1B. What is a vector space?

Definitions

Definitions of scaler multiplication and vector addition:

Vector Addition:

Addition on set $V$ is a function ($V\times V\to V$) that takes a pair to its sum, $(v,u)\to v+u$

Scaler Multiplication

Scaler multiplication on set $V$ is a function ($\mathbb{F}\times V\to V$) that takes $(\lambda ,v)\to \lambda v$. Note in the action above $\lambda \in \mathbb{F},v\in V$

Definition of a vector space:

A vector space is a set $V$ that satisfies given properties around scaler multiplication and addition. We say $V$ is a vector space over $\mathbb{F}$ if it meets these properties.

Properties:

1. Additive commutative property $a+b=b+aa,b\in V$
2. Additive Associative Property $(a+b)+c=a+(b+c)a,b,c\in V$
3. Additive Identity $\exists 0\in V:0+a=a\forall a\in V$
4. Additive Inverse $\forall a\in V\exists !b\in V:a+b=0$
5. Associative property of scaler multiplication $\forall a,b\in \mathbb{F};v\in V(ab)v=a(bv)$
6. Multiplicative identity $1v=v\forall v\in V$
7. Distributive properties $a(u+v)=au+ava\in \mathbb{F};u,v\in V$ and $(a+b)v=av+bva,b\in \mathbb{F};v\in V$ Remember, elements of a vector space are called vectors or points.

Notation

The scaler multiplication in a vector space depends on the underlying field $\mathbb{F}$. Thus, we say $V$ is a vector space over $\mathbb{F}$ when precision is needed. We do not always mention the underlying field if it is not needed.

Couple examples is that you already looked at is ${\mathbb{R}}^n$ is a vector space over $\mathbb{R}$. Another one is ${\mathbb{C}}^n$ is a vector space over $\mathbb{C}$

A real vector space is simply a vector space over $\mathbb{R}$. A complex vector space is a vector space over $\mathbb{C}$.

You can use the above axioms to verify that ${\mathbb{F}}^n$ is a vector space over $\mathbb{F}$ practice Similarly, prove that ${\mathbb{F}}^{\infty }$ is a vector space over $\mathbb{F}$. (Page 13)

Vector space of functions

A vector space could also be a set of functions.

For some set $S$, ${\mathbb{F}}^S$ denotes the set of functions from $S$ to $\mathbb{F}$. That means for any $f\in {\mathbb{F}}^S$ we have a function where $f:S\to \mathbb{F}$

Addition

For $f,g\in {\mathbb{F}}^S$ the sum $f+g\in {\mathbb{F}}^S$ is defined by: $(f+g)(x)=f(x)+g(x)x\in S$

Scaler Multiplication

For some $\lambda \in \mathbb{F}$ and $f\in {\mathbb{F}}^S$, the product $\lambda f\in {\mathbb{F}}^S$ is the function defined by $(\lambda f)(x)=\lambda f(x)x\in S$ As an example of a vector space of functions, if $S$ is the interval $[0,1]$ and $\mathbb{F}=\mathbb{R}$, we can think of ${\mathbb{R}}^{[0,1]}$ as the set of real valued functions on the interval $[0,1]$

${\mathbb{F}}^s$ is a vector space

Now, it is important to note that ${\mathbb{F}}^s$ is a vector space over $\mathbb{F}$ with the operations of addition and scaler multiplication defined above.

Additive identity and inverse

Additive identity:

Additionally, the additive identity is the function $0:S\to \mathbb{F}$ defined by $0(x)=0\forall x\in S$

Additive inverse:

The additive inverse of the function $f$ would be $-f:S\to \mathbb{F}$ defined as: $(-f)(x)=-f(x)\forall x\in S$ We can similarly think of ${\mathbb{F}}^n$ being a special case of functions as for any element $(x_1,x_2,....,x_n)\in {\mathbb{F}}^n$ we can think of it as a function in $x:\{1,2,...,n\}\to \mathbb{F}$. I.E, instad of $x_k$ for we denote it as $x(k)$. So for example, for the element $x=(5,8,3)\in {\mathbb{F}}^3$ instead of sating $x_1=5$ we can say $x(1)=5$ where x is a function taking $1$ to $5$. Similarly, we are thinking of ${\mathbb{F}}^3$ as being ${\mathbb{F}}^{[1,2,3]}$.

Elementary properties of vector spaces:

Unique additive identity

Vector spaces have an unique additive identity. Proof:

1. Pretend this was not the case and both $0$ and $0^{\prime }$ are additive identities.
2. Then notice the following: $0=0+0^{\prime }=0^{\prime }$
3. The left side of the equation is because any number added to $0^{\prime }$ returns itsself. Similarly, the right side is because any number added to $0$ would return itsself. Hence, if there was another additive identity it would have to be $0$ $■$

Note: the book does it by adding commutivity in the middle, I will leave that result here: $0=0+0^{\prime }=0^{\prime }+0=0^{\prime }$

Unique Additive inverse

Every element within a vector space has an unique additive inverse. Proof:

1. Suppose that was not the case, and both $b,b^{\prime }$ were additive inverses for some $a\in V$.
2. This means $a+b=0$ and $a+b^{\prime }=0$.
3. notice: $b=b+0=b+(a+b^{\prime })=b+a+b^{\prime }=(b+a)+b^{\prime }=0+b^{\prime }=b^{\prime }$
4. Hense $b=b^{\prime }$ and they would have to be the same.

$0v=0$ for scaler $0$

For this and the following proofs, remember $v\in V$ is not always in the form of a list, i.e. $(a_1,a_2,...,a_n)$. Thus we can not just say $0v=0(v_1,v_2,...,v_n)=(0v_1,0v_2,....,0v_n)=(0,0,...,0)=0$ as it may be that $v\ne (a_1,a_2,...,a_n)$

The product of the scaler $0$ times any vector $v\in V$ is always the vector $0$. Keep in mind the distinction between the vector $0$ and the scaler $0$ Proof:

1. Notice the following: $0v=(0+0)v=0v+0v$
2. From here, we can see that $0v=0v+0v$
3. Subtracting $0v$ from both sides we get the following: $0=0v-0v=0v+0v-0v=0v$
4. Hense we have it so $0=0v$ $■$

Similarly, you can prove that any scaler multiplied by the vector $0$ is the vector $0$ practice

$(-1)v=-v$ for all vectors $v\in V$

Proof:

1. Notice: $v+(-1)v=1v+(-1)v=(1+(-1))v=0v=0$
2. From here, we can subtract $v$ from both sides to get $(-1)v=-v$
3. Also, we proved that $(-1)v$ is always the additive inverse of $v$