Lecture 2 - Vector Spaces (cont.)

For our class, we'll need to think about complex scalars $λ \in C$ . We'll talk briefly in class about it, but just know what the arithmetic of complex numbers is.

Recap

Suppose we have some set $V$ and a field $F$ , along with operations of vector addition and scalar multiplication. They need to satisfy the following rules:

Associativity of vector addition: $(\vec{u} + \vec{v}) + \vec{w} = \vec{v} + (\vec{u} + \vec{w}$
Commutativity of vector addition: $\vec{u} + \vec{v} = \vec{v} + \vec{u}$
Distributivity of scalar to vector addition: $α (\vec{u} + \vec{v}) = α \vec{u} + α \vec{v}$
Additive Identity: $\vec{0}, \vec{0} + \vec{v} = \vec{v}$
Additive Inverse: $\vec{v} + (- \vec{v}) = \vec{0}$
Multiplicative Identity: $1, 1 \vec{v} = \vec{v}$
Associativity of scalar multiplication: $(α β) \vec{u} = α (β \vec{u})$
Distributivity of vector to scalar multiplication: $(α + β) \vec{v} = α \vec{v} + β \vec{v}$

Consider the following, is $0 \vec{v} = \vec{0}$ . Is this true? We'll later see that we can prove this is true given the above axioms. And that's why it's important to know the definitions above.

Some Notes on the Reading

The definitions for the additive identity don't say persay that the additive identity is unique. However, it is true and a fact.

Uniqueness of the Additive Identity

There is a unique additive identity in a vector space $V$ .

Proof
To prove this, you:

Prove that the additive identity itself exists
You assume that you have two different ones, and show by contradiction that you that they are equal.

For (1), it's given by the definition above. We show uniqueness now. Suppose you have two unique additive identities ${\vec{0}}_{1}, {\vec{0}}_{2}$ that are different ( ${\vec{0}}_{1} \neq {\vec{0}}_{2}$ ). Notice that:

\begin{aligned} {\vec{0}}_{1} & = {\vec{0}}_{1} + {\vec{0}}_{2} \\ = {\vec{0}}_{2} \end{aligned}

Via the definition of additive identity above. Clearly there's a contradiction, so therefore the additive identity must be unique.
☐

Since the vector is unique, we denote $\vec{0}$ as the additive identity, which we've proved is unique. Note too that we can always find this unique vector, as we just do $\vec{v} + (- \vec{v}) = \vec{0}$ , so just take some random vector in $V$ , do the equation, and find $\vec{0}$ .

Some Examples

Let's look at some examples:

F^{n}

We say that:

F^{n} = {[\begin{matrix} x_{1} \\ x_{2} \\ ⋮ \end{matrix}] : x_{1}, . . ., x_{n} \in F}

is a vector space over $F$ , with the operations:

[\begin{matrix} x_{1} \\ x_{2} \\ ⋮ \\ x_{n} \end{matrix}] + [\begin{matrix} y_{1} \\ y_{2} \\ ⋮ \\ y_{n} \end{matrix}] = [\begin{matrix} x_{1} + y_{1} \\ x_{2} + y_{2} \\ ⋮ \\ x_{n} + y_{n} \end{matrix}]

And:

α [\begin{matrix} x_{1} \\ x_{2} \\ ⋮ \\ x_{n} \end{matrix}] = [\begin{matrix} α x_{1} \\ α x_{2} \\ ⋮ \\ α x_{n} \end{matrix}]

Note that the $+$ inside of the brackets is scalar addition, while the $+$ on the outside is vector addition.

Note that the zero vector $\vec{0} = (0, 0, . . ., 0)$ .

A Real Example

$(0, \infty)$ is a vector space over $R$ , where the operations are:

\vec{a} + \vec{b} = \vec{a b}

Where $a b$ is the multiplication of both values. For instance $\vec{3} + \vec{6} = \vec{18}$ .

For scalar multiplication:

α \vec{a} = \vec{a^{α}}

So for instance $(- 2) \vec{5} = \vec{\frac{1}{25}}$

Let's spot check this to check for understanding:

1 \vec{a} = \vec{a^{1}} = \vec{a}

Which is the multiplicative identity. Again, now consider:

a (\vec{x} + \vec{y}) = a (\vec{x y}) = \vec{(x y)^{a}} = \vec{x^{a} y^{a}} = \vec{x^{a}} + \vec{y^{a}} = a \vec{x} + a \vec{y}

So distributivity holds. What about the identity identity, which can find out is $\vec{1}$ :

\vec{x} - \vec{x} = \vec{x} + \vec{\frac{1}{x}} = \vec{\frac{x}{x}} = \vec{1}

Notice this! The vector $\vec{0}$ DOESN'T EXIST here (at least considering $0$ as a number) since we consider the interval $(0, \infty)$ . But the notation is that the additive identity which is ${\vec{0}}_{label} = {\vec{1}}_{value}$ .

F^{m, n}

We define $F^{m, n}$ as the set of $m \times n$ matrices with entries from $F$ , is a vector space over $F$ , which addition and scalar multiplication done componentwise.

The $\vec{0}$ is the $m \times n$ matrix with all 0 entries.

Proposition

Given a nonempty set $S$ and a field $F$ , we define $F^{S}$ is the set of all functions $f : S \to F$ . We claim that $F^{S}$ is a vector space over $F$ , with addition defined as ( $f, g \in F^{S}$ ):

(f + g) (s) = f (s) + g (s), \forall s \in S

And define scalar multiplication as:

α f (s) = α (f (s)), \forall s \in S

Here $\vec{0}$ is the function $\vec{0} : S \to F$ where $\vec{0} (s) = 0$ for all $s \in F$ .