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\begin{document}
\title{314, March 20 '07-QUIZ 3}
\author{Name:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ }
\maketitle
\section{$< 13pts >$}
Given the vectors
\[ 
{\bf x}_1 =
\left[
\begin{array}{r} 1 \\ -2 \\ 4 \end{array}
\right] \ ,\ 
{\bf x}_2 =
\left[
\begin{array}{r} -1 \\  2 \\ 4 \end{array}
\right] \ ,\ 
{\bf x}_3 =
\left[
\begin{array}{r} -1 \\  2 \\ 12 \end{array}
\right] 
\]
Give the dimension of Span$({\bf x}_1,{\bf x}_2,{\bf x}_3 )$ and find
a basis. \\[.1in]
{\bf Solution}\\
Form the matrix
\begin{eqnarray*} 
\begin{array}{r} \ \\2 \\ -4 \end{array}
\left(
\begin{array}{rrr}
1 & -1 & -1\\-2 & 2 & 2 \\4 & 4 & 12 \end{array} \right) 
& & \rightarrow
\begin{array}{r} \ \\2 \\ -4 \end{array}
\left(
\begin{array}{rrr}
1 & -1 & -1\\ 0 & 0 & 0 \\0 & 8 & 16 \end{array} \right)
\end{eqnarray*} 
There are pivots on columns $1$ and $2$, therefore the vectors
${\bf x}_1$ and ${\bf x}_2$ are linearly independent, providing
a basis for Span$({\bf x}_1,{\bf x}_2,{\bf x}_3 )$, which is
thus $2$-dimensional.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{$< 12pts >$}
Given 
\[
{\bf v}_1 =
\left[ \begin{array}{r}
       2 \\ 3 
       \end{array} \right] \ ,\ 
{\bf v}_2 =
\left[ \begin{array}{r}
      -1 \\ 2 
       \end{array} \right] \ ,\ 
S = 
\left[ \begin{array}{rr}
       3 & -1 \\ 1 & 3
       \end{array} \right] \ .
\]
Find vectors ${\bf u}_1\ ,\ {\bf u}_2$ so that S is the
transition matrix from $\left[ {\bf u}_1\ ,\ {\bf u}_2 \right]$
to $\left[ {\bf v}_1\ ,\ {\bf v}_2 \right]$. \\

\noindent {\bf Solution}\\
\[ S = V^{-1}U \Rightarrow U = V S =
\left( \begin{array}{rr}
 2 & -1 \\
 3 &  2 \end{array}\right)
\left( \begin{array}{rr}
 3 & -1 \\
 1 &  3 \end{array}\right)
=
\left( \begin{array}{rr}
 5 & -5 \\
11 &  3 \end{array}\right)
\]
so
\[
{\bf u}_1 =
\left[ \begin{array}{r}
       5 \\ 11 
       \end{array} \right] \ ,\ 
{\bf u}_2 =
\left[ \begin{array}{r}
      -5 \\ 3 
       \end{array} \right]
\]
\newpage

\section{$< 13pts >$}
Find a basis of the row space, column space and null space of
\[ A = 
\left[
\begin{array}{rrrrr}
 1 & -1 & \ 0 &  1 &  2 \\
 2 & -1 & \ 1 &  3 &  1 \\
-1 &  2 & \ 1 & -1 & -3
\end{array}
\right]
\]
and give the dimension of each.\\

\noindent {\bf Solution} (use reduced row echelon form to
find null vectors most easily)\\
\begin{eqnarray*}
\begin{array}{r}
\ \\ -2 \\ 1 \end{array}
\left[
\begin{array}{rrrrr}
 1 & -1 & \ 0 &  1 &  2 \\
 2 & -1 & \ 1 &  3 &  1 \\
-1 &  2 & \ 1 & -1 & -3
\end{array}
\right]
& &\rightarrow
\begin{array}{r}
\ \\    \\-1 \end{array}
\left[
\begin{array}{rrrrr}
 1 & -1 & \ 0 &  1 &  2 \\
 0 &  1 & \ 1 &  1 & -3 \\
 0 &  1 & \ 1 &  0 & -1
\end{array}
\right]
\rightarrow\\
\begin{array}{r}
\ \\ \ \\ -1*\end{array}
\left[
\begin{array}{rrrrr}
 1 & -1 & \ 0 &  1 &  2 \\
 0 &  1 & \ 1 &  1 & -3 \\
 0 &  0 & \ 0 & -1 &  2
\end{array}
\right] & &
\rightarrow 
\begin{array}{r}
-1 \\ -1 \\ \ \end{array}
\left[
\begin{array}{rrrrr}
 1 & -1 & \ 0 &  1 &  2 \\
 0 &  1 & \ 1 &  1 & -3 \\
 0 &  0 & \ 0 &  1 & -2
\end{array}
\right] \rightarrow\\
\begin{array}{r}
 1 \\ \ \\ \ \end{array}
\left[
\begin{array}{rrrrr}
 1 & -1 & \ 0 &  0 &  4 \\
 0 &  1 & \ 1 &  0 & -1 \\
 0 &  0 & \ 0 &  1 & -2
\end{array}
\right] & &
\rightarrow \left[
\begin{array}{rrrrr}
 1 &  0 & \ 1 &  0 &  3 \\
 0 &  1 & \ 1 &  0 & -1 \\
 0 &  0 & \ 0 &  1 & -2
\end{array}
\right]
\end{eqnarray*}
so the pivots are found on columns 1, 2, 4 and 
so $x_3\ ,\ x_5$ are free variables. Therefore
the dimensions of the column and row spaces are
equal to 3, with the rows of $A$ giving the basis of the
row space, ${\cal R}(A^T)$, and columns 1,2, and 4 of $A$ 
giving the basis of the column space,  ${\cal R}(A)$. 
The dimension of the null space is 2. 
To find the basis of the null space:
\[ \left[
\begin{array}{rrrrr}
 1 &  0 &  1 &  0 &  3 \\
 0 &  1 &  1 &  0 & -1 \\
 0 &  0 &  0 &  1 & -2
\end{array}
\right]\left[ \begin{array}{r}
x_1\\x_2\\x_3\\x_4\\x_5 \end{array}\right]
= 0 \rightarrow\left[
\begin{array}{rrr}
1 &  0 & 0\\
0 &  1 &   0 \\
 0 &  0 &  1 \end{array}
\right]\left[ \begin{array}{r} x_1\\x_2\\x_4
\end{array}\right]
= -x_3 \left[ \begin{array}{r}
1 \\ 1 \\ 0 \end{array}\right]
-x_5 \left[ \begin{array}{r}
 3 \\ -1 \\-2 \end{array}
\right]
\]
so that the null vectors (basis of 
${\cal N}(A)$) are:
\[ (x_3 =1\ ,\  x_5 =0:)\ {\bf n}_1
=
\left[ \begin{array}{r}
-1 \\ -1 \\ 1 \\  0 \\ 0 \end{array}\right]
\ , \ 
(x_3 =0\ ,\  x_5 =1:)\ {\bf n}_2
=
\left[ \begin{array}{r}
-3\\ 1\\0\\2\\1 \end{array}\right]
\]
\end{document}
