spanningSetsAndMATLAB.tex

\documentclass{ximera}

\input{../preamble.tex}

\title{Spanning Sets and MATLAB}

\begin{document}
\begin{abstract}
\end{abstract}
\maketitle

 \label{S:5.3}

In this section we discuss:
\begin{itemize}
\item	how to find a spanning set for the subspace of solutions to a
homogeneous system of linear equations using the \Matlab command {\tt null},
and
\item	how to determine when a vector is in the subspace spanned by a
set of vectors using the \Matlab command {\tt rref}.
\end{itemize}

\subsection*{Spanning Sets for Homogeneous Linear Equations}

In Chapter~\ref{lineq} we saw how to use Gaussian elimination,
back substitution, and \Matlab to compute solutions to a system
of linear equations.  For systems of
homogeneous equations\index{homogeneous}, \Matlab
provides a command to find a spanning set for the subspace of solutions.
That command is {\tt null}.  For example, if we type
\begin{verbatim}
A = [2 1 4 0; -1 0 2 1]
B = null(A)
\end{verbatim} \index{\computer!null}
then we obtain
\begin{verbatim}
B =
   -0.4100    0.2555
    0.8130    0.5219
    0.0017   -0.2582
   -0.4134    0.7718
\end{verbatim}
The two columns of the matrix $B$ span the set of solutions of
the equation $Ax=0$.  In particular, the vector $(2,-8,1,0)$ is a
solution to $Ax=0$ and is therefore a
linear combination \index{linear!combination} of the
column vectors of {\tt B}.  Indeed, type
\begin{verbatim}
-7.3224*B(:,1)-3.9221*B(:,2)
\end{verbatim}
and observe that this linear combination is the desired one.

Next we describe how to find the coefficients {\tt -7.3224} and
{\tt -3.9221} by showing that these coefficients themselves are
solutions to another system of linear equations.

\subsection*{When is a Vector in a Span?} \index{span}

Let $w_1,\ldots,w_k$ and $v$ be vectors in $\R^n$.  We now
describe a method that allows us to decide whether $v$ is in
$\Span\{w_1,\ldots,w_k\}$.  To answer this question one has
to solve a system of $n$ linear equations in $k$ unknowns.
The unknowns correspond to the coefficients in the linear
combination of the vectors $w_1,\ldots,w_k$ that gives $v$.

Let us be more precise.  The vector\index{vector} $v$ is in
$\Span\{w_1,\ldots,w_k\}$ if and only if there are constants
$r_1,\ldots,r_k$ such that the equation
\begin{equation}  \label{e:lindepeqn}
     r_1 w_1 + \cdots + r_k w_k = v
\end{equation}
is valid.  Define the $n\times k$ matrix $A$ as the one having
$w_1,\ldots,w_k$ as its columns; that is,
\begin{equation}  \label{E:Abycol}
A = (w_1| \cdots |w_k).
\end{equation}
Let  $r$ be the $k$-vector
\[
r= \left(\begin{array}{c} r_1 \\ \vdots \\ r_k\end{array}\right).
\]
Then we may rewrite equation \eqref{e:lindepeqn} as
\begin{equation}  \label{E:Ar=v}
   Ar=v.
\end{equation}
To summarize:
\begin{lemma}
Let $w_1,\ldots,w_k$ and $v$ be vectors in $\R^n$.  Then $v$
is in $\Span\{w_1,\ldots,w_k\}$ if and only if the system of linear
equations \eqref{E:Ar=v} has a solution where $A$ is the $n\times k$
defined in \eqref{E:Abycol}.
\end{lemma}

To solve \eqref{E:Ar=v} we row reduce the
augmented matrix\index{matrix!augmented} $(A|v)$.
For example, is $v=(2,1)$ in the span of $w_1=(1,1)$ and $w_2=(1,-1)$?
That is, do there exist scalars $r_1,r_2$ such that
\[
r_1\vectwo{1}{1} + r_2\vectwo{1}{-1} = \vectwo{2}{1}?
\]
As noted, we can rewrite this equation as
\[
\mattwo{1}{1}{1}{-1}\vectwo{r_1}{r_2} = \vectwo{2}{1}.
\]
We can solve this equation by row reducing the augmented
matrix
\[
\left(\begin{array}{rr|r}
1 & 1 & 2 \\ 1 & -1 & 1 \end{array}\right)
\]
to obtain
\[
\left(\begin{array}{rr|r}
1 & 0 & \frac{3}{2} \\ 0 & 1 & \frac{1}{2}
\end{array}\right).
\]
So $v = \frac{3}{2}w_1 + \frac{1}{2}w_2$.

Row reduction to reduced echelon form\index{echelon form!reduced}
has been preprogrammed in the
\Matlab command {\tt rref}. \index{\computer!rref}  Consider the
following example.  Let
\begin{equation}  \label{e:w1w2}
     w_1=(2,0,-1,4) \AND w_2=(2,-1,0,2)
\end{equation}
and ask the question whether $v=(-2,4,-3,4)$ is in $\Span\{w_1,w_2\}$.

In \Matlab load the matrix $A$ having $w_1$ and
$w_2$ as its columns and the vector $v$ by typing {\tt e5\_3\_5}
\begin{matlabEquation}  \label{e:Aandv}
A=\left(\begin{array}{rr} 2 & 2 \\ 0 & -1 \\ -1 & 0 \\ 4 & 2
\end{array}\right) \AND
v=\left(\begin{array}{r} -2 \\ 4 \\ -3 \\ 4 \end{array}\right).
\end{matlabEquation}%
We can solve the system of equations using \Matlabp.
First, form the augmented matrix by typing
\begin{verbatim}
aug = [A v]
\end{verbatim}
Then solve the system by typing {\tt rref(aug)} to obtain
\begin{verbatim}
ans =
     1   0   3
     0   1  -4
     0   0   0
     0   0   0
\end{verbatim}
It follows that $(r_1,r_2)=(3,-4)$ is a solution and $v=3w_1-4w_2$.

Now we change the $4^{th}$ entry in $v$ slightly by typing
{\tt v(4) = 4.01}.  There is no solution to the system of equations
\[
Ar = \left(\begin{array}{r} -2 \\ 4 \\ -3 \\ 4.01
\end{array}\right)
\]
as we now show.  Type
\begin{verbatim}
aug = [A v]
rref(aug)
\end{verbatim}
which yields
\begin{verbatim}
ans =
     1    0    0
     0    1    0
     0    0    1
     0    0    0
\end{verbatim}
This matrix corresponds to an inconsistent\index{inconsistent} system;
thus $v$ is no longer in the span\index{span} of $w_1$ and $w_2$.



\includeexercises

\end{document}