1.2 Vector spaces · Matrix Theory

Section 1.1 made precise what we ask of scalars. This section does the same for vectors. In $ℝ^{2}$ we can add two vectors and scale a vector by a real number, and these operations obey a short list of rules. A vector space is a set with an addition and a scalar multiplication that obey the same rules, with scalars taken from any field. Tuples, matrices, functions, and polynomials all form vector spaces, and a fact proved from the rules alone holds in each of them.

1.2.1The axioms

The definition lists the rules that addition and scaling obey in $ℝ^{2}$ , and nothing more. We number them (A1)–(A8) and cite them by number.

Definition 1.2.1 (Vector space). Let $𝐹$ be a field. A vector space over $𝐹$ is a set $𝑉$ together with two operations, vector addition $+ : 𝑉 \times 𝑉 \to 𝑉$ and scalar multiplication $\cdot : 𝐹 \times 𝑉 \to 𝑉$ , that satisfy the following vector space axioms. We write $𝑎 𝐯$ for $𝑎 \cdot 𝐯$ .

(A1) Commutativity. For all $𝐯, 𝐰 \in 𝑉$ , $𝐯 + 𝐰 = 𝐰 + 𝐯$ .
(A2) Associativity. For all $𝐮, 𝐯, 𝐰 \in 𝑉$ , $(𝐮 + 𝐯) +$ $𝐰 =$ $𝐮 +$ $(𝐯 + 𝐰)$ .
(A3) Additive identity. There is an element $𝟎 \in 𝑉$ such that $𝐯 + 𝟎 = 𝐯$ for all $𝐯 \in 𝑉$ .
(A4) Additive inverses. For every $𝐯 \in 𝑉$ there is some $𝐰 \in 𝑉$ with $𝐯 + 𝐰 = 𝟎$ .
(A5) Multiplicative identity. For all $𝐯 \in 𝑉$ , $1 𝐯 = 𝐯$ , where $1$ is the multiplicative identity of $𝐹$ .
(A6) Associativity of scaling. For all $𝑎, 𝑏 \in 𝐹$ and $𝐯 \in 𝑉$ , $(𝑎 𝑏) 𝐯 = 𝑎 (𝑏 𝐯)$ .
(A7) Distributivity over vector addition. For all $𝑎 \in 𝐹$ and $𝐯, 𝐰 \in 𝑉$ , $𝑎 (𝐯 + 𝐰) =$ $𝑎 𝐯 +$ $𝑎 𝐰$ .
(A8) Distributivity over scalar addition. For all $𝑎, 𝑏 \in 𝐹$ and $𝐯 \in 𝑉$ , $(𝑎 + 𝑏) 𝐯 =$ $𝑎 𝐯 +$ $𝑏 𝐯$ .

The elements of $𝑉$ are called vectors, and $𝟎$ is called the zero vector. We also say that $𝑉$ is an $𝐹$ -vector space. A vector space over $ℝ$ is a real vector space, and a vector space over $ℂ$ is a complex vector space.

Remark 1.2.2 (Reading the axioms). Closure is part of the definition. Since $+$ and $\cdot$ are functions into $𝑉$ , sums and scalar multiples of vectors lie in $𝑉$ . When we test whether a candidate is a vector space, closure comes first: if a sum leaves the set, there is no addition on the set whose axioms could be checked (Non-example 1.2.21).

The symbols are overloaded. In (A8), the $+$ on the left adds scalars and the $+$ on the right adds vectors. In (A6), $𝑎 𝑏$ is a product in $𝐹$ , while the other two products are scalar multiplications. Bold type separates the scalar $0 \in 𝐹$ from the vector $𝟎 \in 𝑉$ . No axiom multiplies two vectors or adds a scalar to a vector, because neither operation exists.

Quantifier order matters. (A3) has the form “there is $𝟎$ such that for all $𝐯$ ”: one vector serves every $𝐯$ at once. The statement with the quantifiers swapped, that each $𝐯$ has some $𝐳$ with $𝐯 + 𝐳 = 𝐯$ , is weaker; part (c) of Example 1.2.25 satisfies it and fails (A3). In (A4) the order is “for every $𝐯$ there is $𝐰$ ”, so the inverse $𝐰$ may depend on $𝐯$ . Also, (A4) mentions $𝟎$ , the vector promised by (A3). If two different vectors satisfied (A3), then (A4) would be ambiguous. Proposition 1.2.27 rules this out, and its proof uses only (A1) and (A3), so there is no circularity.

The field is part of the data. The same set with the same addition can be a vector space over one field and not over another (Example 1.2.20). When the field matters, we say explicitly that $𝑉$ is a vector space over $𝐹$ .

1.2.2Examples

To show that a candidate is a vector space, we specify the two operations, check closure, and check the eight axioms. In the standard examples below, the operations act one coordinate (or one input) at a time, and each axiom reduces to the field axiom of the same name.

Example 1.2.3 (The zero space). Let

𝑉 = {𝟎}

be a set with one element, with

𝟎 + 𝟎 = 𝟎

and

𝑎 𝟎 = 𝟎

for all

𝑎 \in 𝐹

. Each axiom is either an equation between elements of

𝑉

, which holds because

𝑉

has only one element, or asks for an element of

𝑉

, which

𝟎

supplies. So

{𝟎}

is a vector space over every field, called the zero space. By (A3), every vector space has at least one element, so the zero space is as small as a vector space can be.

Tuples and matrices

Definition 1.2.4 (The space $𝐹^{𝑛}$ ). Let $𝑛 \geq 1$ . The set $𝐹^{𝑛}$ consists of all columns

𝐱 = (\begin{matrix} 𝑥_{1} \\ ⋮ \\ 𝑥_{𝑛} \end{matrix})

with $𝑥_{1}, \dots, 𝑥_{𝑛} \in 𝐹$ , called the coordinates of $𝐱$ . Two columns are equal when all their coordinates are equal. The operations are coordinatewise:

(\begin{matrix} 𝑥_{1} \\ ⋮ \\ 𝑥_{𝑛} \end{matrix}) + (\begin{matrix} 𝑦_{1} \\ ⋮ \\ 𝑦_{𝑛} \end{matrix}) = (\begin{matrix} 𝑥_{1} + 𝑦_{1} \\ ⋮ \\ 𝑥_{𝑛} + 𝑦_{𝑛} \end{matrix}), 𝑎 (\begin{matrix} 𝑥_{1} \\ ⋮ \\ 𝑥_{𝑛} \end{matrix}) = (\begin{matrix} 𝑎 𝑥_{1} \\ ⋮ \\ 𝑎 𝑥_{𝑛} \end{matrix}) .

In running text we write $(𝑥_{1}, \dots, 𝑥_{𝑛})$ for the column with coordinates $𝑥_{1}, \dots, 𝑥_{𝑛}$ .

Proposition 1.2.5 (

𝐹^{𝑛}

is a vector space).

𝐹^{𝑛}

is a vector space over

𝐹

. The column

(0, \dots, 0)

is a zero vector, and

(- 𝑥_{1}, \dots, - 𝑥_{𝑛})

is an additive inverse of

(𝑥_{1}, \dots, 𝑥_{𝑛})

Proof.routine

Uses. Definition 1.2.4 (The space $𝐹^{𝑛}$ ), Definition 1.1.1 (Field)

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. Two columns are equal exactly when their

𝑖

-th coordinates are equal for every

𝑖

. So compute the

𝑖

-th coordinate of each side of each axiom.

Hint 2. Coordinatewise, each axiom becomes a field axiom. For example, the

𝑖

-th coordinate of (A7) reads

𝑎 (𝑥_{𝑖} + 𝑦_{𝑖}) =

𝑎 𝑥_{𝑖} +

𝑎 𝑦_{𝑖}

Sketch. Sums and multiples of columns with entries in

𝐹

have entries in

𝐹

. Comparing

𝑖

-th coordinates, each of (A1)–(A8) becomes the field axiom of the same name applied to

𝑥_{𝑖}, 𝑦_{𝑖}, 𝑧_{𝑖}

, with

(0, \dots, 0)

and

(- 𝑥_{1}, \dots, - 𝑥_{𝑛})

supplying (A3) and (A4).

The coordinates of $𝐱 + 𝐲$ and $𝑎 𝐱$ are sums and products in $𝐹$ , so they lie in $𝐹$ , and the operations map into $𝐹^{𝑛}$ . Let $𝐱 = (𝑥_{1}, \dots, 𝑥_{𝑛})$ , $𝐲 = (𝑦_{1}, \dots, 𝑦_{𝑛})$ , and $𝐳 = (𝑧_{1}, \dots, 𝑧_{𝑛})$ be in $𝐹^{𝑛}$ , let $𝑎, 𝑏 \in 𝐹$ , and fix $𝑖 \in {1, \dots, 𝑛}$ . Two columns are equal when their $𝑖$ -th coordinates agree for every $𝑖$ . By the definition of the operations, the $𝑖$ -th coordinates of the two sides of each axiom are as follows, and they agree by the field axiom named on the right:

\begin{aligned} (A1) & 𝑥_{𝑖} + 𝑦_{𝑖} = 𝑦_{𝑖} + 𝑥_{𝑖} & commutativity, \\ (A2) & (𝑥_{𝑖} + 𝑦_{𝑖}) + 𝑧_{𝑖} = 𝑥_{𝑖} + (𝑦_{𝑖} + 𝑧_{𝑖}) & associativity, \\ (A3) & 𝑥_{𝑖} + 0 = 𝑥_{𝑖} & additive identity, \\ (A4) & 𝑥_{𝑖} + (- 𝑥_{𝑖}) = 0 & additive inverses, \\ (A5) & 1 𝑥_{𝑖} = 𝑥_{𝑖} & multiplicative identity, \\ (A6) & (𝑎 𝑏) 𝑥_{𝑖} = 𝑎 (𝑏 𝑥_{𝑖}) & associativity, \\ (A7) & 𝑎 (𝑥_{𝑖} + 𝑦_{𝑖}) = 𝑎 𝑥_{𝑖} + 𝑎 𝑦_{𝑖} & distributivity, \\ (A8) & (𝑎 + 𝑏) 𝑥_{𝑖} = 𝑎 𝑥_{𝑖} + 𝑏 𝑥_{𝑖} & distributivity . \end{aligned}

In (A3) we compare $𝐱 + (0, \dots, 0)$ with $𝐱$ , and in (A4) we compare $𝐱 + (- 𝑥_{1}, \dots, - 𝑥_{𝑛})$ with $(0, \dots, 0)$ . So all eight axioms hold. □

Example 1.2.6 (Small fields). $𝔽_{2}^{3}$ has $2^{3} = 8$ elements, and in it $(1, 1, 0) +$ $(0, 1, 1) =$ $(1, 0, 1)$ , because $1 + 1 = 0$ in $𝔽_{2}$ . For the same reason, $𝐯 + 𝐯 = 𝟎$ for every $𝐯 \in 𝔽_{2}^{𝑛}$ . In $ℂ^{2}$ , $𝑖 (1, 𝑖) = (𝑖, - 1)$ and $(1 + 𝑖) (1, - 𝑖) =$ $(1 + 𝑖, 1 - 𝑖)$ .

For $𝑛 = 1$ , $𝐹^{1}$ is $𝐹$ with its own addition and multiplication. So every field is a vector space over itself.

Definition 1.2.7 (Matrices). Let $𝑚, 𝑛 \geq 1$ . An $𝑚 \times 𝑛$ matrix over $𝐹$ is a rectangular array

𝐴 = (\begin{matrix} 𝑎_{11} & \dots & 𝑎_{1 𝑛} \\ ⋮ & ⋱ & ⋮ \\ 𝑎_{𝑚 1} & \dots & 𝑎_{𝑚 𝑛} \end{matrix})

of elements $𝑎_{𝑖 𝑗} \in 𝐹$ , the entries of $𝐴$ . The entry $𝑎_{𝑖 𝑗}$ sits in row $𝑖$ and column $𝑗$ , and we also write $𝐴 = (𝑎_{𝑖 𝑗})$ . Two matrices are equal when all their entries are equal. The set of $𝑚 \times 𝑛$ matrices over $𝐹$ is denoted $𝐹^{𝑚 \times 𝑛}$ , and its operations are entrywise:

(𝑎_{𝑖 𝑗}) + (𝑏_{𝑖 𝑗}) = (𝑎_{𝑖 𝑗} + 𝑏_{𝑖 𝑗}), 𝑐 (𝑎_{𝑖 𝑗}) = (𝑐 𝑎_{𝑖 𝑗}) .

A matrix is $𝑚 𝑛$ entries arranged in a rectangle, and the operations ignore the arrangement. So the proof of Proposition 1.2.5, run entry by entry, shows that $𝐹^{𝑚 \times 𝑛}$ is a vector space over $𝐹$ (Remark 1.2.11 below makes this precise). Its zero vector is the zero matrix, all of whose entries are $0$ . The rectangular shape starts to matter when matrices are multiplied (Section 2.4). An $𝑛 \times 1$ matrix is a column, so $𝐹^{𝑛 \times 1}$ is $𝐹^{𝑛}$ . For example, in $ℝ^{2 \times 3}$ ,

2 (\begin{matrix} 1 & 0 & - 1 \\ 3 & 2 & 0 \end{matrix}) + (\begin{matrix} 0 & 4 & 1 \\ - 1 & 1 & 1 \end{matrix}) = (\begin{matrix} 2 & 4 & - 1 \\ 5 & 5 & 1 \end{matrix}) .

Exercise 1.2.8.routine

(a)In $𝔽_{3}^{3}$ , find $𝐱$ with $2 𝐱 +$ $(1, 2, 0) =$ $(2, 0, 1)$ .
(b)In $𝔽_{5}^{2 \times 2}$ , find $𝑋$ with $3 𝑋 + 𝐴 = 𝐵$ , where $𝐴 = (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix})$ and $𝐵 = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ .
(c)In $ℂ^{2}$ , find $𝐯$ with $(1 + 𝑖) 𝐯 +$ $(1, 0) =$ $(0, 2 𝑖)$ .

Solution.

Uses. Definition 1.2.4 (The space $𝐹^{𝑛}$ ), Definition 1.2.7 (Matrices), Proposition 1.1.13 (Solving $𝑎 + 𝑥 = 𝑏$ and $𝑎 𝑥 = 𝑏$ ), Proposition 1.1.30 ( $ℂ$ is a field)

Hint 1. The operations are coordinatewise, so each equation is a separate equation in

𝐹

for each coordinate or entry.

Hint 2. Solve each one with Proposition 1.1.13: subtract, then multiply by the inverse of the coefficient. In

𝔽_{3}

2^{- 1} = 2

; in

𝔽_{5}

3^{- 1} = 2

; in

ℂ

{(1 + 𝑖)}^{- 1} = (1 - 𝑖) / 2

Sketch. (a)

2 𝐱 = (1, 1, 1)

, so

𝐱 = (2, 2, 2)

. (b)

3 𝑋 = 𝐵 - 𝐴 = (\begin{matrix} 4 & 4 \\ 3 & 1 \end{matrix})

, so

𝑋 = (\begin{matrix} 3 & 3 \\ 1 & 2 \end{matrix})

. (c)

(1 + 𝑖) 𝐯 = (- 1, 2 𝑖)

, so

𝐯 =

((- 1 + 𝑖) / 2, 1 + 𝑖)

In each part, both sides are equal exactly when they agree in every coordinate (or entry), so we solve one equation in $𝐹$ per coordinate, using Proposition 1.1.13.

(a) In $𝔽_{3}$ , the coordinates of $(2, 0, 1)$ minus those of $(1, 2, 0)$ are $2 - 1 = 1$ , $0 - 2 = - 2 = 1$ , and $1 - 0 = 1$ , so $2 𝐱 = (1, 1, 1)$ . Since $2 \cdot 2 = 4 = 1$ in $𝔽_{3}$ , $2^{- 1} = 2$ , and $𝐱 = (2, 2, 2)$ . Check: $2 (2, 2, 2) +$ $(1, 2, 0) =$ $(1, 1, 1) +$ $(1, 2, 0) =$ $(2, 0, 1)$ in $𝔽_{3}$ .

(b) Entrywise in $𝔽_{5}$ ,

3 𝑋 = (\begin{matrix} 0 - 1 & 1 - 2 \\ 1 - 3 & 0 - 4 \end{matrix}) = (\begin{matrix} 4 & 4 \\ 3 & 1 \end{matrix}) .

Since $3 \cdot 2 = 6 = 1$ in $𝔽_{5}$ , $3^{- 1} = 2$ , so

𝑋 = 2 (\begin{matrix} 4 & 4 \\ 3 & 1 \end{matrix}) = (\begin{matrix} 8 & 8 \\ 6 & 2 \end{matrix}) = (\begin{matrix} 3 & 3 \\ 1 & 2 \end{matrix}) .

Check: $3 𝑋 = (\begin{matrix} 9 & 9 \\ 3 & 6 \end{matrix}) = (\begin{matrix} 4 & 4 \\ 3 & 1 \end{matrix})$ , and adding $𝐴$ gives $(\begin{matrix} 5 & 6 \\ 6 & 5 \end{matrix}) = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) = 𝐵$ in $𝔽_{5}$ .

(c) We need $(1 + 𝑖) 𝑣_{1} = - 1$ and $(1 + 𝑖) 𝑣_{2} = 2 𝑖$ . By Proposition 1.1.30, ${(1 + 𝑖)}^{- 1} = (1 - 𝑖) / 2$ , so

𝑣_{1} = \frac{- (1 - 𝑖)}{2} = \frac{- 1 + 𝑖}{2}, 𝑣_{2} = \frac{2 𝑖 (1 - 𝑖)}{2} = 𝑖 - 𝑖^{2} = 1 + 𝑖 .

Check: $(1 + 𝑖) (- 1 + 𝑖) / 2 =$ $(- 1 + 𝑖 - 𝑖 + 𝑖^{2}) / 2 =$ $- 1$ , and $(1 + 𝑖) (1 + 𝑖) = 2 𝑖$ .

Functions

Matrices, functions, and polynomials keep their usual names $𝐴$ , $𝑓$ , $𝑝$ even when we treat them as vectors, and so do numbers in the few examples whose vectors are numbers. Bold letters are reserved for vectors of a general vector space and of $𝐹^{𝑛}$ . The zero vector is written $𝟎$ , whether it is the zero matrix, the zero function, or the zero polynomial; only when the vectors are numbers do we write it as a number.

Definition 1.2.9 (Function spaces). Let $𝑋$ be a set. The set $𝐹^{𝑋}$ consists of all functions $𝑓 : 𝑋 \to 𝐹$ . For $𝑓, 𝑔 \in 𝐹^{𝑋}$ and $𝑎 \in 𝐹$ , the functions $𝑓 + 𝑔$ and $𝑎 𝑓$ are defined pointwise:

(𝑓 + 𝑔) (𝑡) = 𝑓 (𝑡) + 𝑔 (𝑡), (𝑎 𝑓) (𝑡) = 𝑎 𝑓 (𝑡) (𝑡 \in 𝑋) .

Proposition 1.2.10 (

𝐹^{𝑋}

is a vector space).

𝐹^{𝑋}

is a vector space over

𝐹

. The zero function

𝟎

, given by

𝟎 (𝑡) = 0

for all

𝑡 \in 𝑋

, is a zero vector, and the function

𝑡 \mapsto - 𝑓 (𝑡)

is an additive inverse of

𝑓

Proof.routine

Uses. Definition 1.2.9 (Function spaces), Definition 1.1.1 (Field), Proposition 1.2.5 ( $𝐹^{𝑛}$ is a vector space)

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. Two functions

𝑋 \to 𝐹

are equal exactly when they take the same value at every

𝑡 \in 𝑋

Hint 2. Evaluate both sides of each axiom at

𝑡

. This is the proof of Proposition 1.2.5, with the value

𝑓 (𝑡)

in place of the coordinate

𝑥_{𝑖}

Sketch. Pointwise operations give functions

𝑋 \to 𝐹

again. Evaluating both sides of each axiom at

𝑡 \in 𝑋

turns it into the field axiom of the same name for the values

𝑓 (𝑡),

𝑔 (𝑡),

ℎ (𝑡)

, and the zero function and

𝑡 \mapsto - 𝑓 (𝑡)

supply (A3) and (A4).

For $𝑓, 𝑔 \in 𝐹^{𝑋}$ and $𝑎 \in 𝐹$ , the rules above assign an element of $𝐹$ to each $𝑡 \in 𝑋$ , so $𝑓 + 𝑔$ and $𝑎 𝑓$ are functions $𝑋 \to 𝐹$ , that is, elements of $𝐹^{𝑋}$ . Two functions $𝑋 \to 𝐹$ are equal when they agree at every $𝑡 \in 𝑋$ . Let $𝑓, 𝑔, ℎ \in 𝐹^{𝑋}$ , $𝑎, 𝑏 \in 𝐹$ , and $𝑡 \in 𝑋$ . Evaluating both sides of each axiom at $𝑡$ gives

\begin{aligned} (A1) & 𝑓 (𝑡) + 𝑔 (𝑡) = 𝑔 (𝑡) + 𝑓 (𝑡), \\ (A2) & (𝑓 (𝑡) + 𝑔 (𝑡)) + ℎ (𝑡) = 𝑓 (𝑡) + (𝑔 (𝑡) + ℎ (𝑡)), \\ (A3) & 𝑓 (𝑡) + 0 = 𝑓 (𝑡), \\ (A4) & 𝑓 (𝑡) + (- 𝑓 (𝑡)) = 0, \\ (A5) & 1 𝑓 (𝑡) = 𝑓 (𝑡), \\ (A6) & (𝑎 𝑏) 𝑓 (𝑡) = 𝑎 (𝑏 𝑓 (𝑡)), \\ (A7) & 𝑎 (𝑓 (𝑡) + 𝑔 (𝑡)) = 𝑎 𝑓 (𝑡) + 𝑎 𝑔 (𝑡), \\ (A8) & (𝑎 + 𝑏) 𝑓 (𝑡) = 𝑎 𝑓 (𝑡) + 𝑏 𝑓 (𝑡), \end{aligned}

where in (A3) the second summand is $𝟎 (𝑡) = 0$ , and in (A4) it is the value at $𝑡$ of $𝑡 \mapsto - 𝑓 (𝑡)$ . Each line is a field axiom, as in the proof of Proposition 1.2.5. Since $𝑡$ was arbitrary, both sides of each axiom are the same function. □

Remark 1.2.11 (Tuples and matrices are functions). A column

(𝑥_{1}, \dots, 𝑥_{𝑛})

carries the same information as the function

{1, \dots, 𝑛} \to 𝐹

𝑖 \mapsto 𝑥_{𝑖}

, and adding or scaling columns coordinatewise is adding or scaling these functions pointwise. So

𝐹^{𝑛}

𝐹^{𝑋}

for

𝑋 = {1, \dots, 𝑛}

, under another name. In the same way,

𝐹^{𝑚 \times 𝑛}

𝐹^{𝑋}

for

𝑋 =

{1, \dots, 𝑚} \times

{1, \dots, 𝑛}

, a matrix being the function

(𝑖, 𝑗) \mapsto 𝑎_{𝑖 𝑗}

. In particular, Proposition 1.2.10 shows that

𝐹^{𝑚 \times 𝑛}

is a vector space. Other choices of

𝑋

give the space

ℝ^{ℝ}

of all functions

ℝ \to ℝ

, and the space of sequences

(𝑥_{0}, 𝑥_{1}, 𝑥_{2}, \dots)

𝐹

, which is

𝐹^{𝑋}

for

𝑋 = {0, 1, 2, \dots}

Many vector spaces of interest sit inside some $𝐹^{𝑋}$ and use its operations. Then only closure needs a real argument.

Example 1.2.12 (Continuous functions). Let $𝐶 [0, 1]$ be the set of continuous functions $[0, 1] \to ℝ$ , with the operations of $ℝ^{[0, 1]}$ . Sums and constant multiples of continuous functions are continuous, so the operations map into $𝐶 [0, 1]$ . The equations in (A1), (A2), and (A5)–(A8) hold for all elements of $ℝ^{[0, 1]}$ , so in particular for continuous ones. The zero function is continuous, which gives (A3), and if $𝑓$ is continuous then so is $- 𝑓$ , which gives (A4). So $𝐶 [0, 1]$ is a real vector space.

The same argument applies to

{𝑓 \in 𝐶 [0, 1] | \int_{0}^{1} 𝑓 (𝑡) d 𝑡 = 0},

since $\int_{0}^{1} (𝑓 + 𝑔) =$ $\int_{0}^{1} 𝑓 +$ $\int_{0}^{1} 𝑔$ and $\int_{0}^{1} 𝑎 𝑓 = 𝑎 \int_{0}^{1} 𝑓$ : sums and multiples of functions with integral $0$ have integral $0$ , and so do $𝟎$ and $- 𝑓$ . On the other hand, the functions with integral $1$ do not form a vector space under these operations: if $\int_{0}^{1} 𝑓 = \int_{0}^{1} 𝑔 = 1$ , then $\int_{0}^{1} (𝑓 + 𝑔) = 2$ , so addition is not an operation on that set.

The pattern of this example, a subset of a known vector space that contains $𝟎$ and is closed under the operations, is common enough that Section 1.4 turns it into a criterion.

Polynomials

In school, a polynomial is an expression $𝑎_{0} +$ $𝑎_{1} 𝑥 +$ $\dots +$ $𝑎_{𝑑} 𝑥^{𝑑}$ . As a definition this is not quite right: $5$ and $5 + 0 𝑥$ are different expressions, and we want them to be the same polynomial. What determines a polynomial is its list of coefficients, all but finitely many of which are $0$ . So we define a polynomial to be that list, and recover the expressions afterwards.

Definition 1.2.13 (Polynomials). A polynomial over $𝐹$ is a sequence $𝑝 =$ $(𝑎_{0}, 𝑎_{1}, 𝑎_{2}, \dots)$ of elements of $𝐹$ such that only finitely many $𝑎_{𝑘}$ are nonzero. The $𝑎_{𝑘}$ are the coefficients of $𝑝$ , and two polynomials are equal when all their coefficients are equal. The set of polynomials over $𝐹$ is denoted $𝐹 [𝑥]$ . Its operations are coefficientwise:

(𝑎_{0}, 𝑎_{1}, \dots) + (𝑏_{0}, 𝑏_{1}, \dots) = (𝑎_{0} + 𝑏_{0}, 𝑎_{1} + 𝑏_{1}, \dots), 𝑐 (𝑎_{0}, 𝑎_{1}, \dots) = (𝑐 𝑎_{0}, 𝑐 𝑎_{1}, \dots) .

Proposition 1.2.14 (

𝐹 [𝑥]

is a vector space).

𝐹 [𝑥]

is a vector space over

𝐹

. Its zero vector is the zero polynomial

𝟎 = (0, 0, \dots)

Proof.routine

Uses. Definition 1.2.13 (Polynomials), Proposition 1.2.10 ( $𝐹^{𝑋}$ is a vector space), Remark 1.2.11 (Tuples and matrices are functions), Proposition 1.1.15 (Multiplying by zero)

Hint 1. A sequence is a function on

{0, 1, 2, \dots}

, and the operations of

𝐹 [𝑥]

are those of

𝐹^{𝑋}

for

𝑋 = {0, 1, 2, \dots}

. Argue as for

𝐶 [0, 1]

in Example 1.2.12.

Hint 2. The only real work is closure. If

𝑎_{𝑘} = 0

for

𝑘 > 𝑚

and

𝑏_{𝑘} = 0

for

𝑘 > 𝑛

, which coefficients of the sum must vanish?

Sketch.

𝐹 [𝑥]

is the set of finitely supported elements of

𝐹^{𝑋}

𝑋 = {0, 1, 2, \dots}

, with the same operations. Sums and multiples of finitely supported sequences are finitely supported, as are the zero sequence and

(- 𝑎_{𝑘})

; the equational axioms are inherited from

𝐹^{𝑋}

A sequence $(𝑎_{0}, 𝑎_{1}, \dots)$ is the function $𝑘 \mapsto 𝑎_{𝑘}$ on $𝑋 = {0, 1, 2, \dots}$ , and the operations of $𝐹 [𝑥]$ are those of $𝐹^{𝑋}$ . So $𝐹 [𝑥] \subseteq 𝐹^{𝑋}$ , and we argue as in Example 1.2.12.

Closure. Let $𝑝 = (𝑎_{𝑘})$ and $𝑞 = (𝑏_{𝑘})$ be polynomials, and choose $𝑚$ and $𝑛$ with $𝑎_{𝑘} = 0$ for $𝑘 > 𝑚$ and $𝑏_{𝑘} = 0$ for $𝑘 > 𝑛$ . For $𝑘 > max (𝑚, 𝑛)$ we have $𝑎_{𝑘} + 𝑏_{𝑘} = 0 + 0 = 0$ and $𝑐 𝑎_{𝑘} = 𝑐 0 = 0$ (Proposition 1.1.15). So $𝑝 + 𝑞$ and $𝑐 𝑝$ have only finitely many nonzero coefficients.

The axioms. The equations in (A1), (A2), and (A5)–(A8) hold for all elements of $𝐹^{𝑋}$ by Proposition 1.2.10, so in particular for polynomials. The zero sequence has no nonzero coefficients, so it lies in $𝐹 [𝑥]$ and gives (A3). For $𝑝 = (𝑎_{𝑘})$ , the sequence $𝑞 = (- 𝑎_{0}, - 𝑎_{1}, \dots)$ has its nonzero coefficients in the same positions as $𝑝$ (as $- 𝑎_{𝑘} = 0$ only when $𝑎_{𝑘} = 0$ ), so $𝑞 \in 𝐹 [𝑥]$ , and $𝑝 + 𝑞 = 𝟎$ gives (A4). □

Notation 1.2.15. For $𝑘 \geq 0$ , $𝑥^{𝑘}$ denotes the polynomial whose $𝑘$ -th coefficient is $1$ and whose other coefficients are $0$ . We also write $1$ for $𝑥^{0}$ and $𝑥$ for $𝑥^{1}$ . If $𝑎_{𝑘} = 0$ for all $𝑘 > 𝑑$ , then

(𝑎_{0}, 𝑎_{1}, 𝑎_{2}, \dots) = 𝑎_{0} + 𝑎_{1} 𝑥 + \dots + 𝑎_{𝑑} 𝑥^{𝑑} = \sum_{𝑘 = 0}^{𝑑} 𝑎_{𝑘} 𝑥^{𝑘},

where the right side is computed with the operations of $𝐹 [𝑥]$ : its $𝑗$ -th coefficient is $𝑎_{𝑗}$ for $𝑗 \leq 𝑑$ and $0$ for $𝑗 > 𝑑$ .

So the familiar expressions are honest sums of vectors in $𝐹 [𝑥]$ , and the problem with $5$ and $5 + 0 𝑥$ disappears: both are the sequence $(5, 0, 0, \dots)$ . A polynomial can be written with any $𝑑$ that is at least as large as the index of its last nonzero coefficient. To decide whether two expressions are the same polynomial, compare coefficients. We multiply polynomials in Section 7.1; here only the vector space operations matter.

Definition 1.2.16 (Degree). Let $𝑝 =$ $\sum_{𝑘 = 0}^{𝑑} 𝑎_{𝑘} 𝑥^{𝑘} \in$ $𝐹 [𝑥]$ . If $𝑝 \neq 𝟎$ , the degree $deg 𝑝$ is the largest $𝑘$ with $𝑎_{𝑘} \neq 0$ , and $𝑎_{deg 𝑝}$ is the leading coefficient of $𝑝$ . The zero polynomial has degree $deg 𝟎 = - \infty$ , with the convention that $- \infty < 𝑛$ for every integer $𝑛$ . For $𝑛 \geq 0$ ,

{𝐹 [𝑥]}_{\leq 𝑛} = {𝑝 \in 𝐹 [𝑥] | deg 𝑝 \leq 𝑛}

is the set of polynomials $𝑎_{0} +$ $𝑎_{1} 𝑥 +$ $\dots +$ $𝑎_{𝑛} 𝑥^{𝑛}$ with $𝑎_{0}, \dots, 𝑎_{𝑛} \in 𝐹$ (any of which may be $0$ ).

Definition 1.2.17 (Evaluation). Let $𝑝 =$ $\sum_{𝑘 = 0}^{𝑑} 𝑎_{𝑘} 𝑥^{𝑘} \in$ $𝐹 [𝑥]$ and $𝑐 \in 𝐹$ . The value of $𝑝$ at $𝑐$ is

𝑝 (𝑐) = \sum_{𝑘 = 0}^{𝑑} 𝑎_{𝑘} 𝑐^{𝑘} \in 𝐹 .

The function $𝐹 \to 𝐹$ , $𝑐 \mapsto 𝑝 (𝑐)$ , is the polynomial function of $𝑝$ .

The value does not depend on which $𝑑$ we use to write $𝑝$ , since the extra terms are $0 \cdot 𝑐^{𝑘} = 0$ . For example, over $ℝ$ , $deg (3 - 𝑥^{2}) = 2$ and $(3 - 𝑥^{2}) (2) = - 1$ .

Proposition 1.2.18 (Degree and evaluation versus the operations). Let $𝑝, 𝑞 \in 𝐹 [𝑥]$ and $𝑎, 𝑐 \in 𝐹$ .

(a) $deg (𝑝 + 𝑞) \leq$ $max (deg 𝑝, deg 𝑞)$ , with equality if $deg 𝑝 \neq deg 𝑞$ .
(b)If $𝑎 \neq 0$ , then $deg (𝑎 𝑝) = deg 𝑝$ .
(c) $(𝑝 + 𝑞) (𝑐) =$ $𝑝 (𝑐) +$ $𝑞 (𝑐)$ and $(𝑎 𝑝) (𝑐) =$ $𝑎 𝑝 (𝑐)$ .

Proof.routine

Uses. Definition 1.2.16 (Degree), Definition 1.2.17 (Evaluation), Proposition 1.1.17 (No zero divisors), Definition 1.1.1 (Field)

Hint 1. Write

𝑝 = (𝑎_{𝑘})

and

𝑞 = (𝑏_{𝑘})

. For (a), which coefficients of

𝑝 + 𝑞

are forced to vanish? If

deg 𝑝 > deg 𝑞

, look at the coefficient of

𝑝 + 𝑞

in position

deg 𝑝

Hint 2. For (b), when is

𝑎 𝑎_{𝑘}

zero? For (c), write

𝑝

and

𝑞

with the same

𝑑

and expand.

Sketch. (a) Beyond

max (deg 𝑝, deg 𝑞)

both coefficient sequences vanish, so the sum’s does; if

deg 𝑝 > deg 𝑞

, the coefficient of

𝑝 + 𝑞

in position

deg 𝑝

𝑎_{deg 𝑝} + 0 \neq 0

. (b)

𝑎 𝑎_{𝑘} = 0

if and only if

𝑎_{𝑘} = 0

. (c) Distributivity and reordering of finite sums.

Write $𝑝 = (𝑎_{𝑘})$ and $𝑞 = (𝑏_{𝑘})$ .

(a) Let $𝑛 =$ $max (deg 𝑝, deg 𝑞)$ . For every integer $𝑘 > 𝑛$ we have $𝑎_{𝑘} = 𝑏_{𝑘} = 0$ , so the $𝑘$ -th coefficient $𝑎_{𝑘} + 𝑏_{𝑘}$ of $𝑝 + 𝑞$ is $0$ . Hence $deg (𝑝 + 𝑞) \leq 𝑛$ ; this includes the case $𝑝 + 𝑞 = 𝟎$ , and the case $𝑛 = - \infty$ , where $𝑝 = 𝑞 = 𝟎$ . Now suppose $deg 𝑝 \neq deg 𝑞$ ; by symmetry we may assume $deg 𝑝 > deg 𝑞$ . Then $𝑝 \neq 𝟎$ , and with $𝑚 = deg 𝑝$ we have $𝑏_{𝑚} = 0$ because $𝑚 > deg 𝑞$ . So the $𝑚$ -th coefficient of $𝑝 + 𝑞$ is $𝑎_{𝑚} + 0 = 𝑎_{𝑚} \neq 0$ , and $deg (𝑝 + 𝑞) \geq 𝑚$ . With the first part, $deg (𝑝 + 𝑞) =$ $𝑚 =$ $max (deg 𝑝, deg 𝑞)$ .

(b) The $𝑘$ -th coefficient of $𝑎 𝑝$ is $𝑎 𝑎_{𝑘}$ . Since $𝑎 \neq 0$ , Proposition 1.1.17 gives $𝑎 𝑎_{𝑘} = 0$ if and only if $𝑎_{𝑘} = 0$ . So $𝑎 𝑝$ and $𝑝$ have their nonzero coefficients in the same positions, and in particular the same degree.

(c) Choose $𝑑$ with $𝑎_{𝑘} = 𝑏_{𝑘} = 0$ for all $𝑘 > 𝑑$ . Then, by distributivity and by reordering the finite sum,

(𝑝 + 𝑞) (𝑐) = \sum_{𝑘 = 0}^{𝑑} (𝑎_{𝑘} + 𝑏_{𝑘}) 𝑐^{𝑘} = \sum_{𝑘 = 0}^{𝑑} 𝑎_{𝑘} 𝑐^{𝑘} + \sum_{𝑘 = 0}^{𝑑} 𝑏_{𝑘} 𝑐^{𝑘} = 𝑝 (𝑐) + 𝑞 (𝑐),

and $(𝑎 𝑝) (𝑐) =$ $\sum_{𝑘 = 0}^{𝑑} (𝑎 𝑎_{𝑘}) 𝑐^{𝑘} =$ $𝑎 \sum_{𝑘 = 0}^{𝑑} 𝑎_{𝑘} 𝑐^{𝑘} =$ $𝑎 𝑝 (𝑐)$ . □

Parts (a) and (b) say that sums and nonzero multiples of polynomials of degree at most $𝑛$ again have degree at most $𝑛$ (and $0 𝑝 = 𝟎$ has degree $- \infty$ ). So ${𝐹 [𝑥]}_{\leq 𝑛}$ is closed under the operations of $𝐹 [𝑥]$ , and it is a vector space by the argument of Example 1.2.12. Part (c) says that taking polynomial functions is compatible with the operations of $𝐹 [𝑥]$ and of $𝐹^{𝐹}$ .

Aside (Polynomials versus polynomial functions). Different polynomials can have the same polynomial function. Over $𝔽_{2}$ , the polynomial $𝑓 = 𝑥^{2} + 𝑥$ has coefficients $(0, 1, 1, 0, \dots)$ , so $𝑓 \neq 𝟎$ and $deg 𝑓 = 2$ . But $𝑓 (0) = 0$ and $𝑓 (1) = 1 + 1 = 0$ , so the polynomial function of $𝑓$ is the zero function $𝔽_{2} \to 𝔽_{2}$ , the same as that of $𝟎$ . Similarly, for every prime $𝑝$ , the polynomial $𝑥^{𝑝} - 𝑥 \in 𝔽_{𝑝} [𝑥]$ is nonzero but vanishes at every element of $𝔽_{𝑝}$ (Exercise 1.2.51).

This is one reason to define polynomials by their coefficients. If a polynomial were its function, then $𝑥^{2} + 𝑥$ and $𝟎$ would be the same polynomial over $𝔽_{2}$ , and degree would not be well defined. Later we will also substitute matrices and linear maps into polynomials (Section 7.2), which a function $𝐹 \to 𝐹$ cannot accommodate.

Over an infinite field such as $ℚ$ , $ℝ$ , or $ℂ$ , nothing is lost by thinking of functions. A nonzero polynomial of degree $𝑑$ has at most $𝑑$ roots (Section 7.1). So if $𝑝$ and $𝑞$ have the same polynomial function, then $(𝑝 - 𝑞) (𝑐) =$ $𝑝 (𝑐) -$ $𝑞 (𝑐) =$ $0$ for every $𝑐 \in 𝐹$ (Proposition 1.2.18(c)); a nonzero $𝑝 - 𝑞$ would have infinitely many roots, so $𝑝 - 𝑞 = 𝟎$ . This is why some books, LADR among them, can define polynomials over $ℝ$ and $ℂ$ as functions.

Changing the field

The field is part of the data of a vector space, and shrinking the field is always allowed.

Proposition 1.2.19 (Restriction of scalars). Let

𝐾

be a subfield of

𝐹

, and let

𝑉

be a vector space over

𝐹

. Then

𝑉

, with the same addition and with scalar multiplication restricted to

𝐾 \times 𝑉

, is a vector space over

𝐾

Proof.routine

Uses. Definition 1.2.1 (Vector space), Definition 1.1.32 (Subfield), Proposition 1.1.33 (A subfield is a field), Theorem 1.1.9 (Uniqueness of identities and inverses)

Hint 1. Sort the axioms into those about addition alone, those that hold “for all scalars”, and (A5).

Hint 2. For (A5), you need to know that the multiplicative identity of the field

𝐾

is the

1

𝐹

Sketch. (A1)–(A4) only involve addition, which is unchanged. (A6)–(A8) hold for all scalars in

𝐹

, hence for those in

𝐾

, whose sums and products are computed as in

𝐹

. The identity of

𝐾

is the

1

𝐹

, by uniqueness of identities in the field

𝐾

, so (A5) holds.

The restricted scalar multiplication is a function

𝐾 \times 𝑉 \to 𝑉

. Axioms (A1)–(A4) involve only the addition of

𝑉

, which has not changed. Axioms (A6)–(A8) hold for all

𝑎, 𝑏 \in 𝐹

, so in particular for all

𝑎, 𝑏 \in 𝐾

; here

𝑎 + 𝑏

and

𝑎 𝑏

mean the same elements whether computed in

𝐾

or in

𝐹

, because

𝐾

carries the operations of

𝐹

. For (A5),

𝐾

is a field (Proposition 1.1.33), and

1 \in 𝐾

by Definition 1.1.32. Since

𝑎 \cdot 1 = 𝑎

for every

𝑎 \in 𝐾

, the element

1

is a multiplicative identity of

𝐾

, so it is the multiplicative identity of

𝐾

by Theorem 1.1.9. As

1 𝐯 = 𝐯

for all

𝐯 \in 𝑉

, (A5) holds over

𝐾

. □

Example 1.2.20 (Real and rational scalars). $ℂ$ is a vector space over itself (Example 1.2.6), hence over its subfield $ℝ$ . For $𝑎 \in ℝ$ , $𝑎 (𝑥 + 𝑦 𝑖) =$ $𝑎 𝑥 +$ $𝑎 𝑦 𝑖$ , so as a real vector space $ℂ$ is $ℝ^{2}$ with the pair $(𝑥, 𝑦)$ written as $𝑥 + 𝑦 𝑖$ . Likewise $ℂ^{𝑛}$ is a real vector space.

$ℝ$ is a vector space over $ℚ$ , with scalar multiplication the usual product. The change of field changes the geometry. Over $ℝ$ , every real number is a scalar multiple of $1$ . Over $ℚ$ , $\sqrt{2}$ is not a rational multiple of $1$ , so $1$ and $\sqrt{2}$ point in different directions. Section 1.5 makes this precise.

In the other direction, $ℚ$ with the usual product is not a vector space over $ℝ$ : $\sqrt{2} \cdot 1 \notin ℚ$ , so the product is not a function $ℝ \times ℚ \to ℚ$ . For the same reason, $ℝ$ with the usual product is not a vector space over $ℂ$ . Whether some other scalar multiplication works is a different question (Exercise 1.2.47).

1.2.3Non-examples and repairs

To show that a candidate is not a vector space, one failure suffices: a sum or multiple that leaves the set, or a specific choice of vectors and scalars for which an axiom fails. The candidates below fail in different ways, and the first can be repaired by changing its operations.

Figure 1.2.1. The line

𝐿

𝑥 + 2 𝑦 = 2

(solid) and the line

𝐿_{0}

𝑥 + 2 𝑦 = 0

through the origin (dashed). The vectors

𝐮 = (2, 0)

and

𝐰 = (0, 1)

lie on

𝐿

, but their sum

(2, 1)

does not.

Non-example 1.2.21 (A line not through the origin). Let $𝐿 =$ ${(𝑥, 𝑦) \in ℝ^{2} | 𝑥 + 2 𝑦 = 2}$ , with the operations of $ℝ^{2}$ . This is not a real vector space, and the first problem is that the operations of $ℝ^{2}$ are not operations on $𝐿$ . The vectors $𝐮 = (2, 0)$ and $𝐰 = (0, 1)$ lie on $𝐿$ , but $𝐮 + 𝐰 = (2, 1)$ does not, since $2 + 2 \cdot 1 = 4$ (Figure 1.2.1). Similarly $0 𝐮 = (0, 0) \notin 𝐿$ .

Even setting closure aside, (A3) fails. Notice what showing this requires. Observing that $(0, 0) \notin 𝐿$ is not enough on its own, because the zero vector of a structure on $𝐿$ need not be $(0, 0)$ (see Example 1.2.23). We must show that no $𝐳 \in 𝐿$ satisfies $𝐯 + 𝐳 = 𝐯$ for all $𝐯 \in 𝐿$ . Here that is quick: if $𝐯 + 𝐳 = 𝐯$ in $ℝ^{2}$ for even one $𝐯$ , then subtracting $𝐯$ coordinatewise gives $𝐳 = (0, 0)$ , which is not in $𝐿$ .

Intuition.

𝐿

is the line

𝐿_{0}

through the origin, shifted by

𝐩 = (2, 0)

: every point of

𝐿

𝐩 + 𝐯

for exactly one

𝐯 \in 𝐿_{0}

(Figure 1.2.1). The line

𝐿_{0}

is closed under the operations of

ℝ^{2}

, so the only defect of

𝐿

is where its origin is. To repair

𝐿

, we can add and scale as if

𝐩

were the origin. The cleanest way to carry this out is to copy the operations of a known vector space across a bijection.

Proposition 1.2.22 (Transport of structure). Let $𝑉$ be a vector space over $𝐹$ , let $𝑊$ be a set, and let $𝜑 : 𝑉 \to 𝑊$ be a bijection. Define operations on $𝑊$ by

𝐰_{1} \oplus 𝐰_{2} = 𝜑 (𝜑^{- 1} (𝐰_{1}) + 𝜑^{- 1} (𝐰_{2})), 𝑎 ⊙ 𝐰 = 𝜑 (𝑎 𝜑^{- 1} (𝐰)) .

Then $𝑊$ with $\oplus$ and $⊙$ is a vector space over $𝐹$ in which $𝜑 (𝟎)$ is a zero vector, and for all $𝐯, 𝐯^{'} \in 𝑉$ and $𝑎 \in 𝐹$ ,

𝜑 (𝐯 + 𝐯^{'}) = 𝜑 (𝐯) \oplus 𝜑 (𝐯^{'}), 𝜑 (𝑎 𝐯) = 𝑎 ⊙ 𝜑 (𝐯) .

(1.2.1)

Proof.routine

Uses. Definition 1.2.1 (Vector space), a bijection $𝜑$ has an inverse $𝜑^{- 1}$ with $𝜑^{- 1} (𝜑 (𝐯)) = 𝐯$ and $𝜑 (𝜑^{- 1} (𝐰)) = 𝐰$

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. First prove (1.2.1) directly from the definitions. Then every element of

𝑊

has the form

𝜑 (𝐯)

Hint 2. With (1.2.1), each side of an axiom in

𝑊

, written for elements

𝜑 (𝐯)

, becomes

𝜑

applied to the corresponding side of the same axiom in

𝑉

Sketch. The definitions give

𝜑 (𝐯) \oplus

𝜑 (𝐯^{'}) =

𝜑 (𝐯 + 𝐯^{'})

and

𝑎 ⊙ 𝜑 (𝐯) =

𝜑 (𝑎 𝐯)

. Since

𝜑

is surjective, it suffices to check each axiom on elements

𝜑 (𝐯)

; pushing

\oplus

and

⊙

inside

𝜑

turns it into

𝜑

applied to the same axiom in

𝑉

. The zero vector is

𝜑 (𝟎)

, and

𝜑 (𝐯^{'})

is an inverse of

𝜑 (𝐯)

whenever

𝐯 + 𝐯^{'} = 𝟎

The operations take values in $𝑊$ because $𝜑$ does. Substituting $𝐰_{1} = 𝜑 (𝐯)$ , $𝐰_{2} = 𝜑 (𝐯^{'})$ , and $𝐰 = 𝜑 (𝐯)$ into the definitions and using $𝜑^{- 1} (𝜑 (𝐯)) = 𝐯$ gives (1.2.1). Since $𝜑$ is surjective, every element of $𝑊$ is $𝜑 (𝐯)$ for some $𝐯 \in 𝑉$ , so it suffices to verify each axiom for elements of this form. Let $𝐮, 𝐯, 𝐰 \in 𝑉$ and $𝑎, 𝑏 \in 𝐹$ . Each of the following chains uses (1.2.1) to move the operations inside $𝜑$ , applies the axiom of $𝑉$ named on the right, and moves the operations back out:

\begin{aligned} 𝜑 (𝐯) \oplus 𝜑 (𝐰) & = 𝜑 (𝐯 + 𝐰) = 𝜑 (𝐰 + 𝐯) \\ = 𝜑 (𝐰) \oplus 𝜑 (𝐯) & (A1), \\ (𝜑 (𝐮) \oplus 𝜑 (𝐯)) \oplus 𝜑 (𝐰) & = 𝜑 ((𝐮 + 𝐯) + 𝐰) = 𝜑 (𝐮 + (𝐯 + 𝐰)) \\ = 𝜑 (𝐮) \oplus (𝜑 (𝐯) \oplus 𝜑 (𝐰)) & (A2), \\ 1 ⊙ 𝜑 (𝐯) & = 𝜑 (1 𝐯) = 𝜑 (𝐯) & (A5), \\ (𝑎 𝑏) ⊙ 𝜑 (𝐯) & = 𝜑 ((𝑎 𝑏) 𝐯) = 𝜑 (𝑎 (𝑏 𝐯)) = 𝑎 ⊙ (𝑏 ⊙ 𝜑 (𝐯)) & (A6), \\ 𝑎 ⊙ (𝜑 (𝐯) \oplus 𝜑 (𝐰)) & = 𝜑 (𝑎 (𝐯 + 𝐰)) = 𝜑 (𝑎 𝐯 + 𝑎 𝐰) \\ = (𝑎 ⊙ 𝜑 (𝐯)) \oplus (𝑎 ⊙ 𝜑 (𝐰)) & (A7), \\ (𝑎 + 𝑏) ⊙ 𝜑 (𝐯) & = 𝜑 ((𝑎 + 𝑏) 𝐯) = 𝜑 (𝑎 𝐯 + 𝑏 𝐯) \\ = (𝑎 ⊙ 𝜑 (𝐯)) \oplus (𝑏 ⊙ 𝜑 (𝐯)) & (A8) . \end{aligned}

For (A3), $𝜑 (𝐯) \oplus$ $𝜑 (𝟎) =$ $𝜑 (𝐯 + 𝟎) =$ $𝜑 (𝐯)$ for every $𝐯$ , so $𝜑 (𝟎)$ is a zero vector of $𝑊$ . For (A4), given $𝐯$ , choose $𝐯^{'}$ with $𝐯 + 𝐯^{'} = 𝟎$ ; then $𝜑 (𝐯) \oplus$ $𝜑 (𝐯^{'}) =$ $𝜑 (𝟎)$ . □

The map $𝜑$ is a dictionary: it renames each vector of $𝑉$ , and $𝑊$ inherits the arithmetic of $𝑉$ under the new names. In the language of Section 4.5, (1.2.1) says that $𝜑$ is an isomorphism from $𝑉$ to $𝑊$ with its new operations.

Example 1.2.23 (Repairing the line). Parametrize $𝐿$ by $𝜑 : ℝ \to 𝐿$ , $𝜑 (𝑡) = (2 - 2 𝑡, 𝑡)$ . This is a bijection: each $(𝑥, 𝑦) \in 𝐿$ has $𝑥 = 2 - 2 𝑦$ , so $(𝑥, 𝑦) = 𝜑 (𝑦)$ , and $𝜑^{- 1} (𝑥, 𝑦) = 𝑦$ . Transport the structure of $ℝ$ , a real vector space by Example 1.2.6, along $𝜑$ . By Proposition 1.2.22, $𝐿$ becomes a real vector space with

\begin{aligned} (𝑥, 𝑦) \oplus (𝑥^{'}, 𝑦^{'}) & = 𝜑 (𝑦 + 𝑦^{'}) = (2 - 2 (𝑦 + 𝑦^{'}), 𝑦 + 𝑦^{'}), \\ 𝑎 ⊙ (𝑥, 𝑦) & = 𝜑 (𝑎 𝑦) = (2 - 2 𝑎 𝑦, 𝑎 𝑦) . \end{aligned}

Its zero vector is $𝜑 (0) = (2, 0) = 𝐩$ , and an additive inverse of $(𝑥, 𝑦)$ is $𝜑 (- 𝑦) =$ $(2 + 2 𝑦, - 𝑦)$ .

These formulas can be written without $𝜑$ . For $𝐮 = (𝑥, 𝑦)$ and $𝐮^{'} = (𝑥^{'}, 𝑦^{'})$ in $𝐿$ we have $𝑥 = 2 - 2 𝑦$ and $𝑥^{'} = 2 - 2 𝑦^{'}$ , and substituting shows

𝐮 \oplus 𝐮^{'} = 𝐮 + 𝐮^{'} - 𝐩, 𝑎 ⊙ 𝐮 = 𝐩 + 𝑎 (𝐮 - 𝐩),

where the right sides use the operations of $ℝ^{2}$ . For instance, the first coordinate of $𝐮 + 𝐮^{'} - 𝐩$ is $𝑥 +$ $𝑥^{'} -$ $2 =$ $2 -$ $2 (𝑦 + 𝑦^{'})$ . So $\oplus$ and $⊙$ are the operations of $ℝ^{2}$ recentered at $𝐩$ : subtract $𝐩$ , compute as usual, and add $𝐩$ back. Any other point of $𝐿$ could serve as the origin instead and gives another valid structure; the line itself has no preferred origin. Such sets are studied as affine spaces in Section 4.7.

Exercise 1.2.24.routine Find the error in the following argument, and show that its conclusion is false.

Claim. No operations make the interval $(1, 2) =$ ${𝑡 \in ℝ | 1 < 𝑡 < 2}$ a real vector space.

Proof. Every vector space contains a zero vector, and the zero vector of a set of real numbers is $0$ . Since $0 \notin (1, 2)$ , the interval cannot be a vector space.

Solution.

Uses. Proposition 1.2.22 (Transport of structure), Example 1.2.23 (Repairing the line)

Hint 1. The zero vector is defined by (A3) relative to the addition. Which addition is the argument assuming?

Hint 2. To refute the claim, transport the structure of

ℝ

along a bijection

ℝ \to (1, 2)

. The function

𝑔 (𝑡) =

𝑡 / (1 + | 𝑡 |)

is a bijection

ℝ \to (- 1, 1)

Sketch. The zero vector depends on the operations;

0

is the zero of

ℝ

for its usual addition, which is not an operation on

(1, 2)

anyway. Transporting along the bijection

𝜑 (𝑡) =

(3 + 𝑔 (𝑡)) / 2

𝑔 (𝑡) =

𝑡 / (1 + | 𝑡 |)

, makes

(1, 2)

a real vector space with zero vector

𝜑 (0) = 3 / 2

The error is the assertion that the zero vector must be $0$ . The zero vector is whatever element satisfies (A3) for the addition in question, and $0$ plays that role only for the usual addition of $ℝ$ , which is not even an operation on $(1, 2)$ . A different addition on $(1, 2)$ can have a different zero vector.

The conclusion is false. Let $𝑔 (𝑡) =$ $𝑡 / (1 + | 𝑡 |)$ . For $𝑡 \geq 0$ , $𝑔 (𝑡) =$ $𝑡 / (1 + 𝑡) \in$ $[0, 1)$ , and for $𝑠 \in [0, 1)$ the equation $𝑠 = 𝑡 / (1 + 𝑡)$ has the unique solution $𝑡 = 𝑠 / (1 - 𝑠) \geq 0$ . For $𝑡 < 0$ , $𝑔 (𝑡) =$ $𝑡 / (1 - 𝑡) \in$ $(- 1, 0)$ , and for $𝑠 \in (- 1, 0)$ the equation $𝑠 = 𝑡 / (1 - 𝑡)$ has the unique solution $𝑡 = 𝑠 / (1 + 𝑠) < 0$ . So $𝑔$ is a bijection $ℝ \to (- 1, 1)$ with inverse $𝑠 \mapsto 𝑠 / (1 - | 𝑠 |)$ . Then $𝜑 (𝑡) =$ $(3 + 𝑔 (𝑡)) / 2$ is a bijection $ℝ \to (1, 2)$ , and by Proposition 1.2.22, transporting the structure of $ℝ$ along $𝜑$ makes $(1, 2)$ a real vector space. Its zero vector is $𝜑 (0) = 3 / 2$ .

The next example changes one operation of $ℝ^{2}$ at a time. Each variant fails the axioms listed and satisfies the others, which shows what those axioms are for.

Example 1.2.25 (Changing one operation on

ℝ^{2}

(a)Usual addition, $𝑎 ⊙ (𝑥, 𝑦) =$ $(𝑎 𝑥, 0)$ . Axioms (A1)–(A4) are those of $ℝ^{2}$ . Axioms (A6)–(A8) hold, because both sides agree in the first coordinate as in $ℝ^{2}$ and both have second coordinate $0$ . But (A5) fails: $1 ⊙ (0, 1) =$ $(0, 0) \neq$ $(0, 1)$ . So (A5) does not follow from the other axioms; it is what stops scalar multiplication from discarding part of a vector. (The extreme case $𝑎 ⊙ 𝐯 = 𝟎$ for all $𝑎$ and $𝐯$ also satisfies every axiom except (A5).)
(b)Usual addition, $𝑎 ⊙ (𝑥, 𝑦) =$ $(𝑎 𝑥, 𝑦)$ . Now (A5), (A6), and (A7) hold; for instance $(𝑎 𝑏) ⊙ (𝑥, 𝑦) =$ $(𝑎 𝑏 𝑥, 𝑦) =$ $𝑎 ⊙ (𝑏 𝑥, 𝑦)$ . But (A8) fails:
$(1 + 1) ⊙ (0, 1) = (0, 1), while 1 ⊙ (0, 1) + 1 ⊙ (0, 1) = (0, 2) .$
Together with (A5), (A8) forces $2 ⊙ 𝐯 = 𝐯 + 𝐯$ , so scalar multiplication cannot leave a coordinate untouched.
(c)Midpoint addition, $(𝑥, 𝑦) \oplus$ $(𝑥^{'}, 𝑦^{'}) =$ $((𝑥 + 𝑥^{'}) / 2, (𝑦 + 𝑦^{'}) / 2)$ , usual scalar multiplication. (A1) holds, but (A2) fails: with $𝐮 = 𝐯 = (0, 0)$ and $𝐰 = (4, 0)$ ,
$(𝐮 \oplus 𝐯) \oplus 𝐰 = (2, 0), while 𝐮 \oplus (𝐯 \oplus 𝐰) = (1, 0) .$
(A3) fails too, and this failure shows the quantifier order at work (Remark 1.2.2). We have $𝐯 \oplus 𝐳 = 𝐯$ exactly when $𝐳 = 𝐯$ . So every $𝐯$ has a vector that acts like zero for it, namely $𝐯$ itself, but no single $𝐳$ works for every $𝐯$ . Without a zero vector, (A4) cannot even be stated. Finally, (A8) fails: $(1 + 1) (1, 0) = (2, 0)$ , while $(1, 0) \oplus (1, 0) = (1, 0)$ . Axioms (A5)–(A7) hold.

The last non-example has operations that stay inside the set and satisfy most of the axioms.

Non-example 1.2.26 (Addition modulo $1$ ). For $𝑡 \in ℝ$ , write $𝑡 mod 1 =$ $𝑡 -$ $⌊ 𝑡 ⌋ \in$ $[0, 1)$ for the fractional part of $𝑡$ . Then $𝑠 mod 1 = 𝑡 mod 1$ exactly when $𝑠 - 𝑡 \in ℤ$ . Let $𝑉 = [0, 1)$ with

𝑠 \oplus 𝑡 = (𝑠 + 𝑡) mod 1, 𝑎 ⊙ 𝑡 = (𝑎 𝑡) mod 1 (𝑎 \in ℝ) .

For instance $0.7 \oplus 0.5 = 0.2$ . Think of $𝑉$ as a circle of circumference $1$ , on which adding means walking around the circle. Both operations land in $𝑉$ .

Replacing a real number by its fractional part changes it by an integer, and numbers that differ by an integer have the same fractional part. So, as in Lemma 1.1.21, both sides of (A2) equal $(𝑠 + 𝑡 + 𝑢) mod 1$ , and (A1) holds because $𝑠 + 𝑡 = 𝑡 + 𝑠$ . The number $0$ is a zero vector, an inverse of $𝑡 \neq 0$ is $1 - 𝑡$ , and $0$ is its own inverse. (A5) holds since $1 ⊙ 𝑡 = 𝑡 mod 1 = 𝑡$ for $𝑡 \in [0, 1)$ . (A8) holds since $(𝑎 + 𝑏) 𝑡 =$ $𝑎 𝑡 +$ $𝑏 𝑡$ differs from $(𝑎 ⊙ 𝑡) + (𝑏 ⊙ 𝑡)$ by an integer, so the fractional part of the latter, which is $(𝑎 ⊙ 𝑡) \oplus (𝑏 ⊙ 𝑡)$ , equals that of the former, which is $(𝑎 + 𝑏) ⊙ 𝑡$ . But (A7) fails: with $𝑎 = 1 / 2$ and $𝑠 = 𝑡 = 1 / 2$ ,

\frac{1}{2} ⊙ (\frac{1}{2} \oplus \frac{1}{2}) = \frac{1}{2} ⊙ 0 = 0, while (\frac{1}{2} ⊙ \frac{1}{2}) \oplus (\frac{1}{2} ⊙ \frac{1}{2}) = \frac{1}{4} \oplus \frac{1}{4} = \frac{1}{2} .

(A6) fails as well: with $𝑎 = 1 / 2$ , $𝑏 = 2$ , and $𝑡 = 1 / 2$ , we get $(𝑎 𝑏) ⊙ 𝑡 =$ $1 ⊙ 𝑡 =$ $1 / 2$ , while $𝑎 ⊙ (𝑏 ⊙ 𝑡) = 𝑎 ⊙ 0 = 0$ .

One might hope that a cleverer scalar multiplication repairs this. It cannot, and the reason (Example 1.2.40) is a consequence of the axioms that involves only the addition.

1.2.4Consequences of the axioms

From now on, $𝑉$ is a vector space over a field $𝐹$ . The axioms are all we know about $𝑉$ , so as with fields, we prove each elementary fact from them (Technique 1.1.7). Each fact then holds in every example of the previous subsections.

Uniqueness

(A3) and (A4) assert existence only. As for fields, uniqueness is proved by comparing two candidates (Technique 1.1.8).

Proposition 1.2.27 (Uniqueness of the zero vector). A vector space has exactly one zero vector. That is, if

𝟎, 𝟎^{'} \in 𝑉

both satisfy (A3), then

𝟎 = 𝟎^{'}

Proof.routine

Uses. Definition 1.2.1 (Vector space)

Techniques. Uniqueness via two candidates (Technique 1.1.8), Proving from axioms (Technique 1.1.7)

Hint 1. Suppose

𝟎

and

𝟎^{'}

both satisfy (A3). Find one expression that involves both.

Hint 2. Evaluate

𝟎 + 𝟎^{'}

twice: once using that

𝟎^{'}

satisfies (A3), and once, after (A1), using that

𝟎

does.

Sketch. Each absorbs the other:

𝟎 =

𝟎 +

𝟎^{'} =

𝟎^{'} +

𝟎 =

𝟎^{'}

Suppose $𝟎$ and $𝟎^{'}$ both satisfy (A3). Then

\begin{aligned} 𝟎 & = 𝟎 + 𝟎^{'} & (as 𝟎^{'} satisfies (A3)) \\ = 𝟎^{'} + 𝟎 & (A1) \\ = 𝟎^{'} & (as 𝟎 satisfies (A3)), \end{aligned}

as desired. □

The proof used only (A1) and (A3), so (A4) refers to a definite vector. When several vector spaces are in play, we write $𝟎_{𝑉}$ for the zero vector of $𝑉$ . The proof is word for word that of Theorem 1.1.9(a), because (A1)–(A4) are the additive axioms of a field with vectors in place of scalars.

“Exactly one additive inverse” is two claims. Existence is (A4). Uniqueness becomes precise once we name two candidates.

Proposition 1.2.28 (Uniqueness of additive inverses). Every vector has exactly one additive inverse. That is, if

𝐯, 𝐰, 𝐰^{'} \in 𝑉

satisfy

𝐯 + 𝐰 = 𝟎

and

𝐯 + 𝐰^{'} = 𝟎

, then

𝐰 = 𝐰^{'}

Proof.routine

Uses. Definition 1.2.1 (Vector space), Proposition 1.2.27 (Uniqueness of the zero vector)

Techniques. Uniqueness via two candidates (Technique 1.1.8), Proving from axioms (Technique 1.1.7)

Hint 1. Start from

𝐰 = 𝐰 + 𝟎

and replace

𝟎

by something involving

𝐰^{'}

Hint 2. Replace

𝟎

𝐯 + 𝐰^{'}

, then regroup so that

𝐰

meets

𝐯

Sketch. Insert

𝟎 = 𝐯 + 𝐰^{'}

into

𝐰 = 𝐰 + 𝟎

and regroup:

𝐰 =

(𝐰 + 𝐯) +

𝐰^{'} =

𝟎 +

𝐰^{'} =

𝐰^{'}

Suppose $𝐯 + 𝐰 = 𝟎$ and $𝐯 + 𝐰^{'} = 𝟎$ . Then

\begin{aligned} 𝐰 & = 𝐰 + 𝟎 & (A3) \\ = 𝐰 + (𝐯 + 𝐰^{'}) & (as 𝐯 + 𝐰^{'} = 𝟎) \\ = (𝐰 + 𝐯) + 𝐰^{'} & (A2) \\ = 𝟎 + 𝐰^{'} & (as 𝐰 + 𝐯 = 𝐯 + 𝐰 = 𝟎 by (A1)) \\ = 𝐰^{'} & (A1 and A3), \end{aligned}

as desired. □

Notation 1.2.29. The additive inverse of

𝐯

is written

- 𝐯

, and subtraction is defined by

𝐮 - 𝐯 = 𝐮 + (- 𝐯)

The additive facts about fields carry over with the same proofs, and we record the ones we use.

Proposition 1.2.30 (Additive arithmetic). Let $𝐮, 𝐯, 𝐰 \in 𝑉$ .

(a)The equation $𝐮 + 𝐱 = 𝐰$ has exactly one solution $𝐱 \in 𝑉$ , namely $𝐱 = 𝐰 - 𝐮$ .
(b)If $𝐮 + 𝐯 = 𝐮 + 𝐰$ , then $𝐯 = 𝐰$ .
(c) $- (- 𝐯) = 𝐯$ , and $- 𝟎 = 𝟎$ .

Proof.routine

Uses. Definition 1.2.1 (Vector space), Proposition 1.2.28 (Uniqueness of additive inverses), Notation 1.2.29, Proposition 1.1.13 (Solving $𝑎 + 𝑥 = 𝑏$ and $𝑎 𝑥 = 𝑏$ )

Techniques. Uniqueness via two candidates (Technique 1.1.8), Proving from axioms (Technique 1.1.7)

Hint 1. For (a), check that

𝐰 - 𝐮

is a solution. For uniqueness, add

- 𝐮

on the left of

𝐮 + 𝐱

and regroup.

Hint 2. (b) is (a) with

𝐰

replaced by

𝐮 + 𝐯

. For (c), to show a vector equals

- 𝐲

, show it is an additive inverse of

𝐲

Sketch. This is Proposition 1.1.13(a) and its consequences, whose proofs used only the additive axioms.

𝐮 +

(𝐰 - 𝐮) =

(𝐮 + (- 𝐮)) +

𝐰 =

𝐰

, and any solution satisfies

𝐱 =

(- 𝐮 + 𝐮) +

𝐱 =

- 𝐮 +

𝐰

. For (c),

𝐯

is an inverse of

- 𝐯

, and

𝟎 + 𝟎 = 𝟎

(a) First, $𝐰 - 𝐮$ is a solution:

\begin{aligned} 𝐮 + (𝐰 + (- 𝐮)) & = 𝐮 + ((- 𝐮) + 𝐰) & (A1) \\ = (𝐮 + (- 𝐮)) + 𝐰 & (A2) \\ = 𝟎 + 𝐰 = 𝐰 & (A4, then A1 and A3) . \end{aligned}

Conversely, suppose $𝐮 + 𝐱 = 𝐰$ . Then

\begin{aligned} 𝐱 & = 𝟎 + 𝐱 & (A1 and A3) \\ = ((- 𝐮) + 𝐮) + 𝐱 & (A4 and A1) \\ = (- 𝐮) + (𝐮 + 𝐱) & (A2) \\ = (- 𝐮) + 𝐰 = 𝐰 - 𝐮 & (hypothesis, then A1), \end{aligned}

so $𝐰 - 𝐮$ is the only solution.

(b) Both $𝐯$ and $𝐰$ solve $𝐮 + 𝐱 = 𝐮 + 𝐯$ , which has only one solution by (a). So $𝐯 = 𝐰$ .

(c) By (A1) and (A4), $(- 𝐯) +$ $𝐯 =$ $𝐯 +$ $(- 𝐯) =$ $𝟎$ , so $𝐯$ is an additive inverse of $- 𝐯$ , and $𝐯 = - (- 𝐯)$ by Proposition 1.2.28. Since $𝟎 + 𝟎 = 𝟎$ by (A3), $𝟎$ is an additive inverse of $𝟎$ , so $- 𝟎 = 𝟎$ . □

Remark 1.2.31. A set with an operation satisfying (A1)–(A4) is called an abelian group (Section 14.1). Everything in this subsection is a fact about abelian groups: scalars have not appeared yet, and the same statements hold in

(𝐹, +)

Scalars and vectors together

Intuition. Every fact so far used only (A1)–(A4). A fact relating the additive objects

0

𝟎

to scalar multiplication must use (A7) or (A8), the only axioms that involve addition and scalar multiplication together. For the scalar

0

, the additive fact to start from is

0 + 0 = 0

𝐹

, and (A8) carries it over to vectors, just as distributivity did for

0 𝑎 = 0

in a field (Proposition 1.1.15).

Proposition 1.2.32 (The scalar zero). For every

𝐯 \in 𝑉

0 𝐯 = 𝟎

Proof.routine

Uses. Definition 1.2.1 (Vector space), Proposition 1.2.30 (Additive arithmetic)

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. Which axiom can turn the scalar equation

0 + 0 = 0

into a statement about vectors?

Hint 2. Apply (A8) to

(0 + 0) 𝐯

. You get

0 𝐯 + 0 𝐯 = 0 𝐯

. Compare with

0 𝐯 + 𝟎 = 0 𝐯

and cancel.

Sketch. (A8) turns

0 + 0 = 0

into

0 𝐯 +

0 𝐯 =

0 𝐯 =

0 𝐯 +

𝟎

, and cancelling

0 𝐯

(Proposition 1.2.30(b)) gives

0 𝐯 = 𝟎

Let $𝐯 \in 𝑉$ . Then

\begin{aligned} 0 𝐯 + 0 𝐯 & = (0 + 0) 𝐯 & (A8) \\ = 0 𝐯 & (as 0 + 0 = 0 in 𝐹) \\ = 0 𝐯 + 𝟎 & (A3) . \end{aligned}

Cancelling $0 𝐯$ from both sides (Proposition 1.2.30(b)) gives $0 𝐯 = 𝟎$ . □

The companion statement is about the zero vector multiplied by any scalar. It is a different statement, and its first proof below uses the other distributive law. The second proof reduces it to Proposition 1.2.32.

Proposition 1.2.33 (The zero vector). For every

𝑎 \in 𝐹

𝑎 𝟎 = 𝟎

Proof (via (A7)).routine

Uses. Definition 1.2.1 (Vector space), Proposition 1.2.30 (Additive arithmetic)

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. Imitate the proof of Proposition 1.2.32, starting from the vector equation

𝟎 + 𝟎 = 𝟎

Hint 2. Apply (A7) to

𝑎 (𝟎 + 𝟎)

and cancel

𝑎 𝟎

Sketch. (A7) turns

𝟎 + 𝟎 = 𝟎

into

𝑎 𝟎 +

𝑎 𝟎 =

𝑎 𝟎 =

𝑎 𝟎 +

𝟎

; cancel

𝑎 𝟎

Let $𝑎 \in 𝐹$ . Then

\begin{aligned} 𝑎 𝟎 + 𝑎 𝟎 & = 𝑎 (𝟎 + 𝟎) & (A7) \\ = 𝑎 𝟎 & (A3) \\ = 𝑎 𝟎 + 𝟎 & (A3), \end{aligned}

and cancelling $𝑎 𝟎$ (Proposition 1.2.30(b)) gives $𝑎 𝟎 = 𝟎$ . □

Proof (from $0 𝐯 = 𝟎$ ).routine

Uses. Proposition 1.2.32 (The scalar zero), Proposition 1.1.15 (Multiplying by zero), Definition 1.2.1 (Vector space)

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. Proposition 1.2.32 says every vector times the scalar

0

𝟎

. Apply it to the vector

𝟎

to rewrite

𝟎

Hint 2. Write

𝑎 𝟎 = 𝑎 (0 𝟎)

and use (A6) to move the scalar

0

next to

𝑎

Sketch.

𝑎 𝟎 =

𝑎 (0 𝟎) =

(𝑎 0) 𝟎 =

0 𝟎 =

𝟎

, by Proposition 1.2.32, (A6),

𝑎 0 = 0

𝐹

, and Proposition 1.2.32 again.

Let $𝑎 \in 𝐹$ . By Proposition 1.2.32 applied to the vector $𝟎$ , we have $𝟎 = 0 𝟎$ . Using this, then (A6), then $𝑎 0 = 0$ in $𝐹$ (Proposition 1.1.15), and finally Proposition 1.2.32 once more,

𝑎 𝟎 = 𝑎 (0 𝟎) = (𝑎 0) 𝟎 = 0 𝟎 = 𝟎 .

□

The first proof is the template of Proposition 1.2.32 with the other distributive law. The second uses no distributive law directly: it writes $𝟎$ as $0 𝟎$ and moves the scalar $0$ through (A6), so the work is done by the field ( $𝑎 0 = 0$ ) and by the earlier result, through which (A8) enters. Reducing a new claim to one already proved is often shorter than starting again from the axioms.

Intuition. To show that a vector equals

- 𝐯

, it suffices to show that it is an additive inverse of

𝐯

, by Proposition 1.2.28. For

(- 1) 𝐯

, write

𝐯 = 1 𝐯

and combine the two terms with (A8).

Proposition 1.2.34 (Negatives and the scalar $- 1$ ). For all $𝑎 \in 𝐹$ and $𝐯 \in 𝑉$ :

(a) $(- 1) 𝐯 = - 𝐯$ ;
(b) $(- 𝑎) 𝐯 =$ $- (𝑎 𝐯) =$ $𝑎 (- 𝐯)$ .

Proof.routine

Uses. Proposition 1.2.28 (Uniqueness of additive inverses), Proposition 1.2.32 (The scalar zero), Proposition 1.1.16 (Sign rules), Definition 1.2.1 (Vector space)

Techniques. Uniqueness via two candidates (Technique 1.1.8), Proving from axioms (Technique 1.1.7)

Hint 1. For (a), compute

𝐯 + (- 1) 𝐯

. Write

𝐯

1 𝐯

first.

Hint 2. (A8) gives

(1 + (- 1)) 𝐯 = 0 𝐯

, and Proposition 1.2.32 finishes. Conclude with Proposition 1.2.28.

Hint 3. For (b), write

- 𝑎 = (- 1) 𝑎

(Proposition 1.1.16) and move the factor

- 1

with (A6).

Sketch.

𝐯 +

(- 1) 𝐯 =

1 𝐯 +

(- 1) 𝐯 =

(1 + (- 1)) 𝐯 =

0 𝐯 =

𝟎

, so

(- 1) 𝐯 = - 𝐯

by uniqueness of inverses. Then

(- 𝑎) 𝐯 =

((- 1) 𝑎) 𝐯 =

(- 1) (𝑎 𝐯) =

- (𝑎 𝐯)

, and

𝑎 (- 𝐯) =

𝑎 ((- 1) 𝐯) =

(𝑎 (- 1)) 𝐯 =

(- 𝑎) 𝐯

(a) We compute

\begin{aligned} 𝐯 + (- 1) 𝐯 & = 1 𝐯 + (- 1) 𝐯 & (A5) \\ = (1 + (- 1)) 𝐯 & (A8) \\ = 0 𝐯 = 𝟎, \end{aligned}

where the last step is Proposition 1.2.32. So $(- 1) 𝐯$ is an additive inverse of $𝐯$ , and $(- 1) 𝐯 = - 𝐯$ by Proposition 1.2.28.

(b) By Proposition 1.1.16, $- 𝑎 = (- 1) 𝑎 = 𝑎 (- 1)$ . Using (A6) and (a),

(- 𝑎) 𝐯 = ((- 1) 𝑎) 𝐯 = (- 1) (𝑎 𝐯) = - (𝑎 𝐯),

and

𝑎 (- 𝐯) = 𝑎 ((- 1) 𝐯) = (𝑎 (- 1)) 𝐯 = (- 𝑎) 𝐯 .

□

In particular $𝑎 (𝐯 - 𝐰) =$ $𝑎 𝐯 +$ $𝑎 (- 𝐰) =$ $𝑎 𝐯 -$ $𝑎 𝐰$ by (A7) and (b), and similarly $(𝑎 - 𝑏) 𝐯 =$ $𝑎 𝐯 -$ $𝑏 𝐯$ by (A8) and (b).

Intuition. In a field,

𝑎 𝑏 = 0

forces

𝑎 = 0

𝑏 = 0

(Proposition 1.1.17), and the proof multiplied by

𝑎^{- 1}

. The vector version has the same proof. It is the first place where the multiplicative inverses of

𝐹

matter for vectors.

Proposition 1.2.35 (Zero products). Let

𝑎 \in 𝐹

and

𝐯 \in 𝑉

. If

𝑎 𝐯 = 𝟎

, then

𝑎 = 0

𝐯 = 𝟎

. Consequently,

𝑎 𝐯 = 𝟎

if and only if

𝑎 = 0

𝐯 = 𝟎

Proof.routine

Uses. Definition 1.2.1 (Vector space), Proposition 1.2.33 (The zero vector), Proposition 1.2.32 (The scalar zero), Definition 1.1.1 (Field)

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. To prove “

𝑎 = 0

𝐯 = 𝟎

”, assume

𝑎 \neq 0

and prove

𝐯 = 𝟎

Hint 2. If

𝑎 \neq 0

, then

𝑎^{- 1}

exists. Write

𝐯 = 1 𝐯 = (𝑎^{- 1} 𝑎) 𝐯

and use (A6).

Sketch. If

𝑎 \neq 0

, then

𝐯 =

1 𝐯 =

(𝑎^{- 1} 𝑎) 𝐯 =

𝑎^{- 1} (𝑎 𝐯) =

𝑎^{- 1} 𝟎 =

𝟎

. The converse is Proposition 1.2.32 and Proposition 1.2.33.

Suppose $𝑎 𝐯 = 𝟎$ . If $𝑎 = 0$ we are done, so suppose $𝑎 \neq 0$ . Then $𝑎^{- 1}$ exists, and

\begin{aligned} 𝐯 & = 1 𝐯 & (A5) \\ = (𝑎^{- 1} 𝑎) 𝐯 & (multiplicative inverse in 𝐹) \\ = 𝑎^{- 1} (𝑎 𝐯) & (A6) \\ = 𝑎^{- 1} 𝟎 & (as 𝑎 𝐯 = 𝟎) \\ = 𝟎, \end{aligned}

where the last step is Proposition 1.2.33. Conversely, if $𝑎 = 0$ then $𝑎 𝐯 = 𝟎$ by Proposition 1.2.32, and if $𝐯 = 𝟎$ then $𝑎 𝐯 = 𝟎$ by Proposition 1.2.33. □

Corollary 1.2.36 (Cancelling scalars and vectors). Let $𝑎, 𝑏 \in 𝐹$ and $𝐯, 𝐰 \in 𝑉$ .

(a)If $𝑎 𝐯 = 𝑎 𝐰$ and $𝑎 \neq 0$ , then $𝐯 = 𝐰$ .
(b)If $𝑎 𝐯 = 𝑏 𝐯$ and $𝐯 \neq 𝟎$ , then $𝑎 = 𝑏$ .

Proof.routine

Uses. Proposition 1.2.35 (Zero products), Proposition 1.2.34 (Negatives and the scalar $- 1$ ), Proposition 1.2.30 (Additive arithmetic)

Hint 1. Move everything to one side:

𝑎 𝐯 = 𝑎 𝐰

is equivalent to

𝑎 𝐯 - 𝑎 𝐰 = 𝟎

. Factor.

Hint 2.

𝑎 𝐯 -

𝑎 𝐰 =

𝑎 (𝐯 - 𝐰)

and

𝑎 𝐯 -

𝑏 𝐯 =

(𝑎 - 𝑏) 𝐯

. Now use Proposition 1.2.35.

Sketch.

𝑎 (𝐯 - 𝐰) =

𝑎 𝐯 -

𝑎 𝐰 =

𝟎

with

𝑎 \neq 0

forces

𝐯 - 𝐰 = 𝟎

; and

(𝑎 - 𝑏) 𝐯 = 𝟎

with

𝐯 \neq 𝟎

forces

𝑎 - 𝑏 = 0

(a) By (A7) and Proposition 1.2.34(b), $𝑎 (𝐯 - 𝐰) =$ $𝑎 𝐯 +$ $𝑎 (- 𝐰) =$ $𝑎 𝐯 -$ $𝑎 𝐰$ , which is $𝟎$ because $𝑎 𝐯 = 𝑎 𝐰$ . Since $𝑎 \neq 0$ , Proposition 1.2.35 gives $𝐯 - 𝐰 = 𝟎$ . Hence, by (A3), (A4) with (A1), (A2), and (A1) with (A3),

𝐯 = 𝐯 + ((- 𝐰) + 𝐰) = (𝐯 - 𝐰) + 𝐰 = 𝟎 + 𝐰 = 𝐰 .

(b) By (A8) and Proposition 1.2.34(b), $(𝑎 - 𝑏) 𝐯 =$ $𝑎 𝐯 +$ $(- 𝑏) 𝐯 =$ $𝑎 𝐯 -$ $𝑏 𝐯 =$ $𝟎$ . Since $𝐯 \neq 𝟎$ , Proposition 1.2.35 gives $𝑎 - 𝑏 = 0$ , so $𝑎 = 𝑏$ . □

Part (b) says that if $𝐯 \neq 𝟎$ , the multiples $𝑎 𝐯$ for different scalars $𝑎$ are different vectors. So a vector space other than ${𝟎}$ has at least as many elements as $𝐹$ (Exercise 1.2.42).

Exercise 1.2.37.routine Let $𝑎 \in 𝐹$ and $𝐮, 𝐰 \in 𝑉$ .

(a)Suppose $𝑎 \neq 0$ . Show that the equation $𝑎 𝐱 + 𝐮 = 𝐰$ has exactly one solution $𝐱 \in 𝑉$ , and find it.
(b)Suppose $𝑎 = 0$ . For which $𝐮$ and $𝐰$ does the equation have a solution, and how many?

Solution.

Uses. Proposition 1.2.30 (Additive arithmetic), Corollary 1.2.36 (Cancelling scalars and vectors), Proposition 1.2.32 (The scalar zero), Definition 1.2.1 (Vector space)

Hint 1. First isolate

𝑎 𝐱

with Proposition 1.2.30(a), as you would isolate

𝑎 𝑥

in a field.

Hint 2. For existence, try

𝐱 = 𝑎^{- 1} (𝐰 - 𝐮)

. For uniqueness, two solutions

𝐱, 𝐱^{'}

have

𝑎 𝐱 = 𝑎 𝐱^{'}

Sketch. The equation holds if and only if

𝑎 𝐱 = 𝐰 - 𝐮

. If

𝑎 \neq 0

, then

𝐱 = 𝑎^{- 1} (𝐰 - 𝐮)

works, and it is the only solution by Corollary 1.2.36(a). If

𝑎 = 0

, then

𝑎 𝐱 = 𝟎

, so there is a solution exactly when

𝐮 = 𝐰

, and then every

𝐱

is one.

By Proposition 1.2.30(a), applied with the vector $𝑎 𝐱$ as the unknown, $𝑎 𝐱 + 𝐮 = 𝐰$ holds if and only if $𝑎 𝐱 = 𝐰 - 𝐮$ (using (A1) to write $𝑎 𝐱 + 𝐮 = 𝐮 + 𝑎 𝐱$ ).

(a) Let $𝐱 = 𝑎^{- 1} (𝐰 - 𝐮)$ . By (A6), $𝑎 𝐱 =$ $(𝑎 𝑎^{- 1}) (𝐰 - 𝐮) =$ $1 (𝐰 - 𝐮) =$ $𝐰 -$ $𝐮$ , so $𝐱$ is a solution. If $𝐱^{'}$ is another, then $𝑎 𝐱^{'} = 𝐰 - 𝐮 = 𝑎 𝐱$ , and $𝐱^{'} = 𝐱$ by Corollary 1.2.36(a). So $𝑎^{- 1} (𝐰 - 𝐮)$ is the only solution.

(b) If $𝑎 = 0$ , then $𝑎 𝐱 = 𝟎$ for every $𝐱$ by Proposition 1.2.32, so the equation reads $𝟎 + 𝐮 = 𝐰$ , that is, $𝐮 = 𝐰$ . So if $𝐮 = 𝐰$ , every $𝐱 \in 𝑉$ is a solution, and otherwise there is none.

Testing claims on small cases

Facts that are second nature in $ℝ^{2}$ can fail in other vector spaces, and the quickest way to find out is to compute in a small one.

Technique 1.2.38 (Try small cases and small fields). Before proving a general claim, test it on the smallest examples available: the zero space, $𝐹^{1}$ and $𝐹^{2}$ , and small fields, especially $𝔽_{2}$ , where $1 + 1 = 0$ . Small cases are where counterexamples live. When the claim survives them, the computation often shows why it is true and which hypothesis a proof must use.

Reach for it when a claim feels obvious from experience with $ℝ^{2}$ , and whenever a hypothesis such as $1 + 1 \neq 0$ or $𝑉 \neq {𝟎}$ might be needed. The canonical example is Example 1.2.39: $𝐯 = - 𝐯$ forces $𝐯 = 𝟎$ over $ℝ$ , but not over $𝔽_{2}$ .

Used in 21 places and 4 practice exercises

Exercise 1.2.49, Exercise 1.3.30, Exercise 1.3.43, Exercise 1.3.46, Exercise 1.4.32, Exercise 1.4.42, Exercise 1.5.36, Exercise 1.7.6, Exercise 1.7.13, Exercise 1.7.15, Exercise 2.8.5, Problem 2.8, Problem 2.14, Problem 2.20, Problem 3.6, Problem 4.17, Exercise 5.6.10, Exercise 5.6.15, Exercise 7.1.53, Exercise 7.5.31, Exercise 7.6.41; 4 practice exercises.

Example 1.2.39 (When does $𝐯 = - 𝐯$ force $𝐯 = 𝟎$ ?). Over $ℝ$ , a vector equal to its own negative is $𝟎$ . Over $𝔽_{2}$ this fails: in $𝔽_{2}^{2}$ , $(1, 0) +$ $(1, 0) =$ $(1 + 1, 0) =$ $(0, 0)$ , so $(1, 0) = - (1, 0)$ , although $(1, 0) \neq 𝟎$ . So the answer depends on the field, and the computation suggests why. By Proposition 1.2.28, $𝐯 = - 𝐯$ holds exactly when $𝐯 + 𝐯 = 𝟎$ , and by (A5) and (A8),

𝐯 + 𝐯 = 1 𝐯 + 1 𝐯 = (1 + 1) 𝐯 .

If $1 + 1 \neq 0$ in $𝐹$ , then $(1 + 1) 𝐯 = 𝟎$ forces $𝐯 = 𝟎$ by Proposition 1.2.35. If $1 + 1 = 0$ in $𝐹$ , then $𝐯 + 𝐯 = 0 𝐯 = 𝟎$ for every $𝐯$ , so every vector is its own negative. The test over $𝔽_{2}$ produced both the counterexample and the hypothesis $1 + 1 \neq 0$ under which the claim holds (compare Aside (Characteristic)).

Example 1.2.40 (Addition modulo $1$ , again). We show that no scalar multiplication makes $𝑉 = [0, 1)$ a real vector space with the addition $\oplus$ of Non-example 1.2.26. In every real vector space, $𝐯 + 𝐯 = 𝟎$ forces $𝐯 = 𝟎$ , by Example 1.2.39, because $1 + 1 = 2 \neq 0$ in $ℝ$ . But in $𝑉$ ,

\frac{1}{2} \oplus \frac{1}{2} = 1 mod 1 = 0,

and $1 / 2$ is not the zero vector $0$ (the zero vector is unique, Proposition 1.2.27, and $0$ satisfies (A3) for $\oplus$ ). The argument uses only consequences of the axioms that every candidate scalar multiplication must satisfy, so it rules them all out at once.

This is a different way to show that something is not a vector space. Instead of finding an axiom that fails for given operations, find a consequence of the axioms that involves only the part we cannot change, here the addition, and check that it fails.

Exercises

Exercise 1.2.41.routine Let $𝑝, 𝑞 \in 𝐹 [𝑥]$ be nonzero. Prove or disprove each statement.

(a) $deg (𝑝 + 𝑞) =$ $max (deg 𝑝, deg 𝑞)$ .
(b)If $deg (𝑝 + 𝑞) <$ $deg 𝑝$ , then $deg 𝑝 = deg 𝑞$ , and the leading coefficients of $𝑝$ and $𝑞$ add to $0$ .
(c)The set of polynomials in $ℝ [𝑥]$ of degree exactly $2$ , with the operations of $ℝ [𝑥]$ , is a real vector space.

Solution.

Uses. Proposition 1.2.18 (Degree and evaluation versus the operations), Definition 1.2.16 (Degree)

Hint 1. For (a), find two polynomials whose leading terms cancel.

Hint 2. (b) is the contrapositive of part of Proposition 1.2.18(a), plus a look at one coefficient. For (c), check closure.

Sketch. (a) False:

(𝑥^{2} + 1) + (- 𝑥^{2}) = 1

. (b) True: if

deg 𝑝 \neq deg 𝑞

, then

deg (𝑝 + 𝑞) =

max (deg 𝑝, deg 𝑞) \geq

deg 𝑝

; so

deg 𝑝 = deg 𝑞 = 𝑛

, and the

𝑛

-th coefficient

𝑎_{𝑛} + 𝑏_{𝑛}

𝑝 + 𝑞

must be

0

. (c) False:

𝑥^{2} + (𝑥 - 𝑥^{2}) = 𝑥

has degree

1

(a) False. Take $𝑝 = 𝑥^{2} + 1$ and $𝑞 = - 𝑥^{2}$ , both of degree $2$ . Then $𝑝 + 𝑞 = 1$ has degree $0 < 2$ .

(b) True. If $deg 𝑝 \neq deg 𝑞$ , then $deg (𝑝 + 𝑞) =$ $max (deg 𝑝, deg 𝑞) \geq$ $deg 𝑝$ by Proposition 1.2.18(a). So $deg (𝑝 + 𝑞) <$ $deg 𝑝$ forces $deg 𝑝 = deg 𝑞$ ; call it $𝑛$ . Write $𝑝 = (𝑎_{𝑘})$ and $𝑞 = (𝑏_{𝑘})$ . If $𝑎_{𝑛} + 𝑏_{𝑛} \neq 0$ , then the $𝑛$ -th coefficient of $𝑝 + 𝑞$ is nonzero and $deg (𝑝 + 𝑞) \geq$ $𝑛 =$ $deg 𝑝$ . So $𝑎_{𝑛} + 𝑏_{𝑛} = 0$ .

(c) False. The polynomials $𝑥^{2}$ and $𝑥 - 𝑥^{2}$ both have degree exactly $2$ , but their sum $𝑥$ has degree $1$ . So addition is not an operation on this set. (The zero polynomial, of degree $- \infty$ , is not in the set either.)

Exercise 1.2.42.routine Suppose

𝑉 \neq {𝟎}

. Show that

𝑉

has at least as many elements as

𝐹

. In particular, a vector space over an infinite field is either

{𝟎}

or infinite, and if

𝑉

is finite, then so is

𝐹

Solution.

Uses. Corollary 1.2.36 (Cancelling scalars and vectors)

Hint 1. Pick a nonzero vector

𝐯

and look at all of its multiples.

Hint 2. By Corollary 1.2.36(b), the map

𝐹 \to 𝑉

𝑎 \mapsto 𝑎 𝐯

, is injective.

Sketch. For

𝐯 \neq 𝟎

𝑎 \mapsto 𝑎 𝐯

is an injective map

𝐹 \to 𝑉

, so

𝑉

has at least

| 𝐹 |

elements.

Since

𝑉 \neq {𝟎}

, there is some

𝐯 \in 𝑉

with

𝐯 \neq 𝟎

. Define

𝑚 : 𝐹 \to 𝑉

𝑚 (𝑎) = 𝑎 𝐯

. If

𝑚 (𝑎) = 𝑚 (𝑏)

, then

𝑎 𝐯 = 𝑏 𝐯

, so

𝑎 = 𝑏

by Corollary 1.2.36(b). So

𝑚

is injective, and its image is a subset of

𝑉

with as many elements as

𝐹

. Hence

𝑉

has at least as many elements as

𝐹

. If

𝐹

is infinite, so is

𝑉

; and if

𝑉

is finite, then

𝐹

, which injects into

𝑉

, is finite.

Exercise 1.2.43.standard

(a)Find the error in the following argument.
Claim. If $𝐮, 𝐯 \in 𝑉$ and $𝐮 + 𝐯 = 𝐮$ , then $𝐯 = 𝟎$ .
Proof. By Proposition 1.2.27, $𝟎$ is the only vector $𝐳$ with $𝐮 + 𝐳 = 𝐮$ . Since $𝐮 + 𝐯 = 𝐮$ , we get $𝐯 = 𝟎$ .
(b)Prove the claim correctly.
(c)Give an example of a set with an addition satisfying (A1)–(A3), in which $𝐮 + 𝐯 = 𝐮$ for some $𝐮$ and some $𝐯$ that is not the identity element of (A3).

Solution.

Uses. Proposition 1.2.27 (Uniqueness of the zero vector), Proposition 1.2.30 (Additive arithmetic), Remark 1.2.2 (Reading the axioms)

Techniques. Uniqueness via two candidates (Technique 1.1.8)

Hint 1. Read Proposition 1.2.27 with its quantifiers. What property does it say only

𝟎

has?

Hint 2. For (b), compare

𝐮 + 𝐯 = 𝐮

with

𝐮 + 𝟎 = 𝐮

. For (c), the argument in (b) must use (A4), so drop it. Try

max

[0, \infty)

Sketch. (a) Proposition 1.2.27 is about vectors

𝐳

with

𝐰 + 𝐳 = 𝐰

for every

𝐰

; the hypothesis gives this for one

𝐮

. (b)

𝐮 + 𝐯 = 𝐮 + 𝟎

, so

𝐯 = 𝟎

by cancellation. (c) On

[0, \infty)

with

𝑠 + 𝑡 = max (𝑠, 𝑡)

, the identity is

0

, and

max (5, 3) = 5

(a) Proposition 1.2.27 says that only one vector $𝐳$ satisfies $𝐰 + 𝐳 = 𝐰$ for every $𝐰 \in 𝑉$ . It says nothing about vectors $𝐳$ that satisfy $𝐮 + 𝐳 = 𝐮$ for one particular $𝐮$ , and that is all the hypothesis gives for $𝐯$ . This is the quantifier distinction of Remark 1.2.2. The claim is true, but this argument does not prove it.

(b) By (A3), $𝐮 + 𝐯 = 𝐮 = 𝐮 + 𝟎$ . Cancelling $𝐮$ (Proposition 1.2.30(b)) gives $𝐯 = 𝟎$ .

(c) Let $𝑆 = [0, \infty)$ with $𝑠 + 𝑡 = max (𝑠, 𝑡)$ . This is commutative and associative, and $0$ satisfies (A3), since $max (𝑠, 0) = 𝑠$ for $𝑠 \geq 0$ . But $max (5, 3) = 5$ , so $5 + 3 = 5$ although $3 \neq 0$ . The proof in (b) breaks down because cancellation relied on (A4), which fails here: $max (5, 𝑡) \geq 5 > 0$ for every $𝑡$ .

Exercise 1.2.44.standard

(a)For which fields $𝐹$ is there a vector space over $𝐹$ with exactly two elements?
(b)Show that no vector space over $𝔽_{2}$ has exactly three elements, and no vector space over $𝔽_{3}$ has exactly four.

Solution.

Uses. Corollary 1.2.36 (Cancelling scalars and vectors), Proposition 1.2.30 (Additive arithmetic), Proposition 1.2.28 (Uniqueness of additive inverses), Proposition 1.2.34 (Negatives and the scalar $- 1$ ), Example 1.2.6 (Small fields)

Hint 1. For (a), a two-element space is

{𝟎, 𝐯}

with

𝐯 \neq 𝟎

. How many multiples does

𝐯

have?

Hint 2. For (b), take a vector space

𝑉

of the given size, and a vector outside the multiples of some nonzero

𝐯

. Where can its sum with

𝐯

lie?

Hint 3. Over

𝔽_{2}

- 𝐯 = (- 1) 𝐯 = 𝐯

; over

𝔽_{3}

- 𝐯 = 2 𝐯

. Rule out each possible value of the sum with cancellation.

Sketch. (a) If

𝑉 = {𝟎, 𝐯}

, then

𝑎 \mapsto 𝑎 𝐯

injects

𝐹

into

𝑉

, so

𝐹

has exactly two elements; conversely such an

𝐹

is a two-element vector space over itself. (b) In a three-element

𝔽_{2}

-space

{𝟎, 𝐮, 𝐰}

, the sum

𝐮 + 𝐰

can be none of

𝟎

𝐮

𝐰

. In a four-element

𝔽_{3}

-space,

𝟎, 𝐯, 2 𝐯

are distinct; for the fourth element

𝐰

, the sum

𝐰 + 𝐯

can be none of the four.

(a) Exactly the fields with two elements. Suppose $𝑉$ is a vector space over $𝐹$ with exactly two elements, so $𝑉 = {𝟎, 𝐯}$ with $𝐯 \neq 𝟎$ . By Corollary 1.2.36(b), $𝑎 \mapsto 𝑎 𝐯$ is an injective map $𝐹 \to 𝑉$ , so $𝐹$ has at most two elements. A field has at least two elements, $0 \neq 1$ , so $𝐹$ has exactly two. Conversely, if $𝐹$ has exactly two elements, then $𝐹 = 𝐹^{1}$ is a vector space over itself (Example 1.2.6) with two elements.

(b) Over $𝔽_{2}$ . Suppose $𝑉 = {𝟎, 𝐮, 𝐰}$ with three distinct elements. Consider $𝐮 + 𝐰 \in 𝑉$ . If $𝐮 + 𝐰 = 𝐮 = 𝐮 + 𝟎$ , then $𝐰 = 𝟎$ by cancellation (Proposition 1.2.30(b)). If $𝐮 + 𝐰 = 𝐰$ , then likewise $𝐮 = 𝟎$ . If $𝐮 + 𝐰 = 𝟎$ , then $𝐰 = - 𝐮 = (- 1) 𝐮$ by Proposition 1.2.28 and Proposition 1.2.34, and $- 1 = 1$ in $𝔽_{2}$ , so $𝐰 = 1 𝐮 = 𝐮$ . Each case contradicts distinctness.

Over $𝔽_{3}$ . Suppose $𝑉$ has exactly four elements, and pick $𝐯 \neq 𝟎$ in $𝑉$ . By Corollary 1.2.36(b), the vectors $0 𝐯 = 𝟎$ , $𝐯$ , and $2 𝐯$ are distinct; let $𝐰$ be the fourth element of $𝑉$ . Consider $𝐰 + 𝐯$ . If it equals $𝐰 = 𝐰 + 𝟎$ , then $𝐯 = 𝟎$ . If it equals $𝟎$ , then $𝐰 = - 𝐯 = (- 1) 𝐯 = 2 𝐯$ , as $- 1 = 2$ in $𝔽_{3}$ . If it equals $𝐯 = 𝟎 + 𝐯$ , then $𝐰 = 𝟎$ . If it equals $2 𝐯 = 𝐯 + 𝐯$ , then $𝐰 = 𝐯$ . Each case is a contradiction.

Exercise 1.2.45.standard For a positive integer $𝑛$ and $𝐯 \in 𝑉$ , write $𝑛 * 𝐯 = 𝐯 + \dots + 𝐯$ ( $𝑛$ terms), and similarly $𝑛 * 1 = 1 + \dots + 1 \in 𝐹$ (written $𝑛 \cdot 1$ in Definition 1.1.28; we use $*$ here because juxtaposition already means scalar multiplication). Suppose $𝑉 \neq {𝟎}$ . Show that the following are equivalent:

(i) $𝑛 * 1 = 0$ in $𝐹$ ;
(ii) $𝑛 * 𝐯 = 𝟎$ for every $𝐯 \in 𝑉$ ;
(iii) $𝑛 * 𝐯 = 𝟎$ for some $𝐯 \neq 𝟎$ in $𝑉$ .

Deduce that $char 𝐹$ (Definition 1.1.28) can be read off from the addition of $𝑉$ : it is the least $𝑛 \geq 1$ with $𝑛 * 𝐯 = 𝟎$ for all $𝐯 \in 𝑉$ , or $0$ if there is none.

Solution.

Uses. Definition 1.2.1 (Vector space), Proposition 1.2.32 (The scalar zero), Proposition 1.2.35 (Zero products), Definition 1.1.28 (Characteristic)

Hint 1. First show

𝑛 * 𝐯 = (𝑛 * 1) 𝐯

by induction on

𝑛

, using (A5) and (A8).

Hint 2. Then (i)

\Rightarrow

(ii) is Proposition 1.2.32, (ii)

\Rightarrow

(iii) uses

𝑉 \neq {𝟎}

, and (iii)

\Rightarrow

(i) is Proposition 1.2.35.

Sketch. By (A5), (A8), and induction,

𝑛 * 𝐯 = (𝑛 * 1) 𝐯

. If

𝑛 * 1 = 0

this is

𝟎

for all

𝐯

; some nonzero

𝐯

exists; and

(𝑛 * 1) 𝐯 = 𝟎

with

𝐯 \neq 𝟎

forces

𝑛 * 1 = 0

. So the

𝑛

killing all of

𝑉

are exactly those with

𝑛 * 1 = 0

First, $𝑛 * 𝐯 = (𝑛 * 1) 𝐯$ for every $𝑛 \geq 1$ , by induction on $𝑛$ . For $𝑛 = 1$ this is (A5). If it holds for $𝑛$ , then

(𝑛 + 1) * 𝐯 = 𝑛 * 𝐯 + 𝐯 = (𝑛 * 1) 𝐯 + 1 𝐯 = (𝑛 * 1 + 1) 𝐯 = ((𝑛 + 1) * 1) 𝐯,

by the inductive hypothesis, (A5), and (A8).

(i) $\Rightarrow$ (ii). If $𝑛 * 1 = 0$ , then $𝑛 * 𝐯 = 0 𝐯 = 𝟎$ for every $𝐯$ , by Proposition 1.2.32.

(ii) $\Rightarrow$ (iii). Since $𝑉 \neq {𝟎}$ , some $𝐯 \neq 𝟎$ exists, and $𝑛 * 𝐯 = 𝟎$ by (ii).

(iii) $\Rightarrow$ (i). If $𝐯 \neq 𝟎$ and $(𝑛 * 1) 𝐯 = 𝑛 * 𝐯 = 𝟎$ , then $𝑛 * 1 = 0$ by Proposition 1.2.35.

By the equivalence of (i) and (ii), the positive integers $𝑛$ with $𝑛 * 𝐯 = 𝟎$ for all $𝐯$ are exactly those with $𝑛 * 1 = 0$ . So the least such $𝑛$ is $char 𝐹$ , and there is none exactly when $char 𝐹 = 0$ .

Exercise 1.2.46.standard

(a)Is there a scalar multiplication that makes $ℤ$ , with its usual addition, a vector space over $𝔽_{𝑝}$ ?
(b)Is there a field $𝐹$ and a scalar multiplication that make $ℤ / 4 ℤ$ , with its usual addition, a vector space over $𝐹$ ?

Solution.

Uses. Exercise 1.2.45, Example 1.2.39 (When does $𝐯 = - 𝐯$ force $𝐯 = 𝟎$ ?), Proposition 1.2.27 (Uniqueness of the zero vector)

Hint 1. Whatever the scalar multiplication is, Exercise 1.2.45 constrains the addition. What does it say about

𝑝 \cdot 𝐯

over

𝔽_{𝑝}

Hint 2. For (b), split into the cases

1 + 1 = 0

and

1 + 1 \neq 0

𝐹

, and look at the elements

1

and

2

ℤ / 4 ℤ

Sketch. (a) No: over

𝔽_{𝑝}

𝑝 * 𝐯 = 𝟎

for every

𝐯

, but

𝑝 * 1 = 𝑝 \neq 0

ℤ

. (b) No: if

1 + 1 = 0

𝐹

, then

𝐯 + 𝐯 = 𝟎

for all

𝐯

, but

1 + 1 = 2 \neq 0

ℤ / 4 ℤ

; if

1 + 1 \neq 0

, then

𝐯 + 𝐯 = 𝟎

forces

𝐯 = 𝟎

, but

2 + 2 = 0

with

2 \neq 0

(a) No. Suppose some scalar multiplication made $ℤ$ a vector space over $𝔽_{𝑝}$ . Its zero vector is the identity element $0$ of the addition, by Proposition 1.2.27. In $𝔽_{𝑝}$ we have $𝑝 * 1 = 0$ , so by Exercise 1.2.45, adding any vector to itself $𝑝$ times gives the zero vector. But adding $1 \in ℤ$ to itself $𝑝$ times gives $𝑝 \neq 0$ .

(b) No. Suppose $𝐹$ and a scalar multiplication make $ℤ / 4 ℤ$ an $𝐹$ -vector space; again its zero vector is $0$ . If $1 + 1 = 0$ in $𝐹$ , then $𝐯 + 𝐯 = 𝟎$ for every $𝐯$ (Example 1.2.39), but $1 + 1 = 2 \neq 0$ in $ℤ / 4 ℤ$ . If $1 + 1 \neq 0$ in $𝐹$ , then $𝐯 + 𝐯 = 𝟎$ forces $𝐯 = 𝟎$ (Example 1.2.39), but $2 + 2 = 0$ in $ℤ / 4 ℤ$ although $2 \neq 0$ . Both cases are impossible.

Exercise 1.2.47.standard Is there a scalar multiplication

⊙ : ℂ \times ℝ \to ℝ

that makes

ℝ

, with its usual addition, a complex vector space, and that agrees with the usual product for real scalars, that is,

𝑎 ⊙ 𝑥 = 𝑎 𝑥

for

𝑎 \in ℝ

Solution.

Uses. Definition 1.2.1 (Vector space)

Hint 1. Suppose such a

⊙

exists, and let

𝑟 = 𝑖 ⊙ 1 \in ℝ

. Compute

𝑖 ⊙ (𝑖 ⊙ 1)

in two ways.

Hint 2. One way uses (A6):

𝑖 ⊙ (𝑖 ⊙ 1) = (𝑖^{2}) ⊙ 1

. For the other, write

𝑟 = 𝑟 ⊙ 1

and use (A6) and commutativity in

ℂ

to move

𝑖

past

𝑟

Sketch. No. With

𝑟 = 𝑖 ⊙ 1

, (A6) gives

𝑖 ⊙ 𝑟 =

𝑖 ⊙ (𝑟 ⊙ 1) =

(𝑟 𝑖) ⊙ 1 =

𝑟 ⊙ (𝑖 ⊙ 1) =

𝑟^{2}

, and also

𝑖 ⊙ 𝑟 =

𝑖 ⊙ (𝑖 ⊙ 1) =

(- 1) ⊙ 1 =

- 1

. So

𝑟^{2} = - 1

with

𝑟 \in ℝ

, which is impossible.

No. Suppose such a $⊙$ exists, and let $𝑟 = 𝑖 ⊙ 1$ , a real number. Since $𝑟$ is real, $𝑟 = 𝑟 \cdot 1 = 𝑟 ⊙ 1$ . By (A6) and commutativity of multiplication in $ℂ$ ,

𝑖 ⊙ 𝑟 = 𝑖 ⊙ (𝑟 ⊙ 1) = (𝑖 𝑟) ⊙ 1 = (𝑟 𝑖) ⊙ 1 = 𝑟 ⊙ (𝑖 ⊙ 1) = 𝑟 ⊙ 𝑟 = 𝑟^{2} .

On the other hand, by (A6),

𝑖 ⊙ 𝑟 = 𝑖 ⊙ (𝑖 ⊙ 1) = (𝑖 𝑖) ⊙ 1 = (- 1) ⊙ 1 = - 1 .

So $𝑟^{2} = - 1$ , which no real number satisfies.

The requirement that real scalars act as usual is essential. Without it, there is a complex vector space structure on $(ℝ, +)$ , but its construction uses the axiom of choice (Section 15.1).

Exercise 1.2.48.standard Keep the usual addition on $ℂ^{2}$ . For $𝑎 = 𝑥 + 𝑦 𝑖 \in ℂ$ with $𝑥, 𝑦 \in ℝ$ , write $𝑎 = 𝑥 - 𝑦 𝑖$ , $| 𝑎 | = \sqrt{𝑥^{2} + 𝑦^{2}}$ , and $Re 𝑎 = 𝑥$ . Which of the following scalar multiplications make $ℂ^{2}$ a complex vector space?

(a) $𝑎 ⊙ 𝐯 = 𝑎 𝐯$ .
(b) $𝑎 ⊙ 𝐯 = | 𝑎 | 𝐯$ .
(c) $𝑎 ⊙ 𝐯 = (Re 𝑎) 𝐯$ .

Solution.

Uses. Proposition 1.2.5 ( $𝐹^{𝑛}$ is a vector space), Definition 1.1.29 (Complex numbers)

Hint 1. Addition is unchanged, so only (A5)–(A8) need checking, and each reduces to a property of the function

𝑎 \mapsto 𝑎

| 𝑎 |

, or

Re 𝑎

Hint 2. (A6) needs the function to be multiplicative, and (A8) needs it to be additive. Check

𝑎 + 𝑏 = 𝑎 + 𝑏

and

𝑎 𝑏 = 𝑎 𝑏

by writing

𝑎

and

𝑏

in coordinates.

Sketch. (a) Yes: conjugation fixes

1

, and is additive and multiplicative. (b) No: (A8) fails for

𝑎 = 1

𝑏 = - 1

, since

| 0 | 𝐯 = 𝟎

but

| 1 | 𝐯 + | - 1 | 𝐯 = 2 𝐯

. (c) No: (A6) fails for

𝑎 = 𝑏 = 𝑖

, since

Re (𝑖^{2}) = - 1

but

Re (𝑖) Re (𝑖) = 0

Since the addition is that of $ℂ^{2}$ , (A1)–(A4) hold in all three cases. For $𝑔 : ℂ \to ℂ$ and $𝑎 ⊙ 𝐯 = 𝑔 (𝑎) 𝐯$ , (A7) also holds, as $𝑔 (𝑎) (𝐯 + 𝐰) =$ $𝑔 (𝑎) 𝐯 +$ $𝑔 (𝑎) 𝐰$ in $ℂ^{2}$ . The remaining axioms read $𝑔 (1) 𝐯 = 𝐯$ , $𝑔 (𝑎 𝑏) 𝐯 =$ $𝑔 (𝑎) 𝑔 (𝑏) 𝐯$ , and $𝑔 (𝑎 + 𝑏) 𝐯 =$ $𝑔 (𝑎) 𝐯 +$ $𝑔 (𝑏) 𝐯$ .

(a) Yes. First, $1 = 1$ , which gives (A5). Write $𝑎 = 𝑥 + 𝑦 𝑖$ and $𝑏 = 𝑢 + 𝑤 𝑖$ . Then $𝑎 + 𝑏 =$ $(𝑥 + 𝑢) -$ $(𝑦 + 𝑤) 𝑖 =$ $𝑎 +$ $𝑏$ , which gives (A8). Also $𝑎 𝑏 =$ $(𝑥 𝑢 - 𝑦 𝑤) +$ $(𝑥 𝑤 + 𝑦 𝑢) 𝑖$ , so $𝑎 𝑏 =$ $(𝑥 𝑢 - 𝑦 𝑤) -$ $(𝑥 𝑤 + 𝑦 𝑢) 𝑖$ , while

𝑎 𝑏 = (𝑥 - 𝑦 𝑖) (𝑢 - 𝑤 𝑖) = (𝑥 𝑢 - 𝑦 𝑤) + (- 𝑥 𝑤 - 𝑦 𝑢) 𝑖 .

These agree, so $𝑎 𝑏 𝐯 = 𝑎 (𝑏 𝐯)$ , which gives (A6).

(b) No. Take $𝑎 = 1$ , $𝑏 = - 1$ , and $𝐯 = (1, 0)$ . Then $(𝑎 + 𝑏) ⊙ 𝐯 =$ $| 0 | 𝐯 =$ $𝟎$ , while $𝑎 ⊙ 𝐯 +$ $𝑏 ⊙ 𝐯 =$ $| 1 | 𝐯 +$ $| -$ $1 | 𝐯 =$ $(2, 0)$ . So (A8) fails.

(c) No. Take $𝑎 = 𝑏 = 𝑖$ and $𝐯 = (1, 0)$ . Then $(𝑎 𝑏) ⊙ 𝐯 =$ $Re (- 1) 𝐯 =$ $(- 1, 0)$ , while $𝑎 ⊙ (𝑏 ⊙ 𝐯) =$ $Re (𝑖) (Re (𝑖) 𝐯) =$ $𝟎$ . So (A6) fails. (Here (A5) and (A8) do hold, since $Re 1 = 1$ and $Re$ is additive.)

Exercise 1.2.49.standard Let

𝐹

be a field. On

𝐹^{2}

, keep the usual addition and define

𝑎 ⊙ (𝑥, 𝑦) =

(𝑎 𝑥, 𝑎^{2} 𝑦)

. For which fields

𝐹

is this a vector space over

𝐹

Solution.

Uses. Proposition 1.2.5 ( $𝐹^{𝑛}$ is a vector space)

Techniques. Try small cases and small fields (Technique 1.2.38)

Hint 1. Test small cases first: try

𝑎 = 𝑏 = 1

in (A8) over

ℝ

, and then over

𝔽_{2}

Hint 2. All axioms except (A8) hold for every

𝐹

. (A8) holds if and only if

{(𝑎 + 𝑏)}^{2} = 𝑎^{2} + 𝑏^{2}

for all

𝑎, 𝑏 \in 𝐹

Sketch. (A1)–(A7) always hold:

1^{2} = 1

and

{(𝑎 𝑏)}^{2} = 𝑎^{2} 𝑏^{2}

. (A8) in the second coordinate needs

{(𝑎 + 𝑏)}^{2} = 𝑎^{2} + 𝑏^{2}

, that is,

(1 + 1) 𝑎 𝑏 = 0

for all

𝑎, 𝑏

. This holds for all

𝑎, 𝑏

exactly when

1 + 1 = 0

. So the answer is: the fields with

1 + 1 = 0

Exactly the fields with $1 + 1 = 0$ .

Over $ℝ$ , (A8) fails: $(1 + 1) ⊙ (0, 1) = (0, 4)$ , while $1 ⊙ (0, 1) +$ $1 ⊙ (0, 1) =$ $(0, 2)$ . Over $𝔽_{2}$ , $𝑎^{2} = 𝑎$ for both elements, so $⊙$ is the usual scalar multiplication and all axioms hold. These two small cases suggest that $1 + 1$ is what matters.

In general, (A1)–(A4) hold because the addition is that of $𝐹^{2}$ . (A5) holds since $1^{2} = 1$ . (A6) holds since $(𝑎 𝑏) ⊙ (𝑥, 𝑦) =$ $(𝑎 𝑏 𝑥, 𝑎^{2} 𝑏^{2} 𝑦) =$ $𝑎 ⊙ (𝑏 𝑥, 𝑏^{2} 𝑦)$ . (A7) holds since $𝑎 ⊙ (𝑥 + 𝑥^{'}, 𝑦 + 𝑦^{'}) =$ $(𝑎 𝑥 + 𝑎 𝑥^{'}, 𝑎^{2} 𝑦 + 𝑎^{2} 𝑦^{'})$ . For (A8), the first coordinates of $(𝑎 + 𝑏) ⊙ (𝑥, 𝑦)$ and $𝑎 ⊙ (𝑥, 𝑦) +$ $𝑏 ⊙ (𝑥, 𝑦)$ agree, and the second coordinates are ${(𝑎 + 𝑏)}^{2} 𝑦$ and $(𝑎^{2} + 𝑏^{2}) 𝑦$ . Since ${(𝑎 + 𝑏)}^{2} =$ $𝑎^{2} +$ $(1 + 1) 𝑎 𝑏 +$ $𝑏^{2}$ , (A8) holds for all $𝑎, 𝑏, 𝑥, 𝑦$ if and only if $(1 + 1) 𝑎 𝑏 𝑦 = 0$ for all $𝑎, 𝑏, 𝑦 \in 𝐹$ . If $1 + 1 = 0$ , this holds. If $1 + 1 \neq 0$ , it fails for $𝑎 = 𝑏 = 𝑦 = 1$ . So $𝐹^{2}$ with $⊙$ is a vector space exactly when $1 + 1 = 0$ in $𝐹$ .

Over $𝔽_{2}$ the new structure is the usual one. Over a larger field with $1 + 1 = 0$ , such as the four-element field of Exercise 1.1.56, $𝑎 \mapsto 𝑎^{2}$ is not the identity, so $⊙$ is a different scalar multiplication from the usual one.

Exercise 1.2.50.standard Let

𝑃 =

{(𝑥, 𝑦) \in ℝ^{2} | 𝑦 = 𝑥^{2}}

be the parabola. Find operations that make

𝑃

a real vector space in which

(1, 1) \oplus (2, 4) = (3, 9)

. In your structure, find the zero vector, the additive inverse of

(2, 4)

, and

3 ⊙ (- 1, 1)

Solution.

Uses. Proposition 1.2.22 (Transport of structure)

Hint 1. The parabola is in bijection with

ℝ

. Transport the structure of

ℝ

(Proposition 1.2.22).

Hint 2. Use

𝜑 (𝑡) = (𝑡, 𝑡^{2})

, whose inverse is

(𝑥, 𝑦) \mapsto 𝑥

Sketch. Transport along

𝜑 (𝑡) = (𝑡, 𝑡^{2})

(𝑠, 𝑠^{2}) \oplus

(𝑡, 𝑡^{2}) =

(𝑠 + 𝑡, {(𝑠 + 𝑡)}^{2})

and

𝑎 ⊙ (𝑡, 𝑡^{2}) =

(𝑎 𝑡, 𝑎^{2} 𝑡^{2})

. Then

(1, 1) \oplus

(2, 4) =

𝜑 (3) =

(3, 9)

, the zero vector is

(0, 0)

- (2, 4) = (- 2, 4)

, and

3 ⊙ (- 1, 1) = (- 3, 9)

Define $𝜑 : ℝ \to 𝑃$ by $𝜑 (𝑡) = (𝑡, 𝑡^{2})$ . It is a bijection: each point of $𝑃$ is $(𝑥, 𝑥^{2}) = 𝜑 (𝑥)$ , and $𝜑 (𝑠) = 𝜑 (𝑡)$ forces $𝑠 = 𝑡$ by comparing first coordinates. So $𝜑^{- 1} (𝑥, 𝑦) = 𝑥$ . By Proposition 1.2.22, $𝑃$ is a real vector space with

(𝑠, 𝑠^{2}) \oplus (𝑡, 𝑡^{2}) = 𝜑 (𝑠 + 𝑡) = (𝑠 + 𝑡, {(𝑠 + 𝑡)}^{2}), 𝑎 ⊙ (𝑡, 𝑡^{2}) = 𝜑 (𝑎 𝑡) = (𝑎 𝑡, 𝑎^{2} 𝑡^{2}) .

In particular $(1, 1) \oplus$ $(2, 4) =$ $𝜑 (1 + 2) =$ $(3, 9)$ . The zero vector is $𝜑 (0) = (0, 0)$ . The inverse of $(2, 4) = 𝜑 (2)$ is $𝜑 (- 2) = (- 2, 4)$ , since $𝜑 (2) \oplus$ $𝜑 (- 2) =$ $𝜑 (0)$ . Finally $3 ⊙ (- 1, 1) =$ $𝜑 (3 \cdot (- 1)) =$ $(- 3, 9)$ .

The set $𝑃$ is curved, but that is no obstacle: the structure lives in the operations, and only the operations of $ℝ^{2}$ fail to preserve $𝑃$ .

Exercise 1.2.51.standard

(a)Find all $𝑓 \in 𝔽_{2} {[𝑥]}_{\leq 3}$ whose polynomial function $𝔽_{2} \to 𝔽_{2}$ is the zero function. How many polynomials does $𝔽_{2} {[𝑥]}_{\leq 3}$ contain, and how many functions $𝔽_{2} \to 𝔽_{2}$ are there? When do two polynomials in $𝔽_{2} {[𝑥]}_{\leq 3}$ have the same polynomial function?
(b)Let $𝑝$ be a prime. Show that $𝑥^{𝑝} - 𝑥 \in 𝔽_{𝑝} [𝑥]$ is a nonzero polynomial whose polynomial function is zero.

Solution.

Uses. Definition 1.2.17 (Evaluation), Proposition 1.2.18 (Degree and evaluation versus the operations), Exercise 1.1.51

Hint 1. Write

𝑓 =

𝑎_{0} +

𝑎_{1} 𝑥 +

𝑎_{2} 𝑥^{2} +

𝑎_{3} 𝑥^{3}

and compute

𝑓 (0)

and

𝑓 (1)

in terms of the coefficients.

Hint 2. Two polynomials have the same function exactly when their difference has the zero function, by Proposition 1.2.18(c). For (b), use Fermat’s little theorem (Exercise 1.1.51).

Sketch. (a)

𝑓 (0) = 𝑎_{0}

and

𝑓 (1) =

𝑎_{0} +

𝑎_{1} +

𝑎_{2} +

𝑎_{3}

, so the zero function needs

𝑎_{0} = 0

and an even number of

𝑎_{1}, 𝑎_{2}, 𝑎_{3}

equal to

1

𝑓 \in

{𝟎, 𝑥 + 𝑥^{2}, 𝑥 + 𝑥^{3}, 𝑥^{2} + 𝑥^{3}}

. There are

16

polynomials and

4

functions;

𝑓

and

𝑔

agree as functions iff

𝑓 - 𝑔

is one of these four. (b) Coefficient

1

in position

𝑝

, so nonzero;

𝑐^{𝑝} = 𝑐

for all

𝑐 \in 𝔽_{𝑝}

(a) Write $𝑓 =$ $𝑎_{0} +$ $𝑎_{1} 𝑥 +$ $𝑎_{2} 𝑥^{2} +$ $𝑎_{3} 𝑥^{3}$ with $𝑎_{𝑘} \in 𝔽_{2}$ . Then $𝑓 (0) = 𝑎_{0}$ and $𝑓 (1) =$ $𝑎_{0} +$ $𝑎_{1} +$ $𝑎_{2} +$ $𝑎_{3}$ . So the polynomial function of $𝑓$ is zero exactly when $𝑎_{0} = 0$ and $𝑎_{1} + 𝑎_{2} + 𝑎_{3} = 0$ in $𝔽_{2}$ , that is, when an even number of $𝑎_{1}, 𝑎_{2}, 𝑎_{3}$ equal $1$ . This gives

𝑓 \in {𝟎, 𝑥 + 𝑥^{2}, 𝑥 + 𝑥^{3}, 𝑥^{2} + 𝑥^{3}} .

$𝔽_{2} {[𝑥]}_{\leq 3}$ has $2^{4} = 16$ elements, one for each choice of $(𝑎_{0}, 𝑎_{1}, 𝑎_{2}, 𝑎_{3})$ , while there are $2^{2} = 4$ functions $𝔽_{2} \to 𝔽_{2}$ . By Proposition 1.2.18(c), $𝑓$ and $𝑔$ have the same polynomial function if and only if $(𝑓 - 𝑔) (𝑐) =$ $𝑓 (𝑐) -$ $𝑔 (𝑐) =$ $0$ for all $𝑐$ , that is, if and only if $𝑓 - 𝑔$ is one of the four polynomials above.

(b) The coefficient of $𝑥^{𝑝} - 𝑥$ in position $𝑝 \geq 2$ is $1$ , so $𝑥^{𝑝} - 𝑥 \neq 𝟎$ ; in fact $deg (𝑥^{𝑝} - 𝑥) = 𝑝$ . Its value at $𝑐 \in 𝔽_{𝑝}$ is $𝑐^{𝑝} - 𝑐$ . For $𝑐 = 0$ this is $0$ . For $𝑐 \neq 0$ , Fermat’s little theorem (Exercise 1.1.51) gives $𝑐^{𝑝 - 1} = 1$ , so $𝑐^{𝑝} = 𝑐$ and $𝑐^{𝑝} - 𝑐 = 0$ .

Exercise 1.2.52.standard Let $𝑉$ be a vector space over $ℚ$ .

(a)Show that for every integer $𝑛 \geq 1$ and $𝐰 \in 𝑉$ , the scalar multiple $𝑛 𝐰$ equals $𝐰 + \dots + 𝐰$ ( $𝑛$ terms).
(b)Let $𝑞 = 𝑚 / 𝑛$ with $𝑚, 𝑛 \in ℤ$ and $𝑛 \geq 1$ . Show that $𝑞 𝐯$ is the only vector $𝐰 \in 𝑉$ with $𝑛 𝐰 = 𝑚 𝐯$ .
(c)Deduce that the scalar multiplication of a $ℚ$ -vector space is determined by its addition: if $⊙$ and $⊙^{'}$ both make $(𝑉, +)$ a vector space over $ℚ$ , then $𝑞 ⊙ 𝐯 = 𝑞 ⊙^{'} 𝐯$ for all $𝑞$ and $𝐯$ .

Solution.

Uses. Definition 1.2.1 (Vector space), Corollary 1.2.36 (Cancelling scalars and vectors), Proposition 1.2.34 (Negatives and the scalar $- 1$ ), Proposition 1.2.28 (Uniqueness of additive inverses), Proposition 1.2.27 (Uniqueness of the zero vector), Proposition 1.2.32 (The scalar zero)

Techniques. Uniqueness via two candidates (Technique 1.1.8)

Hint 1. (a) is induction with (A5) and (A8). For (b), check that

𝑞 𝐯

works using (A6), and get uniqueness from Corollary 1.2.36(a).

Hint 2. For (c), show first that

𝑚 ⊙ 𝐯

, for every integer

𝑚

, is expressible using only

+

and negatives. Then use (b) for both structures.

Sketch. (a)

(𝑛 + 1) 𝐰 = 𝑛 𝐰 + 1 𝐰

. (b)

𝑛 (𝑞 𝐯) =

(𝑛 𝑞) 𝐯 =

𝑚 𝐯

, and

𝑛 𝐰 = 𝑛 𝐰^{'}

with

𝑛 \neq 0

forces

𝐰 = 𝐰^{'}

. (c) By (a),

𝑚 ⊙ 𝐯

is an

𝑚

-fold sum for

𝑚 \geq 1

, it is

𝟎

for

𝑚 = 0

, and

- ((- 𝑚) ⊙ 𝐯)

for

𝑚 < 0

; so it agrees with

𝑚 ⊙^{'} 𝐯

. Then

𝑞 ⊙ 𝐯

and

𝑞 ⊙^{'} 𝐯

both solve “

𝐰 + \dots + 𝐰 = 𝑚 ⊙ 𝐯

” (

𝑛

terms), which has at most one solution.

(a) By induction on $𝑛$ . For $𝑛 = 1$ , $1 𝐰 = 𝐰$ by (A5). If $𝑛 𝐰$ is the $𝑛$ -fold sum, then $(𝑛 + 1) 𝐰 =$ $𝑛 𝐰 +$ $1 𝐰 =$ $𝑛 𝐰 +$ $𝐰$ by (A8) and (A5), which is the $(𝑛 + 1)$ -fold sum.

(b) By (A6), $𝑛 (𝑞 𝐯) =$ $(𝑛 𝑞) 𝐯 =$ $𝑚 𝐯$ , so $𝑞 𝐯$ is such a vector. If $𝑛 𝐰 = 𝑚 𝐯 = 𝑛 𝐰^{'}$ , then $𝐰 = 𝐰^{'}$ by Corollary 1.2.36(a), since $𝑛 \neq 0$ in $ℚ$ .

(c) Let $⊙$ and $⊙^{'}$ be as stated; both have the same addition, hence the same zero vector and the same negatives (Proposition 1.2.27 and Proposition 1.2.28). First let $𝑚 \in ℤ$ . If $𝑚 \geq 1$ , then by (a) both $𝑚 ⊙ 𝐯$ and $𝑚 ⊙^{'} 𝐯$ equal the $𝑚$ -fold sum $𝐯 + \dots + 𝐯$ . If $𝑚 = 0$ , both are $𝟎$ by Proposition 1.2.32. If $𝑚 < 0$ , then by Proposition 1.2.34(b), applied to both structures,

𝑚 ⊙ 𝐯 = - ((- 𝑚) ⊙ 𝐯) = - ((- 𝑚) ⊙^{'} 𝐯) = 𝑚 ⊙^{'} 𝐯 .

So $𝑚 ⊙ 𝐯 = 𝑚 ⊙^{'} 𝐯$ for every integer $𝑚$ .

Now let $𝑞 = 𝑚 / 𝑛$ with $𝑛 \geq 1$ , and write $𝑛 * 𝐰$ for the $𝑛$ -fold sum $𝐰 + \dots + 𝐰$ , which by (a) equals both $𝑛 ⊙ 𝐰$ and $𝑛 ⊙^{'} 𝐰$ . By (b) for $⊙$ , the vector $𝑞 ⊙ 𝐯$ satisfies $𝑛 * (𝑞 ⊙ 𝐯) = 𝑚 ⊙ 𝐯$ , and it is the only vector $𝐰$ with $𝑛 * 𝐰 = 𝑚 ⊙ 𝐯$ . By (b) for $⊙^{'}$ , $𝑛 * (𝑞 ⊙^{'} 𝐯) =$ $𝑚 ⊙^{'} 𝐯 =$ $𝑚 ⊙ 𝐯$ . So $𝑞 ⊙^{'} 𝐯$ is also such a vector, and $𝑞 ⊙^{'} 𝐯 = 𝑞 ⊙ 𝐯$ .

For instance, the usual product is the only scalar multiplication that makes $(ℝ, +)$ a vector space over $ℚ$ . Compare Exercise 1.2.47, where the scalars were $ℂ$ .

Exercise 1.2.53.hard Let $𝑉 = (- 1, 1)$ , and define $𝑢 \oplus$ $𝑣 =$ $(𝑢 + 𝑣) / (1 + 𝑢 𝑣)$ . (In units where the speed of light is $1$ , this is how velocities along a line add in special relativity.)

(a)Show that $𝑢 \oplus 𝑣 \in 𝑉$ for $𝑢, 𝑣 \in 𝑉$ .
(b)Find a scalar multiplication $⊙$ that makes $(𝑉, \oplus, ⊙)$ a real vector space. Compute $2 ⊙ 𝑣$ and $(1 / 2) ⊙ 𝑣$ explicitly, and identify the zero vector and the additive inverse of $𝑣$ .

Solution.

Uses. Proposition 1.2.22 (Transport of structure), $tanh : ℝ \to (- 1, 1)$ is a bijection, the addition formula for $tanh$

Hint 1. For (a), compare

1 + 𝑢 𝑣

with

𝑢 + 𝑣

and

- (𝑢 + 𝑣)

; both differences factor.

Hint 2. For (b), look for a bijection

𝜑 : ℝ \to (- 1, 1)

with

𝜑 (𝑠 + 𝑡) =

𝜑 (𝑠) \oplus

𝜑 (𝑡)

. Which function has an addition formula of this shape?

Hint 3. Use the addition formula for

tanh

, and transport the structure of

ℝ

along

tanh

Sketch. (a)

1 + 𝑢 𝑣 > 0

, and

(1 + 𝑢 𝑣) \mp (𝑢 + 𝑣) =

(1 \mp 𝑢) (1 \mp 𝑣) >

0

. (b)

tanh

is a bijection

ℝ \to (- 1, 1)

carrying

+

\oplus

, so transporting along it gives

𝑎 ⊙ 𝑣 =

tanh (𝑎 artanh 𝑣)

, which can be written with powers of

1 \pm 𝑣

. Then

2 ⊙ 𝑣 = 2 𝑣 / (1 + 𝑣^{2})

(1 / 2) ⊙ 𝑣 =

(1 - \sqrt{1 - 𝑣^{2}}) / 𝑣

for

𝑣 \neq 0

, the zero vector is

0

, and the inverse of

𝑣

- 𝑣

(a) Let $𝑢, 𝑣 \in (- 1, 1)$ . Then $| 𝑢 𝑣 | < 1$ , so $1 + 𝑢 𝑣 > 0$ . Also

(1 + 𝑢 𝑣) - (𝑢 + 𝑣) = (1 - 𝑢) (1 - 𝑣) > 0, (1 + 𝑢 𝑣) + (𝑢 + 𝑣) = (1 + 𝑢) (1 + 𝑣) > 0 .

So $- (1 + 𝑢 𝑣) <$ $𝑢 +$ $𝑣 <$ $1 +$ $𝑢 𝑣$ , and dividing by $1 + 𝑢 𝑣 > 0$ gives $- 1 < 𝑢 \oplus 𝑣 < 1$ .

(b) The function $tanh : ℝ \to (- 1, 1)$ is a bijection: it is continuous and strictly increasing, with limits $- 1$ and $1$ at $- \infty$ and $\infty$ . Its inverse is $artanh 𝑣 =$ $(1 / 2) ln ((1 + 𝑣) / (1 - 𝑣))$ . The addition formula

tanh (𝑠 + 𝑡) = \frac{tanh 𝑠 + tanh 𝑡}{1 + tanh 𝑠 tanh 𝑡}

says exactly that $tanh (𝑠 + 𝑡) =$ $tanh (𝑠) \oplus$ $tanh (𝑡)$ . Hence, with $𝑢 = tanh 𝑠$ and $𝑣 = tanh 𝑡$ , we get $𝑢 \oplus$ $𝑣 =$ $tanh (artanh 𝑢 + artanh 𝑣)$ , which is the addition obtained by transporting the structure of $ℝ$ along $tanh$ . By Proposition 1.2.22, $𝑉$ becomes a real vector space with this addition and

𝑎 ⊙ 𝑣 = tanh (𝑎 artanh 𝑣) .

Using $tanh 𝑠 =$ $(𝑒^{2 𝑠} - 1) / (𝑒^{2 𝑠} + 1)$ and $𝑒^{2 𝑎 artanh 𝑣} =$ ${((1 + 𝑣) / (1 - 𝑣))}^{𝑎}$ ,

𝑎 ⊙ 𝑣 = \frac{{(1 + 𝑣)}^{𝑎} - {(1 - 𝑣)}^{𝑎}}{{(1 + 𝑣)}^{𝑎} + {(1 - 𝑣)}^{𝑎}} .

For $𝑎 = 2$ , this is $(4 𝑣) / (2 + 2 𝑣^{2}) =$ $2 𝑣 / (1 + 𝑣^{2})$ , which equals $𝑣 \oplus 𝑣$ , as (A5) and (A8) require. For $𝑎 = 1 / 2$ , multiplying the numerator and denominator by $\sqrt{1 + 𝑣} - \sqrt{1 - 𝑣}$ gives, for $𝑣 \neq 0$ ,

\frac{1}{2} ⊙ 𝑣 = \frac{\sqrt{1 + 𝑣} - \sqrt{1 - 𝑣}}{\sqrt{1 + 𝑣} + \sqrt{1 - 𝑣}} = \frac{2 - 2 \sqrt{1 - 𝑣^{2}}}{2 𝑣} = \frac{1 - \sqrt{1 - 𝑣^{2}}}{𝑣},

and $(1 / 2) ⊙ 0 = 0$ . For example, $(1 / 2) ⊙ (4 / 5) =$ $(1 - 3 / 5) / (4 / 5) =$ $1 / 2$ , and indeed $(1 / 2) \oplus$ $(1 / 2) =$ $1 / (5 / 4) =$ $4 / 5$ . The zero vector is $tanh 0 = 0$ , and the inverse of $𝑣 = tanh 𝑡$ is $tanh (- 𝑡) = - 𝑣$ ; directly, $𝑣 \oplus$ $(- 𝑣) =$ $0 / (1 - 𝑣^{2}) =$ $0$ .

Exercise 1.2.54.hard

(a)Find the error in the following argument.
Claim. (A4) follows from the other axioms.
Proof. Let $𝐯 \in 𝑉$ . By (A5), (A8), and Proposition 1.2.32,
$𝐯 + (- 1) 𝐯 = 1 𝐯 + (- 1) 𝐯 = (1 + (- 1)) 𝐯 = 0 𝐯 = 𝟎 .$
So $(- 1) 𝐯$ is an additive inverse of $𝐯$ .
(b)Show that the claim is false: construct a set with operations that satisfy every axiom except (A4).

Solution.

Uses. Proposition 1.2.32 (The scalar zero), Proposition 1.2.30 (Additive arithmetic), Definition 1.2.1 (Vector space)

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. Trace the proof of Proposition 1.2.32 back to the axioms. Which ones does it use?

Hint 2. For (b), you need an addition with an identity but without inverses. On

{0, 1}

, try

𝑢 + 𝑣 = max (𝑢, 𝑣)

. Then look for a scalar multiplication satisfying (A5)–(A8); a very lazy one works.

Sketch. (a) Proposition 1.2.32 cancels

0 𝐯

, and cancellation (Proposition 1.2.30) was proved with (A4); the argument is circular. (b) On

{0, 1}

let

𝑢 + 𝑣 = max (𝑢, 𝑣)

and

𝑎 ⊙ 𝑣 = 𝑣

for all

𝑎

. (A1)–(A3) hold with identity

0

, and (A5)–(A8) hold because

max (𝑣, 𝑣) = 𝑣

. But

1 + 𝑤 = 1

for every

𝑤

, so

1

has no inverse.

(a) The argument uses Proposition 1.2.32, whose proof cancelled $0 𝐯$ from $0 𝐯 + 0 𝐯 = 0 𝐯 + 𝟎$ . That cancellation is Proposition 1.2.30(b), which was proved by adding $- 𝐮$ , an inverse supplied by (A4). So the argument assumes (A4) in order to prove it. The computation does show something true: in a vector space, $(- 1) 𝐯$ is the inverse of $𝐯$ (Proposition 1.2.34).

(b) Let $𝑉 = {0, 1}$ , let $𝐹$ be any field, and define

𝑢 + 𝑣 = max (𝑢, 𝑣), 𝑎 ⊙ 𝑣 = 𝑣 (𝑎 \in 𝐹, 𝑢, 𝑣 \in 𝑉) .

(A1) and (A2) hold because $max$ is commutative and associative. (A3) holds with zero vector $0$ , as $max (𝑣, 0) = 𝑣$ . (A5): $1 ⊙ 𝑣 = 𝑣$ . (A6): $(𝑎 𝑏) ⊙ 𝑣 =$ $𝑣 =$ $𝑎 ⊙ (𝑏 ⊙ 𝑣)$ . (A7): $𝑎 ⊙ (𝑢 + 𝑣) =$ $𝑢 +$ $𝑣 =$ $(𝑎 ⊙ 𝑢) +$ $(𝑎 ⊙ 𝑣)$ . (A8): $(𝑎 + 𝑏) ⊙ 𝑣 =$ $𝑣 =$ $max (𝑣, 𝑣) =$ $(𝑎 ⊙ 𝑣) +$ $(𝑏 ⊙ 𝑣)$ . But (A4) fails for $𝑣 = 1$ : $1 +$ $𝑤 =$ $max (1, 𝑤) =$ $1 \neq$ $0$ for both $𝑤 \in 𝑉$ .

Consistently with (a), the conclusion of Proposition 1.2.32 fails here: $0 ⊙ 1 = 1$ , which is not the zero vector.

Exercise 1.2.55.hard Suppose $𝑉$ , with operations $+$ and $\cdot$ , satisfies every vector space axiom except possibly (A5). Let

𝑊 = {𝐯 \in 𝑉 | 1 𝐯 = 𝐯}, 𝑁 = {𝐯 \in 𝑉 | 𝑎 𝐯 = 𝟎 for all 𝑎 \in 𝐹} .

(a)Show that every $𝐯 \in 𝑉$ can be written as $𝐯 = 𝐰 + 𝐧$ with $𝐰 \in 𝑊$ and $𝐧 \in 𝑁$ , in exactly one way.
(b)Show that $𝑊$ , with the operations of $𝑉$ , is a vector space over $𝐹$ .
(c)Show that (A5) holds if and only if $𝑁 = {𝟎}$ .

Solution.

Uses. Proposition 1.2.27 (Uniqueness of the zero vector), Proposition 1.2.30 (Additive arithmetic), Proposition 1.2.32 (The scalar zero), Proposition 1.2.33 (The zero vector), Proposition 1.2.28 (Uniqueness of additive inverses)

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. First list the consequences of the axioms that remain available: everything in Proposition 1.2.27, Proposition 1.2.28, Proposition 1.2.30, Proposition 1.2.32, and Proposition 1.2.33, since none of their proofs used (A5). Check that

𝑎 (- 𝐮) = - (𝑎 𝐮)

still holds.

Hint 2. For existence in (a), try

𝐰 = 1 𝐯

and

𝐧 = 𝐯 - 1 𝐯

. Use (A6) to compute

1 (1 𝐯)

and

𝑎 (1 𝐯)

Hint 3. For uniqueness, apply the scalar

1

𝐯 = 𝐰 + 𝐧

Sketch. Without (A5), the results through Proposition 1.2.33 still hold, and

𝑎 (- 𝐮) = - (𝑎 𝐮)

follows from (A7). Then

1 (1 𝐯) = 1 𝐯

and

𝑎 (𝐯 - 1 𝐯) =

𝑎 𝐯 -

(𝑎 1) 𝐯 =

𝟎

, so

𝐯 = 1 𝐯 + (𝐯 - 1 𝐯)

is a decomposition, and applying

1

to any decomposition recovers

𝐰 = 1 𝐯

𝑊

contains

𝟎

and is closed under the operations and negatives. If (A5) holds,

𝐧 = 1 𝐧 = 𝟎

; if

𝑁 = {𝟎}

, every

𝐯 = 1 𝐯

What is still available. The proofs of Proposition 1.2.27, Proposition 1.2.28, and Proposition 1.2.30 used only (A1)–(A4). The proofs of Proposition 1.2.32 and Proposition 1.2.33 (first approach) used in addition only (A8) and (A7). So all of these hold in $𝑉$ . We also need $𝑎 (- 𝐮) = - (𝑎 𝐮)$ , whose earlier proof used (A5). Instead, by (A7) and Proposition 1.2.33, $𝑎 𝐮 +$ $𝑎 (- 𝐮) =$ $𝑎 (𝐮 + (- 𝐮)) =$ $𝑎 𝟎 =$ $𝟎$ , so $𝑎 (- 𝐮) = - (𝑎 𝐮)$ by Proposition 1.2.28.

(a) Existence. Let $𝐰 = 1 𝐯$ and $𝐧 = 𝐯 - 1 𝐯$ . By (A6), $1 𝐰 =$ $1 (1 𝐯) =$ $(1 \cdot 1) 𝐯 =$ $1 𝐯 =$ $𝐰$ , so $𝐰 \in 𝑊$ . For $𝑎 \in 𝐹$ , by (A7), the rule above, and (A6),

𝑎 𝐧 = 𝑎 𝐯 + 𝑎 (- (1 𝐯)) = 𝑎 𝐯 - 𝑎 (1 𝐯) = 𝑎 𝐯 - (𝑎 \cdot 1) 𝐯 = 𝑎 𝐯 - 𝑎 𝐯 = 𝟎,

so $𝐧 \in 𝑁$ . And $𝐰 +$ $𝐧 =$ $1 𝐯 +$ $(𝐯 - 1 𝐯) =$ $𝐯$ by Proposition 1.2.30(a).

Uniqueness. Suppose $𝐯 = 𝐰 + 𝐧$ with $𝐰 \in 𝑊$ and $𝐧 \in 𝑁$ . By (A7), $1 𝐯 =$ $1 𝐰 +$ $1 𝐧 =$ $𝐰 +$ $𝟎 =$ $𝐰$ . So $𝐰 = 1 𝐯$ is determined by $𝐯$ , and then $𝐧 = 𝐯 - 𝐰$ is determined by Proposition 1.2.30(a).

(b) We have $𝟎 \in 𝑊$ , since $1 𝟎 = 𝟎$ by Proposition 1.2.33. For $𝐰, 𝐰^{'} \in 𝑊$ and $𝑎 \in 𝐹$ : $1 (𝐰 + 𝐰^{'}) =$ $1 𝐰 +$ $1 𝐰^{'} =$ $𝐰 +$ $𝐰^{'}$ by (A7); $1 (𝑎 𝐰) =$ $(1 𝑎) 𝐰 =$ $𝑎 𝐰$ by (A6); and $1 (- 𝐰) = - (1 𝐰) = - 𝐰$ by the rule above. So $𝑊$ is closed under addition, scalar multiplication, and negatives, and the operations of $𝑉$ restrict to operations on $𝑊$ . The equations in (A1), (A2), and (A6)–(A8) hold in $𝑊$ because they hold in $𝑉$ ; (A3) and (A4) hold because $𝟎 \in 𝑊$ and $- 𝐰 \in 𝑊$ ; and (A5) holds in $𝑊$ by the definition of $𝑊$ .

(c) If (A5) holds and $𝐧 \in 𝑁$ , then $𝐧 = 1 𝐧 = 𝟎$ , so $𝑁 = {𝟎}$ (and $𝟎 \in 𝑁$ by Proposition 1.2.33). Conversely, if $𝑁 = {𝟎}$ , then in the decomposition of (a), $𝐯 = 1 𝐯 + 𝟎 = 1 𝐯$ for every $𝐯$ , which is (A5).

So without (A5), $𝑉$ splits into a genuine vector space $𝑊$ and a part $𝑁$ that every scalar sends to $𝟎$ ; (A5) is exactly what removes $𝑁$ . Example (a) of Example 1.2.25 is an instance, with $𝑊$ the $𝑥$ -axis and $𝑁$ the $𝑦$ -axis.

Exercise 1.2.56.hard Suppose

𝑉

, with operations

+

and

\cdot

, satisfies (A2)–(A8), but not necessarily (A1). (Note that (A3) and (A4) only say

𝐯 + 𝟎 = 𝐯

and

𝐯 + 𝐰 = 𝟎

, with

𝟎

and

𝐰

on the right. Without (A1) we cannot prove that the zero vector is unique, so fix one element

𝟎

satisfying (A3) for which (A4) holds, and let

𝟎

always mean this element.) Show that

𝐮 + 𝐯 = 𝐯 + 𝐮

for all

𝐮, 𝐯 \in 𝑉

Solution.

Uses. Definition 1.2.1 (Vector space), Exercise 1.1.58

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. Expand

(1 + 1) (𝐮 + 𝐯)

in two ways, once with (A8) first and once with (A7) first.

Hint 2. You get

𝐮 +

𝐯 +

𝐮 +

𝐯 =

𝐮 +

𝐮 +

𝐯 +

𝐯

. To cancel

𝐮

on the left and

𝐯

on the right, you need two-sided inverses and a two-sided zero, which (A3) and (A4) do not provide directly.

Hint 3. Use scalars. Show

0 𝐯 = 𝟎

using only right-hand inverses, then check that

(- 1) 𝐯

is an inverse on both sides and that

𝟎 + 𝐯 = 0 𝐯 + 1 𝐯 = 𝐯

Sketch. Adding a right inverse of

0 𝐯

0 𝐯 = 0 𝐯 + 0 𝐯

gives

0 𝐯 = 𝟎

. Then (A8) gives

𝐯 +

(- 1) 𝐯 =

(- 1) 𝐯 +

𝐯 =

0 𝐯 =

𝟎

and

𝟎 + 𝐯 = (0 + 1) 𝐯 = 𝐯

, so cancellation works on both sides. Expanding

(1 + 1) (𝐮 + 𝐯)

two ways gives

𝐮 +

(𝐯 + 𝐮) +

𝐯 =

𝐮 +

(𝐮 + 𝐯) +

𝐯

; cancel.

By (A2) we may drop parentheses in sums. We cannot reorder terms.

Step 1: $0 𝐯 = 𝟎$ . By (A8), $0 𝐯 =$ $(0 + 0) 𝐯 =$ $0 𝐯 +$ $0 𝐯$ . By (A4) choose $𝐲$ with $0 𝐯 + 𝐲 = 𝟎$ , and add it on the right:

𝟎 = 0 𝐯 + 𝐲 = 0 𝐯 + 0 𝐯 + 𝐲 = 0 𝐯 + 𝟎 = 0 𝐯 .

Step 2: two-sided inverses and zero. By (A5), (A8), and Step 1,

𝐯 + (- 1) 𝐯 = (1 + (- 1)) 𝐯 = 0 𝐯 = 𝟎, (- 1) 𝐯 + 𝐯 = ((- 1) + 1) 𝐯 = 𝟎,

and $𝟎 +$ $𝐯 =$ $0 𝐯 +$ $1 𝐯 =$ $(0 + 1) 𝐯 =$ $𝐯$ .

Step 3: cancellation on both sides. If $𝐱 + 𝐲 = 𝐱 + 𝐳$ , adding $(- 1) 𝐱$ on the left and using Step 2 gives $𝟎 + 𝐲 = 𝟎 + 𝐳$ , so $𝐲 = 𝐳$ . If $𝐲 + 𝐱 = 𝐳 + 𝐱$ , adding $(- 1) 𝐱$ on the right and using Step 2 and (A3) gives $𝐲 = 𝐳$ .

Step 4. By (A8) and (A5),

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

and by (A7), then (A8) and (A5),

(1 + 1) (𝐮 + 𝐯) = (1 + 1) 𝐮 + (1 + 1) 𝐯 = 𝐮 + 𝐮 + 𝐯 + 𝐯 = 𝐮 + (𝐮 + 𝐯) + 𝐯 .

Cancelling $𝐮$ on the left and $𝐯$ on the right (Step 3) gives $𝐯 + 𝐮 = 𝐮 + 𝐯$ , as desired.

Compare Exercise 1.1.58, the same statement for fields. There the two-sided inverses had to be built from the additive axioms alone; here the scalar $- 1$ supplies them.

Exercise 1.2.57.challenge Let $𝑓 : ℝ \to ℝ$ be a function, and on $𝑉 = ℝ$ with its usual addition define $𝑎 ⊙ 𝑥 = 𝑓 (𝑎) 𝑥$ .

(a)Show that $(𝑉, +, ⊙)$ is a real vector space if and only if $𝑓 (1) = 1$ , $𝑓 (𝑎 + 𝑏) =$ $𝑓 (𝑎) +$ $𝑓 (𝑏)$ , and $𝑓 (𝑎 𝑏) =$ $𝑓 (𝑎) 𝑓 (𝑏)$ for all $𝑎, 𝑏 \in ℝ$ .
(b)Show that the only such $𝑓$ is the identity. So the usual product is the only scalar multiplication of this form.

Solution.

Uses. Definition 1.2.1 (Vector space), between any two distinct real numbers there is a rational number

Hint 1. For (a), (A7) always holds, and each of (A5), (A6), (A8) is equivalent to one of the three conditions; for the forward direction, take

𝑥 = 1

Hint 2. For (b), first find

𝑓

ℤ

and then on

ℚ

using additivity and

𝑓 (1) = 1

Hint 3. Multiplicativity gives

𝑓 (𝑎^{2}) = {𝑓 (𝑎)}^{2} \geq 0

, so

𝑓

sends nonnegative numbers to nonnegative numbers. With additivity,

𝑓

is nondecreasing. Now squeeze

𝑓 (𝑎)

between rationals.

Sketch. (a) (A5), (A6), (A8) at

𝑥 = 1

are the three conditions, and conversely the conditions give the axioms for all

𝑥

. (b) Additivity and

𝑓 (1) = 1

give

𝑓 (𝑞) = 𝑞

for rational

𝑞

. Since

𝑓 (𝑏) = {𝑓 (\sqrt{𝑏})}^{2} \geq 0

for

𝑏 \geq 0

𝑓

is nondecreasing. If

𝑓 (𝑎) < 𝑎

, a rational

𝑞

with

𝑓 (𝑎) <

𝑞 <

𝑎

gives

𝑞 = 𝑓 (𝑞) \leq 𝑓 (𝑎)

, a contradiction; similarly

𝑓 (𝑎) > 𝑎

is impossible.

(a) Since the addition is that of $ℝ$ , (A1)–(A4) hold, and (A7) holds because $𝑓 (𝑎) (𝑥 + 𝑦) =$ $𝑓 (𝑎) 𝑥 +$ $𝑓 (𝑎) 𝑦$ . The other axioms read, for all $𝑎, 𝑏, 𝑥 \in ℝ$ ,

(A5) 𝑓 (1) 𝑥 = 𝑥, (A6) 𝑓 (𝑎 𝑏) 𝑥 = 𝑓 (𝑎) 𝑓 (𝑏) 𝑥, (A8) 𝑓 (𝑎 + 𝑏) 𝑥 = 𝑓 (𝑎) 𝑥 + 𝑓 (𝑏) 𝑥 .

Taking $𝑥 = 1$ shows that they imply $𝑓 (1) = 1$ , $𝑓 (𝑎 𝑏) =$ $𝑓 (𝑎) 𝑓 (𝑏)$ , and $𝑓 (𝑎 + 𝑏) =$ $𝑓 (𝑎) +$ $𝑓 (𝑏)$ . Conversely, multiplying these three equations by $𝑥$ gives the three axioms.

(b) Let $𝑓$ satisfy the three conditions.

$𝑓$ fixes $ℚ$ . From $𝑓 (0) =$ $𝑓 (0 + 0) =$ $𝑓 (0) +$ $𝑓 (0)$ we get $𝑓 (0) = 0$ , and then $𝑓 (𝑎) +$ $𝑓 (- 𝑎) =$ $𝑓 (0) =$ $0$ gives $𝑓 (- 𝑎) = - 𝑓 (𝑎)$ . For an integer $𝑛 \geq 1$ , additivity and $𝑓 (1) = 1$ give $𝑓 (𝑛) =$ $𝑓 (1) +$ $\dots +$ $𝑓 (1) =$ $𝑛$ by induction, and so $𝑓 (𝑛) = 𝑛$ for all $𝑛 \in ℤ$ . For $𝑞 = 𝑚 / 𝑛$ with $𝑛 \geq 1$ , $𝑛 𝑓 (𝑞) =$ $𝑓 (𝑛 𝑞) =$ $𝑓 (𝑚) =$ $𝑚$ (again by additivity), so $𝑓 (𝑞) = 𝑞$ .

$𝑓$ is nondecreasing. If $𝑏 \geq 0$ , then $𝑓 (𝑏) =$ $𝑓 (\sqrt{𝑏} \sqrt{𝑏}) =$ ${𝑓 (\sqrt{𝑏})}^{2} \geq$ $0$ . So if $𝑎 \leq 𝑏$ , then $𝑓 (𝑏) -$ $𝑓 (𝑎) =$ $𝑓 (𝑏 - 𝑎) \geq$ $0$ .

$𝑓 (𝑎) = 𝑎$ . Let $𝑎 \in ℝ$ , and suppose $𝑓 (𝑎) < 𝑎$ . Choose a rational $𝑞$ with $𝑓 (𝑎) <$ $𝑞 <$ $𝑎$ . Since $𝑞 < 𝑎$ and $𝑓$ is nondecreasing, $𝑞 = 𝑓 (𝑞) \leq 𝑓 (𝑎)$ , contradicting $𝑓 (𝑎) < 𝑞$ . If $𝑓 (𝑎) > 𝑎$ , choose a rational $𝑞$ with $𝑎 <$ $𝑞 <$ $𝑓 (𝑎)$ ; then $𝑓 (𝑎) \leq 𝑓 (𝑞) = 𝑞$ , again a contradiction. So $𝑓 (𝑎) = 𝑎$ .

Over $ℂ$ the analogous question has at least two answers, the identity and conjugation (Exercise 1.2.48). The proof above used the order of $ℝ$ , which $ℂ$ lacks.

More practice

The practice bank has 52 exercises on this section’s topics.