Finite Dimensional Vector Spaces

Span §

[!math|{“type”:“definition”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Linear Combination”,“label”:“linear-combination”,“_index”:0}] Definition 1 (Linear Combination). A vector of the form

$$\sum _{i=1}^n{a}_i\cdot \overrightarrow{v_i}$$

where $\overrightarrow{v_i}\in$ some vector space $V$ over $\mathbb{F}$ and $a_i\in \mathbb{F}$.

[!math|{“type”:“definition”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Span”,“label”:“span”,“_index”:1}] Definition 2 (Span). For $K$-vector space $V$, $S=\{v_1,v_2,\dots ,v_n\}\subset V$, $\text{span}(S)=\{a_1{\vec{v}}_1+\cdots +a_n{\vec{v}}_n\}$ (set of all linear combinations). $S$ spans $V$ (or generates $V$) if $\text{span}(S)=V$ aka .

[!math|{“type”:“definition”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Finite Dimensional”,“label”:“finite-dimensional”,“_index”:2}] Definition 3 (Finite Dimensional). A vector space is finite dimensional if finite set of vectors spans it.

• example: ${\mathbb{R}}^2$ is finite dimensional
• $\vec{u}\in {\mathbb{R}}^2=a{\vec{v}}_1+b{\vec{v}}_2$ for $a,b\in \mathbb{R}$ if $\overrightarrow{v_1},\overrightarrow{v_2}$ are not on the same line through $(0,0)$.
• this property of $\{\overrightarrow{v_1},\overrightarrow{v_2}\}$ is called

Linear Independence §

[!math|{“type”:“definition”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Linear Independence”,“label”:“linear-independence”,“_index”:3}] Definition 4 (Linear Independence). Given a set $S\subset V$, $S$ is linearly independent if given $a_1\overrightarrow{v_1}+\cdots +a_n\overrightarrow{v_n}=\vec{0}$ with $a_i\in K$ and $\overrightarrow{v_i}\in S$ distinct, then $a_1=a_2=\cdots =a_n=0$.

[!math|{“type”:“theorem”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:4}] Theorem 5. If $S$ is linearly independent then $T\subset S$ is linearly independent.

• opposite would be

[!math|{“type”:“definition”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Linear Dependence”,“label”:“linear-dependence”,“_index”:5}] Definition 6 (Linear Dependence). $S$ is linearly dependent if for $a_1,\dots ,a_k\in K$, $\overrightarrow{v_1},\dots ,\overrightarrow{v_n}\in S$, $a_1\overrightarrow{v_1}+\cdots +a_n\overrightarrow{v_n}=\vec{0}$ w/ some $a_i\ne 0$.

Bases §

[!math|{“type”:“definition”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Basis”,“label”:“basis”,“_index”:6}] Definition 7 (Basis). Basis of $V$ ($\mathcal{B}(V)$) is a list of vectors that is linearly independent and spans $V$.

[!math|{“type”:“proposition”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Basis”,“label”:“basis”,“_index”:7}] Proposition 8 (Basis). $S\subset V$ is a basis $⟺$ every vector $\vec{v}\in V$ can be expressed uniquely as $\vec{v}=a_1\overrightarrow{v_1}+\cdots +a_n\overrightarrow{v_n}$.

\begin{proof} $⟸$: definition of span $⟹$: for $\vec{v}=a_1\overrightarrow{v_1}+\cdots +a_n\overrightarrow{v_n}$ and $\vec{w}=b_1\overrightarrow{v_1}+b_n\overrightarrow{v_n}$, $\vec{v}=\vec{w}⟹\vec{v}-\vec{w}=\vec{0}⟹(a_1-b_1)\overrightarrow{v_1}+\cdots +(a_n-b_n)\overrightarrow{v_n}⟹a=b$ \end{proof}

[!math|{“type”:“example”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Standard Basis”,“label”:“standard-basis”,“_index”:8}] Example 9 (Standard Basis). $V=K^n$ has a standard basis $S=\{e_1,\dots ,e_n\}$: $\overrightarrow{e_1}=(1,0,\dots ,0)$ $\overrightarrow{e_2}=(0,1,\dots ,0)$ … $\overrightarrow{e_n}=(0,0,\dots ,1)$

• example: a polynomial of degree $\le d$ has standard basis $\{1,x,x^2,\dots ,x^d\}$

Dimension §

[!math|{“type”:“definition”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Dimension”,“label”:“dimension”,“_index”:9}] Definition 10 (Dimension). If $V$ is basis $S$, dimension of $V$, or $\text{dim}(V)$ is $|S|$

[!math|{“type”:“claim”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:10}] Claim 11. $\text{dim}$ is well defined: any two bases of a finite dimensional vector space have same length.

\begin{proof}Suppose $S,T$ are bases. Then, if $S$ is linearly independent and $T$i s spanning,

$$|S|\le |T|.$$

Since we can flip around WLOG,

$$|T|\le |S|.$$

Thus, $|S|=|T|$. \end{proof}

Connection to Sets §

• finite set $⟺$ finite dimensional
• $|A|⟺\text{dim}V$
• $A_1\cup A_2⟺V_1+V_2$
• $|A_1\cup A_2|=|A_1|+|A_2|-|A_1\cap A_2|⟺\text{dim}(V_1+V_2)=\text{dim}(V_1)+\text{dim}(V_2)-\text{dim}(V_1\cap V_2)$