|
|
|
Viewing Version
10
of
'vector space'
|
[ view 'vector space'
|
back to history
]
| Title of object: |
vector space |
| Canonical Name: |
VectorSpace |
| Type: |
Definition |
| Created on: |
2001-10-19 01:13:53 |
| Modified on: |
2004-10-24 03:56:22 |
| Classification: |
msc:15-00, msc:20-00, msc:13-00, msc:16-00 |
| Defines: |
zero vector |
| Synonyms: |
vector space=linear space |
Revision comment (for changes between this and next version):
| Changes for correction #5641 ('not a binary operation'). |
Preamble:
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{graphicx}
\usepackage{xypic}
\renewcommand{\u}{\mathbf{u}}
\renewcommand{\v}{\mathbf{v}}
\newcommand{\w}{\mathbf{w}}
\newcommand{\0}{\mathbf{0}} |
Content:
Let $F$ be a field. A {\em vector space} $V$ over $F$ is a set with two binary operations, $+: V \times V \longrightarrow V$ and $\cdot: F \times V \longrightarrow V$, such that
\begin{enumerate}
\item $(\u+\v)+\w = \u+(\v+\w)$ for all $\u,\v,\w \in V$
\item $\u+\v=\v+\u$ for all $\u,\v\in V$
\item There exists an element $\0 \in V$ such that $\u+\0=\u$ for all $\u \in V$
\item For any $\u \in V$, there exists an element $\v \in V$ such that $\u+\v=\0$
\item $a \cdot (b \cdot \u) = (a \cdot b) \cdot \u$ for all $a,b \in F$ and $\u \in V$
\item $1 \cdot \u = \u$ for all $\u \in V$
\item $a \cdot (\u+\v) = (a \cdot \u) + (a \cdot \v)$ for all $a \in F$ and $\u,\v \in V$
\item $(a+b) \cdot \u = (a \cdot \u) + (b \cdot \u)$ for all $a,b \in F$ and $\u \in V$
\end{enumerate}
Equivalently, a vector space is a module $V$ over a field $F$.
The elements of $V$ are called {\em vectors}, and the element $\0 \in V$ is called the {\em zero vector} of $V$. |
|
|
|
|
|