4.1 Real Vector Spaces: Class Notes

Definition: (Vector space)

Let $V$ be an arbitrary nonempty set of objects for which two operations are defined: addition and multiplication by numbers called scalars. By addition we mean a rule for associating with each pair of objects $𝐮$ and $𝐯$ in $V$ an object $𝐮 + 𝐯$ , called the sum of $𝐮$ and $𝐯$ ; by scalar multiplication we mean a rule for associating with each scalar $k$ and each object $𝐮$ in $V$ an object $k 𝐮$ , called the scalar multiple of $𝐮$ by $k$ . If the following axioms are satisfied by all objects $𝐮$ , $𝐯$ , and $𝐰$ in $V$ and all scalars $k$ and $m$ , then we call $V$ a vector space and we call the objects in $V$ vectors.

If $𝐮$ and $𝐯$ are objects in $V$ , then $𝐮 + 𝐯$ is in $V$ .
$𝐮 + 𝐯 = 𝐯 + 𝐮$
$𝐮 + (𝐯 + 𝐰) = (𝐮 + 𝐯) + 𝐰$
There exists an object in $V$ , called the zero vector, that is denoted by $𝟎$ and has the property that $𝟎 + 𝐮 = 𝐮 + 𝟎 = 𝐮$ for all $𝐮$ in $V$ .
For each $𝐮$ in $V$ , there is an object $- 𝐮$ in $V$ , called a negative of $𝐮$ such that $𝐮 + (- 𝐮) = (- 𝐮) + 𝐮 = 𝟎$ .
If $k$ is any scalar and $𝐮$ is an object in $V$ , then $k 𝐮$ is in $V$ .
$k (𝐮 + 𝐯) = k 𝐮 + k 𝐯$
$(k + m) 𝐮 = k 𝐮 + m 𝐮$
$k (m 𝐮) = (k m) (𝐮)$
$1 𝐮 = 𝐮$

Axiom 1 is closure under addition, and Axiom 6 is closure under scalar multiplication.

Notes:

Vector addition and scalar multiplication can be any rules that satisfy these axioms.
A vector is any object in a vector space.

#6

Determine whether the set equipped with the given operations is a vector space. For those that are not vector spaces, identify the vector space axioms that fail.

The set of all $n$ -tuples of real numbers that have the form $(x, x, \dots, x)$ with the standard operations on $R^{n}$ .

Solution

Let $\vec{u} = (u, u, \dots, u)$ and $\vec{v} = (v, v, \dots, v)$ be in the set (so both are in $R^{n}$ ).

Axiom 1: $\vec{u} + \vec{v} = (u + v, u + v, \dots, u + v)$ , which has the same form, so the set is closed under addition.

Axioms 2, 3, and 7 through 10 are satisfied by properties of real numbers and Theorem 3.1.1.

Axiom 6: $k \vec{u} = (k u, k u, \dots, k u)$ , which has the same form.

Axiom 4: $\vec{0} = (0, 0, \dots, 0)$ has the same form, so it is in the set.

Axiom 5: $- \vec{u} = (- u, - u, \dots, - u)$ has the same form, so it is in the set.

Answer: $All ten axioms hold, so the set is a vector space.$

Examples of Vector Spaces

The simplest vector space is ${𝟎}$ .
$R^{n}$ with the usual operations of addition and scalar multiplication of $n$ -tuples
$R^{\infty}$ , the set of infinite sequences of numbers with the usual operations of addition and scalar multiplication performed componentwise
The set of $m \times n$ matrices with matrix addition and scalar multiplication, denoted $M_{m n}$
The set of real-valued functions $f (x)$ that are defined for all $x \in R$ , denoted $F (- \infty, \infty)$ with addition and scalar multiplication
$(𝐟 + 𝐠) (x) = f (x) + g (x)$ and $(k 𝐟) (x) = k f (x)$
The set of polynomials of degree $\leq n$ , denoted $P_{n}$

Example

Show why #5 is a vector space.

Solution

Let $\vec{f}$ and $\vec{g}$ be in $F (- \infty, \infty)$ , and let $k$ be a scalar.

Axiom 1 (closure): $(\vec{f} + \vec{g}) (x) = f (x) + g (x)$ , which is defined for every real $x$ , so $\vec{f} + \vec{g}$ is in $F (- \infty, \infty)$ .

Axiom 6 (closure): $(k \vec{f}) (x) = k f (x)$ , which is also defined for every real $x$ .

At any point $x$ , $\vec{f} (x)$ and $\vec{g} (x)$ are real numbers, so Axioms 2, 3, 7, 8, 9, and 10 are satisfied by properties of real numbers.

Axiom 4: $\vec{0}$ is the function $f (x) = 0$ , which is defined on $(- \infty, \infty)$ . Thus Axiom 4 is satisfied.

Axiom 5: $- \vec{f} (x) = - f (x) \in F (- \infty, \infty)$ for all $\vec{f} \in F (- \infty, \infty)$ .

Example 4.1.7: Not a vector space

Let $𝐮 = (u_{1}, u_{2})$ and $𝐯 = (v_{1}, v_{2})$ . Define $𝐮 + 𝐯 = (u_{1} + v_{1}, u_{2} + v_{2})$ and $k 𝐮 = (k u_{1}, 0)$ .

Solution

This is not a vector space, because Axiom 10 fails: for $k = 1$ ,

$1 \vec{u} = 1 (u_{1}, u_{2}) = (u_{1}, 0) \neq \vec{u}$

(whenever $u_{2} \neq 0$ ).

Example 4.1.8: Unusual vector space

Let $V$ be the set of positive real numbers, so $𝐮 = u$ and $𝐯 = v$ are positive real numbers. Define $𝐮 + 𝐯 = u v$ and $k 𝐮 = u^{k}$ .

Solution

For instance, under these operations $2 + 5 = 10$ (multiply) and $2 (5) = 5^{2} = 25$ (raise to the power).

Axioms 1 and 6: the set is closed under addition and scalar multiplication, since $u v$ and $u^{k}$ are positive whenever $u$ and $v$ are.

Axioms 2 and 3: the vector addition is commutative and associative because multiplication of real numbers is.

Axiom 4: the zero element is $\vec{0} = 1$ , because

$\vec{0} + \vec{u} = 1 u = u .$

Axiom 10 holds as well: $1 \vec{u} = u^{1} = u$ .

Axiom 5: the negative of a vector $\vec{u}$ is its reciprocal:

$- 1 \vec{u} = u^{- 1} = \frac{1}{u}, u + \frac{1}{u} = u \cdot \frac{1}{u} = 1 = \vec{0} .$

Axioms 7, 8, and 9:

$\begin{matrix} k (\vec{u} + \vec{v}) & = (u v)^{k} = u^{k} v^{k} = k \vec{u} + k \vec{v} \\ (k + m) \vec{u} & = u^{k + m} = u^{k} u^{m} = k \vec{u} + m \vec{u} \\ k (m \vec{u}) & = (u^{m})^{k} = u^{m k} = u^{k m} = (k m) \vec{u} \end{matrix}$

Answer: $All ten axioms hold, so V is a vector space.$

Example

Let $V$ be the set of all ordered pairs of real numbers, and consider the following addition and scalar multiplication operations on $𝐮 = (u_{1}, u_{2})$ and $𝐯 = (v_{1}, v_{2})$ :

$𝐮 + 𝐯 = (u_{1} + v_{1} + 2, u_{2} + v_{2})$ , $k 𝐮 = (k u_{1}, k u_{2})$ .

a. Show that Axioms 4 and 5 hold.

b. Find all axioms that fail to hold.

Solution

Axiom 4. Let $\vec{0} = (a, b)$ and require $\vec{0} + \vec{u} = \vec{u}$ :

$(a, b) + (u_{1}, u_{2}) = (a + u_{1} + 2, b + u_{2}) = (u_{1}, u_{2})$

Comparing entries: $a + u_{1} + 2 = u_{1} \Rightarrow a = - 2$ and $b + u_{2} = u_{2} \Rightarrow b = 0$ . Then

$\vec{0} = (- 2, 0) .$

Axiom 5. Let $- \vec{u} = (a, b)$ and require $\vec{u} + (- \vec{u}) = \vec{0}$ :

$(u_{1} + a + 2, u_{2} + b) = (- 2, 0)$

so $a = - u_{1} - 4$ and $b = - u_{2}$ , and

$- \vec{u} = (- u_{1} - 4, - u_{2}) .$

Part b. Axioms 7 and 8 fail. For Axiom 7:

$\begin{matrix} k (\vec{u} + \vec{v}) & = (k u_{1} + k v_{1} + 2 k, k u_{2} + k v_{2}) \\ k \vec{u} + k \vec{v} & = (k u_{1} + k v_{1} + 2, k u_{2} + k v_{2}) \end{matrix}$

These differ whenever $k \neq 1$ . Axiom 8 is similar.

Answer: $Axioms 7 and 8 fail.$

Theorem 4.1.1

Let $V$ be a vector space, $𝐮$ be a vector in $V$ , and $k$ a scalar; then:

a) $0 𝐮 = 𝟎$

b) $k 𝟎 = 𝟎$

c) $(- 1) 𝐮 = - 𝐮$

d) If $k 𝐮 = 𝟎$ , then $k = 0$ or $𝐮 = 𝟎$

Example

Prove that the zero vector in any vector space is unique.

Proof

Let $V$ be a vector space and let $\vec{0}$ be a vector such that

$\vec{0} + \vec{u} = \vec{u} + \vec{0} = \vec{u} for all \vec{u} \in V .$

Suppose there is $\vec{0}' \in V$ such that

$\vec{0}' + \vec{u} = \vec{u} + \vec{0}' = \vec{u} for all \vec{u} \in V .$

Let $\vec{u} = \vec{0}$ in this equation. Then

$\vec{0}' + \vec{0} = \vec{0} .$

But $\vec{0}$ is a zero vector, so the same sum is also

$\vec{0}' + \vec{0} = \vec{0}' .$

The two lines start from the same sum $\vec{0}' + \vec{0}$ , so

Answer: $\vec{0}' = \vec{0}$

The zero vector is unique.