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%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Thu Feb 13 17:10:29 1997}
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\begin{document}


\section{Vector Spaces Associated with a Matrix\label{Row space}}

There are four vector spaces naturally associated with an $m\times n$ matrix 
$A$: the row space, the column space, and the left and right nullspaces. A 
\textsl{basis} for a vector space is a linearly independent set of vectors
that spans the space.

\subsection{The Row Space}

The \textsl{row space} is the 
\index{Matrices@Matrices!row basis@row basis}\label{Matrices, row space}%
vector space spanned by the row vectors of $A$.

\begin{quote}
$\blacktriangleright $ To find a basis for the row space
\end{quote}

\begin{enumerate}
\item[1.]  Leave the insertion point in the matrix.

\item[2.]  From the \textsf{Matrices} submenu, choose \textsf{Row Basis}$.$
\end{enumerate}

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Row Basis}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{rrrr}
-85 & -55 & -37 & -35 \\ 
97 & 50 & 79 & 56 \\ 
49 & 63 & 57 & -59 \\ 
-36 & 8 & 20 & -94
\end{array}
\right] $, row basis:

\[
\left[ 
\begin{array}{cccc}
1 & 0 & 0 & 
\dfrac{133337}{68264}
\end{array}
\right] ,\left[ 
\begin{array}{cccc}
0 & 1 & 0 & -\dfrac{74049}{34132}
\end{array}
\right] ,\left[ 
\begin{array}{cccc}
0 & 0 & 1 & -\dfrac{3085}{9752}
\end{array}
\right] 
\]
\end{quotation}

You can find other bases by choosing \textsf{Reduced Row Echelon Form} from
the \textsf{Matrices} submenu, or by applying \textsf{Fraction-Free Gaussian
Elimination} and then taking the nonzero rows from the result.

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Reduced Row Echelon Form}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{rrrr}
-85 & -55 & -37 & -35 \\ 
97 & 50 & 79 & 56 \\ 
49 & 63 & 57 & -59 \\ 
-36 & 8 & 20 & -94
\end{array}
\right] $, row echelon form: $\left[ 
\begin{array}{rrrr}
1 & 0 & 0 & \dfrac{133337}{68264}\vspace{6pt} \\ 
0 & 1 & 0 & -\dfrac{74049}{34132}\vspace{6pt} \\ 
0 & 0 & 1 & -\dfrac{3085}{9752}\vspace{6pt} \\ 
\multicolumn{1}{c}{0} & \multicolumn{1}{c}{0} & \multicolumn{1}{c}{0} & 
\multicolumn{1}{c}{0}
\end{array}
\right] $
\end{quotation}

The preceding calculation gives the same basis as found in the previous
example.

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Fraction-free Gaussian Elimination}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{rrrr}
-85 & -55 & -37 & -35 \\ 
97 & 50 & 79 & 56 \\ 
49 & 63 & 57 & -59 \\ 
-36 & 8 & 20 & -94
\end{array}
\right] $, fraction-free Gaussian elimination: 
\[
\left[ 
\begin{array}{rrrr}
-85 & -55 & -37 & -35 \\ 
0 & 1085 & -3126 & -1365 \\ 
0 & 0 & 136528 & -43190 \\ 
0 & 0 & 0 & 0
\end{array}
\right] 
\]
$\,$
\end{quotation}

The nonzero rows in the preceding matrix give a basis for the row space:

\begin{center}
$\left[ 
\begin{array}{rrrr}
-85 & -55 & -37 & -35
\end{array}
\right] $, $\left[ 
\begin{array}{rrrr}
0 & 1085 & -3126 & -1365
\end{array}
\right] $, $\left[ 
\begin{array}{rrrr}
0 & 0 & 136528 & -43190
\end{array}
\right] $.
\end{center}

\strut

\subsection{The Column Space\label{Column space}}

The \textsl{column space} is the 
\index{Matrices@Matrices!column basis@column basis}\label%
{Matrices, column space}vector space spanned by the columns of $A$.

\begin{quote}
$\blacktriangleright $ To find a basis for the column space
\end{quote}

\begin{enumerate}
\item[1.]  Leave the insertion point in the matrix.

\item[2.]  From the \textsf{Matrices} submenu, choose \textsf{Column Basis}.
\end{enumerate}

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Column Basis}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{rrrr}
-85 & -55 & -37 & -35 \\ 
97 & 50 & 79 & 56 \\ 
49 & 63 & 57 & -59 \\ 
-36 & 8 & 20 & -94
\end{array}
\right] $, column basis: $\left[ 
\begin{array}{c}
0 \\ 
1 \\ 
0 \\ 
0
\end{array}
\right] ,\left[ 
\begin{array}{c}
1 \\ 
0 \\ 
0 \\ 
1
\end{array}
\right] ,\left[ 
\begin{array}{c}
0 \\ 
0 \\ 
1 \\ 
1
\end{array}
\right] $
\end{quotation}

You can also take the transpose of $A$ and apply \textsf{Fraction-Free
Gaussian Elimination} to the result, because the column space of $A$ is the
row space of $A^{T}$.\medskip

\subsection{The Left and Right Nullspaces\label{Nullspaces}}

The (right) \textsl{nullspace} is the vector space consisting of all $%
n\times 1$ vectors $X$ satisfying $AX=0$. You find a basis for the nullspace
by choosing 
\index{Matrices@Matrices!nullspace basis@nullspace basis}\label%
{Matrices, nullspace basis}\textsf{Nullspace Basis} from the \textsf{Matrices%
} submenu.

\ 

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Nullspace Basis}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{rrrr}
-85 & -55 & -37 & -35 \\ 
97 & 50 & 79 & 56 \\ 
49 & 63 & 57 & -59 \\ 
-36 & 8 & 20 & -94
\end{array}
\right] $, nullspace basis: $\left[ 
\begin{array}{c}
\stackunder{}{-%
\dfrac{133337}{68264}} \\ 
\stackunder{}{\dfrac{74049}{34132}} \\ 
\dfrac{3085}{9752} \\ 
1
\end{array}
\right] $
\end{quotation}

The \textsl{left nullspace} is the vector space consisting of all $1\times m$
vectors $Y$ satisfying $YA=0$. You find a basis for the left nullspace by
first taking the transpose of $A$ and then choosing \textsf{Nullspace Basis}
from the \textsf{Matrices} submenu.

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{rrrr}
-85 & -55 & -37 & -35 \\ 
97 & 50 & 79 & 56 \\ 
49 & 63 & 57 & -59 \\ 
-36 & 8 & 20 & -94
\end{array}
\right] ^{T}=\left[ 
\begin{array}{cccc}
-85 & 97 & 49 & -36 \\ 
-55 & 50 & 63 & 8 \\ 
-37 & 79 & 57 & 20 \\ 
-35 & 56 & -59 & -94
\end{array}
\right] $
\end{quotation}

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Nullspace Basis}
\end{quote}

\begin{quotation}
$\,\left[ 
\begin{array}{rrrr}
-85 & 97 & 49 & -36 \\ 
-55 & 50 & 63 & 8 \\ 
-37 & 79 & 57 & 20 \\ 
-35 & 56 & -59 & -94
\end{array}
\right] $, nullspace basis: $\left\{ \left[ 
\begin{array}{r}
1 \\ 
0 \\ 
1 \\ 
-1
\end{array}
\right] \right\} $
\end{quotation}

To check that this vector is in the left nullspace, take the transpose of
the vector and check the product.

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$\,\left[ 
\begin{array}{r}
1 \\ 
0 \\ 
1 \\ 
-1
\end{array}
\right] ^{T}\left[ 
\begin{array}{rrrr}
-85 & -55 & -37 & -35 \\ 
97 & 50 & 79 & 56 \\ 
49 & 63 & 57 & -59 \\ 
-36 & 8 & 20 & -94
\end{array}
\right] =\,\left[ 
\begin{array}{cccc}
0 & 0 & 0 & 0
\end{array}
\right] $
\end{quotation}

\strut

\subsection{\label{QR factorization}The QR Factorization and Orthonormal
Bases}

Any real matrix $A$ with linearly independent columns can be factored as a
product $QR$, 
\index{QR factorization@QR factorization}where 
\index{Matrices@Matrices!QR@QR}\label{Matrices, QR factorization}the columns
of $Q$ are \textsl{orthonormal} 
\index{Orthonormal@Orthonormal}\label{Orthonormal}(the inner product of any
two different columns is $0$, and the inner product of any column with
itself is $1$) and $R$ is invertible and upper-right triangular. If the
original matrix $A$ is square, then so is $Q$. In this case, $Q$ is an 
\textsl{orthogonal} matrix. 
\index{Matrices@Matrices!orthogonal@orthogonal}\label{Matrices, orthogonal}

\begin{quote}
$\blacktriangleright $ To obtain the $QR$ factorization
\end{quote}

\begin{enumerate}
\item[1.]  Leave the insertion point in a matrix.

\item[2.]  From the \textsf{Matrices} submenu, choose \textsf{QR}.
\end{enumerate}

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + QR}
\end{quote}

\begin{quotation}
$\left( 
\begin{array}{cc}
3 & 0 \\ 
4 & 5
\end{array}
\right) =\left( 
\begin{array}{cc}
-.6 & .8 \\ 
-.8 & -.6
\end{array}
\right) \left( 
\begin{array}{cc}
-5.0 & -4.0 \\ 
0 & -3.0
\end{array}
\right) \,$
\end{quotation}

The two matrices $Q$ and $A=QR$ have the same column spaces. Observe, in the
following example, that the columns of $A$ are linear combinations of the
columns of $Q$. Then, since both column spaces have dimension $2$ and one
contains the other, it follows that they must be the same space.

\begin{example}
The\label{Gram-Schmidt orthogonalization} preceding product comes from the
following linear combinations. 
\[
\left( 
\begin{array}{c}
3 \\ 
4
\end{array}
\right) =-5\left( 
\begin{array}{c}
-.6 \\ 
-.8
\end{array}
\right) +0\left( 
\begin{array}{c}
.8 \\ 
-.6
\end{array}
\right) 
\]
and%
\vspace{-0.1in} 
\[
\left( 
\begin{array}{c}
0 \\ 
5
\end{array}
\right) =-4\left( 
\begin{array}{c}
-.6 \\ 
-.8
\end{array}
\right) -3\left( 
\begin{array}{c}
.8 \\ 
-.6
\end{array}
\right) 
\]
\end{example}

This conversion of the columns of $A$ into the orthonormal columns of $Q$ is
referred to as the 
\index{Gram-Schmidt orthogonalization@Gram-Schmidt orthogonalization}\textsl{%
Gram--Schmidt orthogonalization} process. In general, since $R$ is
upper-right triangular, the subspace spanned by the first $k$ columns of the
matrix $A=QR$ is the same as the subspace spanned by the first $k$ columns
of the matrix $Q$.\medskip

\subsection{\label{Matrices, rank}Rank and Dimension}

The \textsl{rank} of a matrix 
\index{Matrices@Matrices!rank@rank} is the dimension of the column space. It
is the same as the dimension of the row space or the number of nonzero
singular values.

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Rank}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{rrrr}
-8 & -5 & 7 & -2 \\ 
7 & 5 & 9 & 5 \\ 
1 & 0 & -16 & -3 \\ 
8 & 5 & -7 & 2
\end{array}
\right] $, rank: $2$
\end{quotation}

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Row Basis}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{rrrr}
-8 & -5 & 7 & -2 \\ 
7 & 5 & 9 & 5 \\ 
1 & 0 & -16 & -3 \\ 
8 & 5 & -7 & 2
\end{array}
\right] $, row basis: $\left\{ \left[ 
\begin{array}{cccc}
1 & 0 & -16 & -3
\end{array}
\right] ,\left[ 
\begin{array}{cccc}
0 & 1 & 
\frac{121}{5} & \frac{26}{5}
\end{array}
\right] \right\} $
\end{quotation}

\ 

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Column Basis}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{rrrr}
-8 & -5 & 7 & -2 \\ 
7 & 5 & 9 & 5 \\ 
1 & 0 & -16 & -3 \\ 
8 & 5 & -7 & 2
\end{array}
\right] $, column basis: $\left[ 
\begin{array}{r}
0 \\ 
1 \\ 
-1 \\ 
0
\end{array}
\right] ,\left[ 
\begin{array}{r}
1 \\ 
0 \\ 
-1 \\ 
-1
\end{array}
\right] $
\end{quotation}

\strut

\subsection{Related topics}

\begin{itemize}
\item  \hyperref{Working With Matrices}{}{}{DM8-1.tex#Matrix Algebra}

\item  \hyperref{Standard Operations}{}{}{DM8-2.tex#Standard matrix
operations}

\item  \hyperref{Row Operations and Echelon Forms}{}{}{DM8-3.tex#Matrices,
row operations}

\item  \hyperref{Systems of Linear Equations}{}{}{DM8-4.tex#Systems of
equations}

\item  \hyperref{Matrix Equations}{}{}{DM8-4.tex#Matrix equations}

\item  \hyperref{Matrix Operators}{}{}{DM8-5.tex#Trace}

\item  \hyperref{Polynomials and Vectors Associated with a Matrix}{}{}{%
DM8-6.tex#Characteristic, polynomial}

\item  \hyperref{Special Forms of Matrices}{}{}{DM8-8.tex#Smith normal form}

\item  \hyperref{Exercises and Solutions}{}{}{DM8-9.tex#Exercises}
\end{itemize}

\rule{0.67in}{0.01in}

\hyperref{\FRAME{itbpF}{0.5275in}{0.3399in}{0.0519in}{}{}{index.wmf}{\special%
{language "Scientific Word";type "GRAPHIC";display "PICT";valid-file
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"1";cropbottom "0";filename 'index.wmf';file-properties "XNPEU";}}}{}{}{%
DMIndex.tex#Math Index}\quad {\small Index entries: \hyperref{Matrices}{}{}{%
DMIndex.tex#Matrices}, \hyperref{Vector spaces}{}{}{DMIndex.tex#Vector spaces%
}}

\end{document}
