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 ${\mathbb{C}}^n$.

• Note that vectors in $V$ and scalars in $\mathbb{F}$ can belong to different sets; the vectors do not need to be elements of the scalar field. For example, in Problem 7 of LADR Exercises 2A, we see the $\mathbb{C}$ defined as a vector space over $\mathbb{R}$. In this case, complex numbers are vectors while real numbers are scalars. The closure of vector addition still applies, as the sum of any two complex numbers is a complex number. The closure of scalar multiplication still applies, as multiplying a complex number by a real number yields another complex number.

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.

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.