01. Vector Spaces and Linear Maps

1.1 Vector Spaces

Field

Set $\mathbf{F}$ with two binary operations $+$ and $\times$ is a field if

$$(\mathbf{F},+,0)\text{ and }(\mathbf{F}\setminus \{0\},\times ,1)\text{ are abelian groups}$$

with the distributive law holding

$$(a+b)c=ac+bc\text{ for all }a,b,c\in \mathbf{F}$$

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Characteristic of a field $\mathbf{F}$

Smallest integer $p$ such that

$$1+1+\cdots +1(p\text{ times})=0$$

where $p$ is known as the characteristic of $\mathbf{F}$
Otherwise if $p$ doesn’t exist then the characteristic of $F$ is defined to be $0$

Note that $p$ is always prime

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![[02 - Vector Spaces#^2a6eb3|^2a6eb3]]

Linearly Independent Set

Let $V$ be a vector space over $\mathbf{F}$

Set $S\subseteq V$ is linearly independent if for

$$a_1,\cdots ,a_n\in \mathbf{F}\text{ and }{s}_1,\cdots ,s_n\in S$$

then

$$a_1{s}_1+\cdots +a_n{s}_n⟹a_1=\cdots =a_n=0$$

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Spanning Set

Let $V$ be a vector space over $\mathbf{F}$

Set $S\subseteq V$ is spanning if for all $v\in V$ there exists

$$a_1,\cdots ,a_n\in \mathbf{F}\text{ and }{s}_1,\cdots ,s_n\in S$$

such that

$$v=a_1{s}_1+\cdots +a_n{s}_n$$

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Basis Set

Let $V$ be a vector space over $\mathbf{F}$

Set $\mathcal{B}\subseteq V$ is a basis of $V$ if

$$B\text{ is spanning and linearly independent}$$

Dimension of a Vector Space Size of $\mathcal{B}$ is the dimension of $V$

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1.2 Linear Maps

Linear Map / Transformation

Suppose $V$ and $W$ are Vector Spaces over $\mathbf{F}$

Map $T:V\to W$ is a linear transformation (or linear map)
If for all $a\in \mathbf{F}$ and $v,v^{\prime }\in V$ then

$$T(av+v^{\prime })=aT(v)+T(v^{\prime })$$

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Isomorphism of Vector Spaces

Bijective linear map of vector spaces

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Homomorphism between Vector Spaces

Let $V,W$ be vector spaces then
Define $\mathrm{Hom}(V,W)$ be the set of linear maps from $V$ to $W$

For $a\in \mathbf{F}$ and $v\in V$ and $S,T\in \mathrm{Hom}(V,W)$ define

$$\begin{array}{rl}(aT)(v)& :=a(T(v))\\ (T+S)(v)& :=T(v)+S(v)\end{array}$$

Then $\mathrm{Hom}(V,W)$ is a vector space over $\mathbf{F}$

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Linear Maps are determined by a basis

Let $V,W$ be finite dimensional vector spaces

Every linear map $T:V\to W$ is determined by its values on a basis $\mathcal{B}$ for $V$ (as $\mathcal{B}$ is spanning)

and vice versa so

Any map $T:\mathcal{B}\to W$ can be extended to linear map $T:V\to W$ (as $\mathcal{B}$ is linearly independent)

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Matrix Representation of a Linear Map

Let $\mathcal{B}=\{e_1,\cdots ,e_n\}$ and ${\mathcal{B}}^{\prime }=\{e_1^{\prime },\cdots ,e_m^{\prime }\}$
Let ${}_{{\mathcal{B}}^{\prime }}[T{]}_{\mathcal{B}}$ be the matrix with $(i,j)$-entry $a_{ij}$ such that

$$T(e_j)=a_{1j}{e}_1^{\prime }+\cdots +a_{mj}{e}_m^{\prime }$$

where $\mathcal{B}$ is the initial basis and ${\mathcal{B}}^{\prime }$ is the final basis

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Properties of Matrix Representation of a Linear Map

1. Scalar Multiplication

$${}_{{\mathcal{B}}^{\prime }}[aT{]}_{\mathcal{B}}=a({}_{{\mathcal{B}}^{\prime }}[T{]}_{\mathcal{B}})$$

2. Addition

$${}_{{\mathcal{B}}^{\prime }}[T+S{]}_{\mathcal{B}}={}_{{\mathcal{B}}^{\prime }}[T{]}_{\mathcal{B}}+{}_{{\mathcal{B}}^{\prime }}[S{]}_{\mathcal{B}}$$

3. Composition

$${}_{{\mathcal{B}}^{\prime \prime }}[S\circ T{]}_{\mathcal{B}}={}_{{\mathcal{B}}^{\prime \prime }}[S{]}_{{\mathcal{B}}^{\prime }}{}_{{\mathcal{B}}^{\prime }}[T{]}_{\mathcal{B}}$$

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01 - Isomorphism between Linear Maps and Matrices

Isomorphism between Linear Maps and Matrices

Map $T\mapsto {}_{{\mathcal{B}}^{\prime }}[T{]}_{\mathcal{B}}$ is an isomorphism from $\mathrm{Hom}(V,W)$ to ${\mathbf{M}}_{m\times n}(\mathbf{F})$

That is it takes composition of maps to multiplication of matrices

Special Case $T:V\to V$ and $\mathcal{B}$, ${\mathcal{B}}^{\prime }$ are two different bases with ${}_{{\mathcal{B}}^{\prime }}[\mathrm{Id}{]}_{\mathcal{B}}$ the change of basis matrix then

If

$${}_{{\mathcal{B}}^{\prime }}[T{]}_{{\mathcal{B}}^{\prime }}={}_{{\mathcal{B}}^{\prime }}[\mathrm{Id}{]}_{\mathcal{B}}{}_{{\mathcal{B}}^{\prime }}[T{]}_{\mathcal{B}}{}_{\mathcal{B}}[\mathrm{Id}{]}_{{\mathcal{B}}^{\prime }}$$

with

$${}_{\mathcal{B}}[\mathrm{Id}{]}_{{\mathcal{B}}^{\prime }}{}_{{\mathcal{B}}^{\prime }}[\mathrm{Id}{]}_{\mathcal{B}}={}_{\mathcal{B}}[\mathrm{Id}{]}_{\mathcal{B}}=I$$

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