Vector Space

A vector space is a set $V$ along with an addition on $V$ and a scalar multiplication on $V$ such that the following properties hold.

• Commutativity: $u+v=v+u$ for all $u,v\in V$.
• Associativity: $(u+v)+w=u+(v+w)$ and $(ab)v=a(bv)$ for all $u,v,w\in V$ and for all $a,b\in \mathbb{F}$.
• Additive Identity: There exists an element $0\in V$ such that $v+0=v$ for $v\in V$.
• Additive Inverse: For every $v\in V$, there exists $w\in V$ such that $v+w=0$.
• Multiplicative Identity: $1v=v$ for all $v\in V$.
• Distributive Properties: $a(u+v)=au+av$ and $(a+b)v=av+bv$ for all $a,b\in \mathbb{F}$ and all $u,v\in V$.

To avoid re-stating this many times, we declare that $V$ denotes a vector space over $\mathbb{F}$.

Definitions: Addition, Scalar Multiplication

• An addition on a set $V$ is a function that assigns an element $u+v\in V$ to each pair of elements $u,v\in V$.
• A scalar multiplication on a set $V$ is a function that assigns an element $\lambda v\in V$ to each $\lambda \in \mathbb{F}$ and each $v\in V$.

We can use some geometric language to aid our intuition by saying that elements of a vector space are called vectors or points.

The scalar multiplication in a vector space depends on $\mathbb{F}$. Thus when we need to be precise, we will say that $V$ is a vector space over $\mathbb{F}$ instead of saying simply that $V$ is a vector space. For example, ${\mathbb{R}}^n$ is a vector space over $\mathbb{R}$, and $\mathbb{C}$ is a vector space over ${\mathbb{C}}^n$.

Usually the choice of $\mathbb{F}$ is either clear from the context or irrelevant. Thus we often assume that $\mathbb{F}$ is lurking in the background without specifically mentioning it. With the usual operations of addition and scalar multiplication, ${\mathbb{F}}^n$ is a vector space over $\mathbb{F}$, as you should verify. The example of ${\mathbb{F}}^n$ (see here) motivated our definition of vector space.

The simplest vector space is $\{0\}$, which contains only one point.

Vector Spaces as Sets of Functions

Our next example of a vector space involves a set of functions.

${\mathbb{F}}^S$

We define ${\mathbb{F}}^S$ with the following:

• If $S$ is a set, then ${\mathbb{F}}^S$ denotes the set of functions from $S$ to $\mathbb{F}$. In other words, each function in ${\mathbb{F}}^S$ takes an input from $S$ and gives an output in the field $\mathbb{F}$.
• For $f,g\in {\mathbb{F}}^S$, the sum $f+g\in {\mathbb{F}}^S$ is the function defined by

$$(f+g)(x)=f(x)+g(x)$$

• For all $\lambda \in \mathbb{F}$ and $f\in {\mathbb{F}}^S$, the product $\lambda \mathbb{F}\in {\mathbb{F}}^S$ is the function defined by

$$(\lambda f)(x)=\lambda f(x)$$

for all $x\in S$.

This perspective generalizes the concept of vectors. While we usually think of vectors as ordered tuples (like points in ${\mathbb{R}}^n$), in this more abstract setting, we treat entire functions as vectors.

As an example of the notation above, if $S$ is the interval $[0,1]$ and $\mathbb{F}=\mathbb{R}$, then ${\mathbb{R}}^{[0,1]}$ is the set of real-valued functions on the interval $[0,1]$. This means the the domain of each function are $[0,1]$, which is mapped to a real value $\mathbb{R}$.

• The elements of the vector space ${\mathbb{R}}^{[0,1]}$ are real-valued functions on $[0,1]$, not lists. In general, a vector space is an abstract entity whose elements might be lists, functions, or weird objects.

We can show that ${\mathbb{F}}^S$ is a vector space by considering the following:

• If $S$ is a nonempty set, then ${\mathbb{F}}^S$ (with the operations of addition and scalar multiplication as defined above) is a vector space over $\mathbb{F}$.
• The additive identity of ${\mathbb{F}}^S$ is the function $0:S\to \mathbb{F}$ for all $x\in S$, defined by:

$$0(x)=0$$

for all $x\in S$.

• For all $f\in {\mathbb{F}}^S$, the additive inverse of $f$ is the function $-f:S\to F$ defined by:

$$(-f)(x)=-f(x)$$

for all $x\in S$.

The vector space ${\mathbb{F}}^n$ is a special case of the vector space ${\mathbb{F}}^S$ because each $(x_1,\dots ,x_n)\in {\mathbb{F}}^n$ can be thought of as a function $x$ from the set $\{1,2,\dots ,n\}$ to $\mathbb{F}$. We just write $x(k)$ instead of writing $x_k$ for the $k$-th coordinate of $(x_1,\dots ,x_n)$.

• In other words, we can think of ${\mathbb{F}}^n$ as ${\mathbb{F}}^{\{1,2,\dots ,n\}}$.
• Similarly, we can think of ${\mathbb{F}}^{\infty }$ as ${\mathbb{F}}^{\{1,2,\dots \}}$.

Elementary Properties

We need to develop some of the elementary properties of vector spaces.

Unique Additive Identity

A vector space has a unique additive identity.

Proof. Suppose $0$ and $0^{\prime }$ are both additive identities for some vector space $V$. Then

$$0^{\prime }=0^{\prime }+0=0+0^{\prime }=0$$

where the first equality holds because $0$ is the additive identity, the second equality comes from commutativity, and the third equality holds because $0^{\prime }$ is an additive identity. Thus, $0=0^{\prime }$, showing that $V$ only has one additive identity.

Unique Additive Inverse

Every element in a vector space has a has a unique additive inverse.

Proof. Suppose $V$ is a vector space. Let $v\in V$. Suppose $w$ and $w^{\prime }$ are additive inverses of $v$. Then

$$w=w+0=w+(v+w^{\prime })=(w+v)+w^{\prime }=0+w^{\prime }=w^{\prime }$$

Thus $w=w^{\prime }$ as desired.

Then we can use the following notation for $v,w\in V$:

• $-v$ denotes the additive inverse of $v$;
• $w-v$ is defined to be $w+(-v)$.

Multiplication by Scalar 0

When we multiply a vector by the number zero, we have $0v=0$ for every $v\in V$.

Proof. For $v\in V$, we have:

$$0v=(0+0)v=0v+0v$$

Adding the additive inverse of both sides gives:

$$0=0v$$

as desired.

Multiplication by Vector 0

When we multiply a scalar by a vector $0$, we have $a0=0$ for every $a\in \mathbb{F}$. Here, $0$ denotes the additive identity of $V$, not scalar $0$.

Proof. For $a\in \mathbb{F}$, we have

$$a0=a(0+0)=a0+a0$$

Adding the additive inverse of $a0$ to both sides of the equation above gives $0=a0$, as desired.

Multiplication by -1

Multiplying a vector by $-1$ gives $(-1)v=-v$ for every $v\in V$.

Proof. For $v\in V$, we have

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

This tells us that $(-1)v$, when added to $v$, gives $0$. Thus, $(-1)v$ is the additive inverse of $v$, as desired.