Chapter 1 extra note 1

Vector space

Youtube

• What is a Vector Space? (Abstract Algebra)

Selected lecture notes:

Vector space

A vector space

$V$ over a field

$F$ is a collection of vectors together with vector addition and scalar multiplication that satisfy the following

$8$ properties: For all

$\mathbf{u}\mathbf{,}\mathbf{v}\mathbf{,}\mathbf{w}\in V$ and

$\alpha ,\beta \in F$,

1. Commutativity:
   $\mathbf{u}\mathbf{+}\mathbf{v}=\mathbf{v}\mathbf{+}\mathbf{u}$;
2. Associativity:
   $\mathbf{(}\mathbf{u}\mathbf{+}\mathbf{v}\mathbf{)}\mathbf{+}\mathbf{w}=\mathbf{u}\mathbf{+}\mathbf{(}\mathbf{v}\mathbf{+}\mathbf{w}\mathbf{)}$;
3. Zero vector: There exists
   $\mathbf{0}\in V$ such that
   $\mathbf{u}\mathbf{+}\mathbf{0}=\mathbf{u}$;
4. Additive inverse: For every vector
   $\mathbf{u}\in V$, there exists a vector
   ${\mathbf{u}}^{\prime }\in V$ s.t.
   $\mathbf{u}\mathbf{+}{\mathbf{u}}^{\prime }=\mathbf{0}$;
5. Multiplicative identity:
   $1\mathbf{v}=\mathbf{v}$;
6. Multiplicative associativity:
   $(\alpha \beta )\mathbf{v}=\alpha (\beta \mathbf{v})$;
7. $\alpha (\mathbf{u}\mathbf{+}\mathbf{v})=\alpha \mathbf{u}+\alpha \mathbf{v}$;
8. $(\alpha +\beta )\mathbf{u}=\alpha \mathbf{u}+\beta \mathbf{u}$.

Theorems

9. If
   $\mathbf{u}\mathbf{+}\mathbf{v}=\mathbf{u}\mathbf{+}\mathbf{w}$, then
   $\mathbf{v}=\mathbf{w}$.
10. $\mathbf{0}$ is unique.
11. $0\cdot \mathbf{u}=\mathbf{0}$.
12. $\alpha \cdot \mathbf{0}=\mathbf{0}$.
13. Additive inverse is unique, and equals to
    $(-1)\cdot \mathbf{u}$.

Proof of the theorems:

• Proof of 9:

  pf:
  Since

  $\mathbf{u}\in V$, (by 4) there exists

  ${\mathbf{u}}^{\prime }\in V$ such that

  $\mathbf{u}\mathbf{+}{\mathbf{u}}^{\prime }=\mathbf{0}$.
  Then we add

  ${\mathbf{u}}^{\prime }$ to both side of

  $\mathbf{u}\mathbf{+}\mathbf{v}=\mathbf{u}\mathbf{+}\mathbf{w}$.
  The left hand side gives
  

  $$\mathbf{(}\mathbf{u}\mathbf{+}\mathbf{v}\mathbf{)}\mathbf{+}{\mathbf{u}}^{\prime }=\mathbf{(}\mathbf{v}\mathbf{+}\mathbf{u}\mathbf{)}\mathbf{+}{\mathbf{u}}^{\prime }=\mathbf{v}\mathbf{+}\mathbf{(}\mathbf{u}\mathbf{+}{\mathbf{u}}^{\prime }\mathbf{)}=\mathbf{v}\mathbf{+}\mathbf{0}=\mathbf{v},$$
  where we have used 1, 2 and 3.
  Similarly, the right hand side gives

  $\mathbf{(}\mathbf{u}\mathbf{+}\mathbf{w}\mathbf{)}\mathbf{+}{\mathbf{u}}^{\prime }=\mathbf{w}$.
  Making both sides equal gives

  $\mathbf{v}=\mathbf{w}$.

• Proof of 10:

  pf:
  Suppose there exists

  ${\mathbf{0}}_{\mathbf{1}}\in V$ and

  ${\mathbf{0}}_{\mathbf{2}}\in V$ such that

  $\mathbf{u}\mathbf{+}{\mathbf{0}}_{\mathbf{1}}=\mathbf{u}\mathbf{+}{\mathbf{0}}_{\mathbf{2}}=\mathbf{u}$ for all

  $\mathbf{u}\in V$.
  Since

  ${\mathbf{0}}_{\mathbf{1}}$ and

  ${\mathbf{0}}_{\mathbf{2}}\in V$, we must have

  ${\mathbf{0}}_{\mathbf{2}}\mathbf{+}{\mathbf{0}}_{\mathbf{1}}={\mathbf{0}}_{\mathbf{2}}$ and

  ${\mathbf{0}}_{\mathbf{1}}\mathbf{+}{\mathbf{0}}_{\mathbf{2}}={\mathbf{0}}_{\mathbf{1}}$. Therefore, using 1),
  

  $${\mathbf{0}}_{\mathbf{1}}={\mathbf{0}}_{\mathbf{1}}\mathbf{+}{\mathbf{0}}_{\mathbf{2}}={\mathbf{0}}_{\mathbf{2}}\mathbf{+}{\mathbf{0}}_{\mathbf{1}}={\mathbf{0}}_{\mathbf{2}}.$$
  So the zero vector is unique.

• Proof of 11:

  pf:
  Given

  $\mathbf{u}\in V$, using 3 and 8, we have
  

  $$1\cdot \mathbf{u}+\mathbf{0}=1\cdot \mathbf{u}=(1+0)\cdot \mathbf{u}=1\cdot \mathbf{u}+0\cdot \mathbf{u}.$$
  Then by 9), we have

  $0\cdot \mathbf{u}=\mathbf{0}$.

• Proof of 12:

  pf:
  Given

  $\alpha \in F$, let

  $\mathbf{u}\in V$, using 7 and 3,
  

  $$\alpha \cdot \mathbf{u}+\alpha \cdot \mathbf{0}=\alpha (\mathbf{u}\mathbf{+}\mathbf{0})=\alpha \cdot \mathbf{u}=\alpha \cdot \mathbf{u}+\mathbf{0}.$$
  Then by 9), we have

  $\alpha \cdot \mathbf{0}=\mathbf{0}$.

• Proof of 13-1:

  pf:
  Let

  $\mathbf{u}\in V$, suppose there exists

  ${\mathbf{u}}_{\mathbf{1}}^{\prime }\in V$ and

  ${\mathbf{u}}_{\mathbf{2}}^{\prime }\in V$ such that

  $\mathbf{u}\mathbf{+}{\mathbf{u}}_{\mathbf{1}}^{\prime }=\mathbf{0}$ and

  $\mathbf{u}\mathbf{+}{\mathbf{u}}_{\mathbf{2}}^{\prime }=\mathbf{0}$. Since

  $\mathbf{0}$ is unique by 10), we have
  

  $$\mathbf{u}\mathbf{+}{\mathbf{u}}_{\mathbf{1}}^{\prime }=\mathbf{u}\mathbf{+}{\mathbf{u}}_{\mathbf{2}}^{\prime }.$$
  Then by 9), we have

  ${\mathbf{u}}_{\mathbf{1}}^{\prime }={\mathbf{u}}_{\mathbf{2}}^{\prime }$.

• Proof of 13-2:

  pf:
  Let

  $\mathbf{u}\in V$, using 5, 8 and 11,
  

  $$\mathbf{u}+(-1)\cdot \mathbf{u}=1\cdot \mathbf{u}+(-1)\cdot \mathbf{u}=(1+(-1))\mathbf{u}=0\cdot \mathbf{u}=\mathbf{0}.$$
  So its additive inverse is

  $(-1)\cdot \mathbf{u}$.

Examples

Examples of vector space

• With usual vector addition and scalar multiplication
  • ${\mathbb{R}}^n$: n-dimensional Euclidean space.
  • ${\mathbb{P}}_n$: Polynomials of degree at most
    $n$.
  • $M_{m\times n}$:
    $m\times n$ matrices.
  • $C([0,1])$: Continuous function defined on
    $[0,1]$.
  • The set of all the even functions on
    $\mathbb{R}$. (
    $f(x)=f(-x)$,
    $x\in \mathbb{R}$.)
• a)
  $V=\{x|x\in {\mathbb{R}}^{+}\}$ with
  $\oplus$ and
  $\odot$ defined as
  
  $$x\oplus y=xy,\alpha \odot y=y^{\alpha }.$$
• b)
  $V=\{(x_1,x_2,x_3)|x_1,x_2,x_3\in {\mathbb{R}}^{+},x_1+x_2+x_3=1\}$,
  $F=\mathbb{R}$, with
  $\oplus$ and
  $\odot$ defined as
  
  $$\begin{array}{rl}(x_1,x_2,x_3)\oplus (y_1,y_2,y_3)& =\frac{1}{x_1{y}_1+x_2{y}_2+x_3{y}_3}(x_1{y}_1,x_2{y}_2,x_3{y}_3),\\ \alpha \odot (x_1,x_2,x_3)& =\frac{1}{x_1^{\alpha }+x_2^{\alpha }+x_3^{\alpha }}(x_1^{\alpha },x_2^{\alpha },x_3^{\alpha }).\end{array}$$

Examples of non-vector space

• The set of all polynomials of degree exactly
  $n$.
• The set of all continuous and positive functions on
  $[0,1]$.
• c) Let
  $V={\mathbb{R}}^2$ with
  $\oplus$ and
  $\odot$ defined as
  
  $$(a_1,a_2)\oplus (b_1,b_2)=(a_1+b_1,a_2{b}_2),\alpha \odot (a_1,b_1)=(\alpha a_1,b_1).$$
• d) Let
  $V={\mathbb{R}}^2$ with usual
  $\odot$ while
  $\oplus$ is defined as
  
  $$(a_1,a_2)\oplus (b_1,b_2)=(a_1+2b_1,a_2+2b_2).$$
• e) Let
  $V={\mathbb{R}}^2$ with usual vector
  $\oplus$ while
  $\odot$ is defined as
  
  $$\alpha \odot (a_1,b_1)=\{\begin{array}{ll}(\frac{a_1}{\alpha },\frac{a_2}{\alpha }),& \text{if }\alpha \ne 0\text{,}\\ (0,0),& \text{if }\alpha =0\text{.}\end{array}$$