

‖ 2 𝐱 ‖ = 2 ‖ 𝐱 ‖

Hint 2. (b) Apply the parallelogram law to the pairs

(𝐱 + 𝐲, 𝐳)

and

(𝐱 - 𝐲, 𝐳)

, and subtract.

Hint 3. (c) Put

𝐳 = 𝐱

in (b) to get

⟨ 2 𝐱, 𝐲 ⟩ = 2 ⟨ 𝐱, 𝐲 ⟩

. Then choose

𝐱

and

𝐳

with

𝐱 + 𝐳 = 𝐮

and

𝐱 - 𝐳 = 𝐰

Hint 4. (d) Additivity gives homogeneity for integers, then for rationals. For real

𝑡

, show

𝑡 \mapsto ⟨ 𝑡 𝐱, 𝐲 ⟩

is continuous, using

| ‖ 𝐚 ‖ -

‖ 𝐛 ‖ | \leq

‖ 𝐚 -

𝐛 ‖

Sketch. (a) is immediate. (b) Subtracting the parallelogram laws for

(𝐱 + 𝐲, 𝐳)

and

(𝐱 - 𝐲, 𝐳)

gives

4 ⟨ 𝐱 + 𝐳, 𝐲 ⟩ +

4 ⟨ 𝐱 - 𝐳, 𝐲 ⟩ =

8 ⟨ 𝐱, 𝐲 ⟩

. (c)

𝐳 = 𝐱

gives

⟨ 2 𝐱, 𝐲 ⟩ = 2 ⟨ 𝐱, 𝐲 ⟩

, and then

𝐱 = (𝐮 + 𝐰) / 2

𝐳 = (𝐮 - 𝐰) / 2

give additivity. (d)

𝑓 (𝑡) = ⟨ 𝑡 𝐱, 𝐲 ⟩

is additive and continuous, so

𝑓 (𝑡) = 𝑡 𝑓 (1)

. Symmetry, linearity in the first slot, and

⟨ 𝐱, 𝐱 ⟩ = {‖ 𝐱 ‖}^{2}

are the inner product axioms.

(a) Since $‖ -$ $𝐯 ‖ =$ $| -$ $1 | ‖ 𝐯 ‖ =$ $‖ 𝐯 ‖$ , we have $‖ 𝐲 - 𝐱 ‖ = ‖ 𝐱 - 𝐲 ‖$ , which gives symmetry, and $‖ - 𝐱 - 𝐲 ‖ = ‖ 𝐱 + 𝐲 ‖$ and $‖ - 𝐱 + 𝐲 ‖ = ‖ 𝐱 - 𝐲 ‖$ , which give $⟨ - 𝐱, 𝐲 ⟩ = - ⟨ 𝐱, 𝐲 ⟩$ . Also $⟨ 𝐱, 𝐱 ⟩ =$ $({‖ 2 𝐱 ‖}^{2} - {‖ 𝟎 ‖}^{2}) / 4 =$ ${‖ 𝐱 ‖}^{2}$ .

(b) The parallelogram law for the pairs $(𝐱 + 𝐲, 𝐳)$ and $(𝐱 - 𝐲, 𝐳)$ reads

\begin{aligned} {‖ 𝐱 + 𝐲 + 𝐳 ‖}^{2} + {‖ 𝐱 + 𝐲 - 𝐳 ‖}^{2} & = 2 {‖ 𝐱 + 𝐲 ‖}^{2} + 2 {‖ 𝐳 ‖}^{2}, \\ {‖ 𝐱 - 𝐲 + 𝐳 ‖}^{2} + {‖ 𝐱 - 𝐲 - 𝐳 ‖}^{2} & = 2 {‖ 𝐱 - 𝐲 ‖}^{2} + 2 {‖ 𝐳 ‖}^{2} . \end{aligned}

Subtract the second from the first and regroup the left side:

\begin{aligned} ({‖ (𝐱 + 𝐳) + 𝐲 ‖}^{2} - {‖ (𝐱 + 𝐳) - 𝐲 ‖}^{2}) + ({‖ (𝐱 - 𝐳) + 𝐲 ‖}^{2} - {‖ (𝐱 - 𝐳) - 𝐲 ‖}^{2}) \\ = 2 ({‖ 𝐱 + 𝐲 ‖}^{2} - {‖ 𝐱 - 𝐲 ‖}^{2}) . \end{aligned}

By the definition of $⟨ \cdot, \cdot ⟩$ , this says $4 ⟨ 𝐱 + 𝐳, 𝐲 ⟩ +$ $4 ⟨ 𝐱 - 𝐳, 𝐲 ⟩ =$ $8 ⟨ 𝐱, 𝐲 ⟩$ .

(c) First, $⟨ 𝟎, 𝐲 ⟩ =$ $({‖ 𝐲 ‖}^{2} - {‖ - 𝐲 ‖}^{2}) / 4 =$ $0$ . Taking $𝐳 = 𝐱$ in (b) gives $⟨ 2 𝐱, 𝐲 ⟩ +$ $⟨ 𝟎, 𝐲 ⟩ =$ $2 ⟨ 𝐱, 𝐲 ⟩$ , so $⟨ 2 𝐱, 𝐲 ⟩ = 2 ⟨ 𝐱, 𝐲 ⟩$ for all $𝐱, 𝐲$ . Now let $𝐮, 𝐰 \in 𝑉$ , and apply (b) with $𝐱 = (𝐮 + 𝐰) / 2$ and $𝐳 = (𝐮 - 𝐰) / 2$ , so that $𝐱 + 𝐳 = 𝐮$ and $𝐱 - 𝐳 = 𝐰$ :

⟨ 𝐮, 𝐲 ⟩ + ⟨ 𝐰, 𝐲 ⟩ = 2 ⟨ \frac{𝐮 + 𝐰}{2}, 𝐲 ⟩ = ⟨ 𝐮 + 𝐰, 𝐲 ⟩,

the last step by the doubling rule applied to $(𝐮 + 𝐰) / 2$ .

(d) Fix $𝐱, 𝐲$ and let $𝑓 (𝑡) = ⟨ 𝑡 𝐱, 𝐲 ⟩$ for $𝑡 \in ℝ$ . By (c), $𝑓 (𝑠 + 𝑡) =$ $⟨ 𝑠 𝐱 + 𝑡 𝐱, 𝐲 ⟩ =$ $𝑓 (𝑠) +$ $𝑓 (𝑡)$ , so $𝑓$ is additive. We show that $𝑓$ is continuous. The triangle inequality gives $‖ 𝐚 ‖ \leq ‖ 𝐚 - 𝐛 ‖ + ‖ 𝐛 ‖$ and the same with $𝐚, 𝐛$ exchanged, so $| ‖ 𝐚 ‖ -$ $‖ 𝐛 ‖ | \leq$ $‖ 𝐚 -$ $𝐛 ‖$ . Hence

| ‖ 𝑡 𝐱 \pm 𝐲 ‖ - ‖ 𝑠 𝐱 \pm 𝐲 ‖ | \leq ‖ (𝑡 - 𝑠) 𝐱 ‖ = | 𝑡 - 𝑠 | ‖ 𝐱 ‖,

so $𝑡 \mapsto ‖ 𝑡 𝐱 \pm 𝐲 ‖$ is continuous, and so is $𝑓 (𝑡) =$ $({‖ 𝑡 𝐱 + 𝐲 ‖}^{2} - {‖ 𝑡 𝐱 - 𝐲 ‖}^{2}) / 4$ . A continuous additive function $ℝ \to ℝ$ satisfies $𝑓 (𝑡) = 𝑡 𝑓 (1)$ for all $𝑡$ (Exercise 2.1.51). Directly: additivity gives $𝑓 (𝑚 𝑡) = 𝑚 𝑓 (𝑡)$ for positive integers $𝑚$ by induction, $𝑓 (0) = 0$ , and $𝑓 (- 𝑡) = - 𝑓 (𝑡)$ by (a); so for $𝑞 = 𝑚 / 𝑛$ with $𝑛 > 0$ , $𝑛 𝑓 (𝑞) =$ $𝑓 (𝑚) =$ $𝑚 𝑓 (1)$ , that is, $𝑓 (𝑞) = 𝑞 𝑓 (1)$ . Every real $𝑡$ is a limit of rationals $𝑞_{𝑘}$ , and continuity gives $𝑓 (𝑡) =$ $lim 𝑓 (𝑞_{𝑘}) =$ $lim 𝑞_{𝑘} 𝑓 (1) =$ $𝑡 𝑓 (1)$ . So $⟨ 𝑡 𝐱, 𝐲 ⟩ =$ $𝑡 ⟨ 𝐱, 𝐲 ⟩$ .

Together with (c), $⟨ \cdot, \cdot ⟩$ is linear in the first slot. It is symmetric by (a), which over $ℝ$ is conjugate symmetry, and $⟨ 𝐱, 𝐱 ⟩ = {‖ 𝐱 ‖}^{2}$ is nonnegative and vanishes only for $𝐱 = 𝟎$ . So $⟨ \cdot, \cdot ⟩$ is an inner product, and its norm $\sqrt{⟨ 𝐱, 𝐱 ⟩}$ is $‖ 𝐱 ‖$ .

More practice

The practice bank has 41 exercises on this section’s topics.