\documentclass[time,mathptm]{article}

\newcommand{\docid}{Exam 1} \input localmacs


\renewcommand{\today}{October 13, 2000}

\begin{document}
\doctitle


\begin{problem}
  Provide the precise definition of each of the following
concept:
\begin{subproblem}
  The span of a set of vectors $\{{\bf v}_1, \ldots, {\bf v}_n\}$.
\end{subproblem}

\vspace{1.8in}
%All linear combinations of the vectors $\{{\bf v}_1, \ldots, {\bf v}_n
%\}$.  All vectors of the form $c_1{\bf v}_1+ \ldots + c_n{\bf v}_n$.

\begin{subproblem}
  A set of vectors $\{{\bf v}_1, ..., {\bf v}_n\}$ being linearly
  independent.
\end{subproblem}

\vspace{1.8in}
%The only solution of to $c_1{\bf v}_1+ \ldots + c_n{\bf v}_n = {\bf
%  0}$ is $c_i = 0$ for all $i$.

\begin{subproblem}
  The elementary row operations.
\end{subproblem}

\vspace{1.8in}
%Scaling (multiplying a row by a scalar), add a multiple of a row to
%another row, and switch two rows.

\begin{subproblem}
  The determinant of an $ n \times n$ matrix $A$.
\end{subproblem}

%The cofactor expansion along the first row.  $\det(A)= \sum_{j=1}^n
%a_{1,j} (-1)^{j+1}\det(A_{1,j})$.

\end{problem}

\newpage

\begin{problem}
  TRUE/FALSE.  Determine whether the followings statements are true or
  false. Be sure to provide a reason for your answer.
  \begin{subproblem}
    Let $T : {\bf R}^m \to {\bf R}^m$ be a linear transformation. Let
    $S = \{{\bf v}_1, \ldots, {\bf v}_n\}$ be a set of linearly
    independent vectors in ${\bf R}^m$. Then the set of vectors
    $\{T({\bf v}_1), \ldots, T({\bf v}_n)\}$ is also linearly
    independent.
  \end{subproblem}
  \vspace{1.8in}
 % FALSE. Consider $T : {\bf R}^m \to {\bf R}^n$ by $T({\bf x}) = {\bf
%    0}$.


  \begin{subproblem}
    Let $A$ and $B$ be two invertible matrices. Then $ABA^{-1}$ is
    invertible.
  \end{subproblem}
  \vspace{1.8in}
  %TRUE.  $(AB^{-1}A^{-1})(ABA^{-1}) = I$.

  \begin{subproblem}
    If $A$ and $B$ are square and $AB = I$ then $\det(A) = 0$.
  \end{subproblem}
  \vspace{1.8in}
 % FALSE.  $A$ is invertible and hence $\det(A) \neq 0$.

  \begin{subproblem}
    Let $T({\mathbf x})$ be a linear transformation.  $T$ is one to
    one if and only if the kernel of $T$ is $\{0\}$.
  \end{subproblem}
  \vspace{1.8in}
  %TRUE.  This was a lemma from class.  $T({\bf 0}) = {\bf 0}$.  So if
%  $T$ is one to one then the kernel of $T$ is $\{0\}$. Now assume the
%  kernel of $T$ is $\{0\}$ and $T$ is not one to one.  So $T({\bf x})
%  = T({\bf y})$ where ${\bf x } \neq{\bf y}$.  ${\bf 0} = T({\bf x})
%  -T({\bf y})= T({\bf x} - {\bf y})$. Hence ${\bf x} - {\bf y} \neq
%  {\bf 0}$ is in the kernel of $T$.  Contradiction.
\end{problem}
\newpage
\begin{problem}
  Let $A$ be the following matrix:
  $$A = \begin{bmatrix} a&b&c\\d&e&f\\g&h&i\end{bmatrix}$$
  if $\det A =
  1$ then (hint: use the properties of determinant)
\begin{subproblem}
  $\det \begin{bmatrix}
a&b&c\\-2d&-2 e&-2f\\g&h&i
\end{bmatrix} = \;?$
\end{subproblem}

%The above matrix is the result of scaling the second row of $A$ by
%$-2$.  This action multiples determinant of $A$ by $-2$.  So the
%determinant is $-2$.


\begin{subproblem}
  $\det \begin{bmatrix} a&b&c\\a+d&b+e&c+f\\g&h&i\end{bmatrix} = \;?$
\end{subproblem}
\end{problem}

%The above matrix is the result of adding the first row to the second
%row of $A$.  This action does not affect the determinant.  The
%determinant is $1$.
\newpage

\begin{problem}
  \begin{subproblem}
    Describe the null space of the matrix:
    $A = \begin{bmatrix} 1&-1&1&-1\\-2&2&-1&3\\3&-2&3&1\\4&-4&4&-4
\end{bmatrix}.$
  \end{subproblem}
  \begin{subproblem}
    Find all solutions of the equation $A {\bf x} = {\bf b}$, where
    ${\bf b} =
    \begin{bmatrix}-2\\8\\4\\-8\end{bmatrix}$.
  \end{subproblem}
  
 % To save time we will row-reduce the matrix $[A | {\bf b}]$.  It
%  reduces to $\begin{bmatrix} 1 & 0 & 0 & 2 & 4 \\ 0 & 1 & 0 & 4 &
%    10 \\ 0 & 0 & 1 & 1 & 4 \\ 0 & 0 & 0&0 & 0 
%  \end{bmatrix}$.  Hence a spanning set for the null space is 
%  $\begin{bmatrix} -2 \\ -4 \\ -1 \\ 1
%  \end{bmatrix}$.  A solution for $A {\bf x} = {\bf b}$ is ${\bf x}_p
%  \begin{bmatrix}
%    4 \\ 10 \\ 4 \\ 0
%  \end{bmatrix}$ and all the solutions are in the form 
%  ${\bf x}_p + {\bf x}$, where ${\bf x}$ is in the null space of $A$.
\end{problem}

\newpage

\begin{problem}
  \begin{subproblem}
    Write 
    $\begin{bmatrix}
      1 \\ 4
    \end{bmatrix}$ as a linear combination of $\begin{bmatrix}
      1 \\ 1
    \end{bmatrix}$ and $\begin{bmatrix}
      -1 \\ 2
    \end{bmatrix}$.
  \end{subproblem}

%We must solve the equation  $x_1 \begin{bmatrix}
%      1 \\ 1
%    \end{bmatrix} + x_2 \begin{bmatrix}
%      -1 \\ 2
%    \end{bmatrix} = \begin{bmatrix}
%      1 \\ 4
%    \end{bmatrix}$. The augmented matrix of this system is  $\begin{bmatrix}
%      1 & -1 & 1 \\ 1 & 2 & 4
%    \end{bmatrix}$ which row-reduces to  $\begin{bmatrix}
%      1 & 0 & 2 \\ 0 & 1 & 1
%    \end{bmatrix}$.  Hence $2 \begin{bmatrix}
%      1 \\ 1
%    \end{bmatrix} +  \begin{bmatrix}
%      -1 \\ 2
%    \end{bmatrix} = \begin{bmatrix}
%      1 \\ 4
%    \end{bmatrix}$.

  \begin{subproblem}
    Let $T : {\bf R}^2 \to {\bf R}^2$ be a linear transformation. If
    $T(1, 1) = (2, 3), T(-1, 2) = (4, 5)$ then $T(1, 4) = ?$.
  \end{subproblem}
  
%  $T(1, 4) = T(2(1,1)+(-1,2)) = 2T(1,1) + T(-1,2) = 2(2,3)+(4,5)=
%  (4,6)+(4,5)=(8,11)$, since $T$ is a linear transformation.
\end{problem}

\newpage

\begin{problem}
  Let ${\bf v}_1, {\bf v}_2, {\bf v}_3$ be linearly independent
  vectors. Show that ${\bf v}_1, {\bf v}_1 - {\bf v}_2, {\bf v}_1 -
  {\bf v}_2 + {\bf v}_3$ are linearly independent vectors.
\end{problem}

%Assume that $x_1{\bf v}_1 + x_2({\bf v}_1 - {\bf v}_2) + x_3( {\bf
%  v}_2 + {\bf v}_3) = {\bf 0}$.  Then $(x_1 +x_2){\bf v}_1 + (-x_2
%+x_3){\bf v}_3 + x_3{\bf v}_3 = {\bf 0}$.  Since ${\bf v}_1, {\bf
%  v}_2, {\bf v}_3$ are linearly independent vectors, $x_1 + x_2 =0$,
%$-x_2 + x_3 = 0$ and $x_3= 0$.  Hence the $x_0 = x_1 = x_2 = 0$ and
%therefore ${\bf v}_1, {\bf v}_1 - {\bf v}_2, {\bf v}_1 - {\bf v}_2 +
%{\bf v}_3$ are linearly independent vectors

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