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%TCIDATA{<META NAME="Title" CONTENT="Reference/Mathematics/Basis and Dimension">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Fri Mar 28 12:36:49 1997}
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\section{Basis and Dimension\label{Basis and Dimension}}

\subsubsection{Basis\label{basis}}

A set of vectors $v_{1},v_{2},...,v_{n}$ in $V$ is a \textbf{basis} for $V$
if the vectors are \hyperref{linearly independent}{}{}{MA02_03.tex#linearly
independent} and \hyperref{span}{}{}{MA02_02.tex#span} $V$.

\begin{itemize}
\item  The \textbf{standard basis\label{standard basis}} for $%
\UNICODE{0x211d} ^{n}$ is the set of vectors $e_{1},e_{2},...e_{n}$ where $
\begin{array}{l}
e_{1}=\left( 1,0,0,...,0\right)  \\ 
e_{2}=\left( 0,1,0,...,0\right)  \\ 
e_{3}=\left( 0,0,1,...,0\right)  \\ 
\vdots  \\ 
e_{n}=\left( 0,0,0,...,1\right) 
\end{array}
$.
\end{itemize}

\begin{theorem}
If $v_{1},v_{2},...,v_{n}$ and $w_{1},w_{2},...,w_{n}$ are both bases for a
vector space $V$, then $m=n$.
\end{theorem}

\subsubsection{Dimension\label{dimension}}

A vector space $V$ has \textbf{dimension} $n$ if $V$ has a basis consisting
of $n$ vectors. The dimension of $V$ is denoted by $\dim \left( V\right) $.

\begin{theorem}
Suppose that $V$ is a vector space of dimension $n$. Then no set of more
than $n$ vectors can be linearly independent, and no set of fewer than $n$
vectors can span $V$.
\end{theorem}

\begin{theorem}
Suppose that $V$ is a vector space of dimension $n$ and $%
v_{1},v_{2},...,v_{n}$ is a set of vectors in $n$. Then

\begin{enumerate}
\item  If the vectors are linearly independent, then they span the space and
thus form a basis for $V$.

\item  If the vectors span $V$, then they must be linearly independent and
thus form a basis for $V$.
\end{enumerate}
\end{theorem}

\subsection{\ Related Scientific Notebook Topics}

{\small \hyperref{Bases for column space and row space}{}{}{%
../../Help/DM8-7.tex#Matrices, rank}}

{\small \hyperref{Nullspaces and nullspace basis}{}{}{%
../../Help/DM8-7.tex#Nullspaces}}

\noindent \hrulefill

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