03. Vector Spaces

3.1 What is a vector space?

Real Vector Space

1. Non-empty set $V$

With binary operations

2. Addition: Binary operation $V\times V\to V$ defined by $(\mathbf{u},\mathbf{v})\mapsto \mathbf{u}+\mathbf{v}$
3. Scalar Multiplication: Binary Operation (Map) $\mathbb{R}\times V\to V$ defined by $(\lambda ,\mathbf{v})\mapsto \lambda \mathbf{v}$

And finally

4. Satisfies the Vector Space Axioms

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Vector Space Axioms (Refer to the main important ones underneath!)

1. Addition is Commutative

$$\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}\text{ for all }\mathbf{u},\mathbf{v}\in V$$

2. Addition is Associative

$$\mathbf{u}+(\mathbf{v}+\mathbf{w})=(\mathbf{u}+\mathbf{v})+\mathbf{w}\text{ for all }\mathbf{u},\mathbf{v},\mathbf{w}\in V$$

3. Existence of Additive Identity
   There exists $0_V\in V$ such that

$$\mathbf{v}+0_V=\mathbf{v}=0_V+\mathbf{v}\text{ for all }\mathbf{v}\in V$$

4. Existence of Additive Inverses
   For all $\mathbf{v}\in V$ there exists $\mathbf{w}\in V$ such that

$$\mathbf{v}+\mathbf{w}=0_V=\mathbf{w}+\mathbf{v}$$

5. Distributivity of Scalar Multiplication over Vector Addition

$$\lambda (\mathbf{u}+\mathbf{v})=\lambda \mathbf{u}+\lambda \mathbf{v}\text{ for all }\mathbf{u},\mathbf{v}\in V,\lambda \in \mathbb{R}$$

6. Distributivity of Scalar Multiplication over Field Addition

$$(\lambda +\mu )\mathbf{v}=\lambda \mathbf{v}+\mu \mathbf{v}\text{ for all }\mathbf{v}\in V,\lambda ,\mu \in \mathbb{R}$$

7. Scalar Multiplication interacts well with Field Multiplication

$$(\lambda \mu )\mathbf{v}=\lambda (\mu \mathbf{v})\text{ for all }\mathbf{v}\in V,\lambda ,\mu \in \mathbb{R}$$

8. Identity for Scalar Multiplication

$$1\mathbf{v}=\mathbf{v}\text{ for all }\mathbf{v}\in V$$

Generally the field refers to $\mathbb{R}$, so the typical addition and multiplication you’ve dealt with before

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Vector Space Axioms (Important Ones)

• $V$ has a zero vector $0_V$ (Axiom 3)
• $V$ is closed under addition (Axiom $1+2+3?+4$)
• $V$ is closed under scalar multiplication (Axiom $5+6+7+8$)

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Vector Space for $\mathbb{R}$

$${\mathbb{R}}^n\text{ represents the set of }n-\text{tuples }(v_1,\cdots ,v_n)\text{ with }{v}_1,v_n\in \mathbb{R}$$

With addition and scalar multiplication defined component wise such that

$$(x_1,x_2,\cdots ,x_n)+(y_1,y_2,\cdots ,y_n)=(x_1+y_1,x_2+y_2,\cdots ,x_n+y_n)$$

and

$$\lambda (x_1,x_2,\cdots ,x_n)=(\lambda x_1,\lambda x_2,\lambda x_n)$$

These satisfy the Vector Space Axioms

Zero vector is defined as $(0,0,\cdots ,0)$
Additive Inverse of $(v_1,v_2,\cdots ,v_n)$ is $(-v_1,-v_2,\cdots ,-v_n)$

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Other Examples of Vector Spaces

• $\mathbb{C}$ is a real vector space (isomorphic to ${\mathbb{R}}^2$)
• ${\mathcal{M}}_{m\times n}(\mathbb{R})$ for $m,n\ge 1$ - same as Matrix Notation
• $V={\mathbb{R}}^{\mathbb{R}}=f:\mathbb{R}\to \mathbb{R}$ - real-valued function defined on $\mathbb{R}$ with addition and scalar multiplication defined pointwise

$$(f+g)(r):=f(r)+g(r),(\alpha f)(r):=\alpha f(r)\text{ for }f,g\in V\text{ and }\alpha ,r\in \mathbb{R}$$

• $V=\{f:\mathbb{R}\to \mathbb{R},f\text{ is differentiable}\}$

$$(f+g{)}^{\prime }=f^{\prime }+g^{\prime },(\alpha f{)}^{\prime }=\alpha f^{\prime }\text{ for }f,g\in V\text{ and }\alpha \in \mathbb{R}$$

• Vector Spaces can also be defined over any Fields (Examples) (refer to Fields (Defintion))

Isomorphic - Refers to groups but it means “essentially the same as”

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Uniqueness of Additive Identity Element (typically written as $0_V$ or $0$)

Suppose $0,0^{\prime }$ are two elements that satisfy the properties of $0_V$ lemma

$$\begin{array}{rl}0& =0+0^{\prime }[\text{as }{0}^{\prime }\text{ is a zero vector}]\\ & =0^{\prime }[\text{ as }0\text{ is a zero vector}]\end{array}$$

Hence $0=0^{\prime }$ so $0_V$ is unique

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Uniqueness of Additive Inverse Element (typically written as $-v$ (inverse of $v$)) lemma

Suppose $w_1,w_2$ are additive inverses of $v$ then

$$w_1+v=0\text{ and }{w}_2+v=0$$

Hence

$$\begin{array}{rlrl}& & & \\ w_2& =0+w_2& & [\text{ as }0\text{ is a zero vector}]\\ & =(w_1+v)+w_2& & [\text{by statement above}]\\ & =w_1+(v+w_2)& & [\text{by associativity}]\\ & =w_1+0& & [\text{by statement above}]\\ & =w_1& & [\text{ as }0\text{ is a zero vector}]\end{array}$$

So $w_1=w_2$ (hence the additive inverse element is unique)

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Properties of a Vector Space ( $V$) over field $\mathbf{F}$

For $v\in V,\lambda \in \mathbf{F}$

1. $\lambda 0_V=0_V$

2. $0v=0_V$

3. $(-\lambda )v=-(\lambda v)=\lambda (-v)$

4. If $\lambda v=0_V$ then $\lambda =0$ or $v=0_V$

5. $-v=(-1)v$

Proof

$$\begin{array}{rl}\lambda 0_V& =\lambda (0_V+0_V)[\text{definition of additive identity}]\\ & =\lambda 0_V+\lambda 0_V[\text{distributivity of scalar }\cdot \text{ over }+]\end{array}$$

Adding $-(\lambda 0_V)$ to both sides then

$$0_V=\lambda 0_V$$

$$\begin{array}{rl}0v& =(0+0)v[\text{definition of additive identity in fields}]\\ & =0v+0v[\text{distributivity of scalar }\cdot \text{ over vector }+]\end{array}$$

Adding $-(0v)$ to both sides

$$0_V=0v$$

$$\begin{array}{rl}\lambda v+\lambda (-v)& =\lambda (v+(-v))[\text{distributivity of scalar}\cdot \text{ over vector }+]\\ & =\lambda 0_V[\text{defintion of additive inverse}]\\ & =0_V[\text{by (b)}]\end{array}$$

4. Suppose that $\lambda v=0_V$ and $\lambda \ne 0$ so that ${\lambda }^{-1}$ exists in $\mathbf{F}$ then

$${\lambda }^{-1}(\lambda v)={\lambda }^{-1}{0}_V=0_V[\text{ by (a)}]$$

So

$$\begin{array}{rl}({\lambda }^{-1}\lambda )& =0_V[\text{scalar }\cdot \text{ interacts well with field }\cdot ]\\ v& =1v=0_V[\text{identity for scalar multiplication}]\end{array}$$

$$\begin{array}{rl}v+(-1)v& =1v+(-1)v\text{ by a vector space axioms (8)}\\ & =(1+(-1))v[\text{distributivity}]\\ & =0v[\text{defintion of additive inverse in field}]\\ & =0_V[\text{by (A)}]\end{array}$$

Hence by uniqueness of the additive inverse $(-1)v=-v$

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3.2 Subspaces

Subspace $U$ of vector space $V$ over $\mathbf{F}$

Non-empty subset of $V$ (so $U\subseteq V$)

• Closed under addition and scalar multiplication

Satisfies the following

1. $U\ne \varnothing$: $U$ is nonempty
   Generally done by showing $0_V\in U$
2. $u_1+u_2\in U$ for all $u_1,u_2\in U$: $U$ is closed under addition
3. $\lambda u\in U$ for all $u\in U$, $\lambda \in \mathbf{F}$: $U$ is closed under scalar multiplication

Addition and Scalar Multiplication are defined the same as $V$

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Zero Subspace

$$\{0_V\}$$

Just the $0_V$

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Trivial Subspace

$$V$$

Just the vector space itself

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![[17 - Subspace#^97b40a|^97b40a]]

Subspace Notation

$$U\le V⟺U\text{ is a subspace of }V$$

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Properties of Subspaces

Let $V$ be a vector space over $\mathbf{F}$ and let $U\le V$

• $U$ is a vector space over $\mathbf{F}$
  Turns out the only subsets of $V$ that are vector spaces over $\mathbb{F}$ are the subspaces
• If $W\le U$ then $W\le V$
  Subspace of a subspace is also a subspace

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Notation on "addition" of subspaces

Let $U,W\le V$ where $V$ is a vector space

$$U+W=\{u+w\mid u\in U,w\in W\}$$

Where $U+W$ is also a subspace of $V$ ($U+W\le V$)

Proof for Subspace

• Condition 1)
  As $0_V\in U$ and $0_V\in W$ then $0_V+0_V=0_V\in U+W$

• Condition2)
  For $v_1,v_2\in U+W$ then by definition of $U+W$ there exists $u_1,u_2\in U$ and $w_1,w_2\in W$ such that

$$v_1=u_1+w_1\text{ and }{v}_2=u_2+w_2$$

As $V$ is a subspace then $\lambda u_1+u_2\in U$
As $W$ is a subspace then $\lambda w_1+w_2\in W$
So

$$(\lambda u_1+u_2)+(\lambda w_1+w_1)=\lambda (u_1+w_1)+(u_2+w_2)=\lambda v_1+v_2\in U+W$$

Hence $U+W$ is a subspace

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Notation on "intersection" of subspaces

Let $U,W\le V$ where $V$ is a vector space

$$U\cap W=\{v\mid v\in V\text{ if }v\in U\text{ and }v\in W\}$$

Where $U\cap W$ is also a subspace of $V$ ($U\cap W\le V$)
Largest subspace of $V$ that is contained in both $U$ and $W$

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3.3 Further Examples

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