\documentclass{article}

\newcommand{\docid}{Quiz 2}
\renewcommand{\today}{September 15, 2000}

\input localmacs

%\documentclass{amsart}
\usepackage{amssymb,latexsym}                                                                
\date{}
\begin{document}
%\begin{center}
%{\bf MATH 221 Quizz 2}
%\end{center}
\doctitle
\bigskip
\noindent

\begin{problem}
  Express the solutions of the equation $A {\bf x} = {\bf 0}$ in
  parametric form, where
$$A = \begin{bmatrix}
1&1&0&2\\
-2&-2&1&-5\\
1&1&-1&3\\
4&4&-1&9
\end{bmatrix}$$
\end{problem}

%The reduce row echolon form of $A$ is $ \begin{bmatrix}
%  1&1&0&2 \\
%  0&0&1&-1 \\
%  0&0&0&0 \\
%  0&0&0&0
%\end{bmatrix}$.  So $x_1$ and $x_3$ are basic variables and $x_2$ 
%and $x_4$ are free variables.  Hence the general solution is
%$x_2\begin{bmatrix} -1 \\ 1 \\ 0 \\ 0
%\end{bmatrix} + x_4\begin{bmatrix}
%  -2 \\ 0 \\ 1  \\ 1
%\end{bmatrix}$.


\begin{problem}
  For what values of $c$ are the vectors $ {\bf v}_1= \begin{bmatrix}
    -1 \\ 0 \\ -1
\end{bmatrix}, {\bf v}_2=\begin{bmatrix}
  2\\ 1\\ 2
\end{bmatrix}$ and ${\bf v}_3=\begin{bmatrix}
  1 \\ 1 \\ c
\end{bmatrix}$
(a) linearly independent? (b) linearly dependent?
\end{problem}

%We want to consider when the vector equation $x_1{\bf v}_1 + x_2{\bf
%  v}_2 = {\bf v}_3$ has solutions.  $x_1{\bf v}_1 + x_2{\bf v}_2 =
%{\bf v}_3$ has solutions iff ${\bf v}_1$, ${\bf v}_2$, ${\bf v}_3$ are
%linearly dependent.  For this we need to rowreduce the matrix
%$\begin{bmatrix}
%  -1 & 2 & 1\\
%  0 & 1 & 1 \\
%  -1 & 2 & c
%\end{bmatrix}$.  Subtract the first row from the third to get 
%$\begin{bmatrix}
%  -1 & 2 & 1\\
%  0 & 1 & 1 \\
%  0  & 0 & c-1
%\end{bmatrix}$. $c = 1$ iff the last column does not has 
%pivot iff vector equation $x_1{\bf v}_1 + x_2{\bf v}_2 = {\bf v}_3$
%has solutions iff ${\bf v}_1$, ${\bf v}_2$, ${\bf v}_3$ are linearly
%dependent.

\begin{problem}
  Suppose that ${\bf v}_1, {\bf v}_2, {\bf v}_3$ are 3 linearly
  independent vectors show that the vectors ${\bf v}_1, {\bf v}_1 +
  {\bf v}_2, {\bf v}_1 + {\bf v}_2 + {\bf v}_3$ are also linearly
  independent.
\end{problem}

%$c_1{\bf v}_1 + c_2 ({\bf v}_1 + {\bf v}_2), ({\bf v}_1 + c_3 {\bf
%  v}_2 + {\bf v}_3) = {\bf 0}$ iff $(c_1 + c_2 + c_3){\bf v}_1 + (c_2
%+ c_3){\bf v}_2 + c_3 {\bf v}_3) = {\bf 0}$.  Hence ${\bf v}_1, {\bf
%  v}_2, {\bf v}_3$ is linearly independent iff ${\bf v}_1, {\bf v}_1 +
%{\bf v}_2, {\bf v}_1 + {\bf v}_2 + {\bf v}_3$ are linearly
%independent.

\begin{problem}
  Let $S$ be a set of vectors and $T$ be a subset of $S$. Are the
  following statements TRUE or FALSE.  Be sure to a proof or a
  counterexample.
  \begin{subproblem}
    If the vectors in $S$ are linearly independent then the vectors in
    $T$ are also linearly independent.
  \end{subproblem}
  \begin{subproblem}
    If the vectors in $S$ are linearly dependent then the vectors in
    $T$ are also linearly dependent.
  \end{subproblem}
\end{problem}

%(a) True. If there is a non-trivial linear combination of the vectors
%in $T$ whose sum is the zero vector then the same is true for $S$. So
%if $T$ is linearly dependent then so is $S$.

%(b) False.  $T = \{ \begin{bmatrix}
%  1 \\ 0 \\ 0
%\end{bmatrix}, \begin{bmatrix}
%  0 \\ 1 \\ 0
%\end{bmatrix}, \begin{bmatrix}
%  0 \\ 0 \\ 1
%\end{bmatrix}\}$ and $S = T \cup \{ \begin{bmatrix}
%  1 \\ 1 \\ 1
%\end{bmatrix}\}$. $T$ is linearly independent but $S$ is not.

\begin{problem}
  Is the absolute value function $T : {\bf R} \to {\bf R}, T(x) = |x|$
  a linear transformation? Justify your answer.
\end{problem}

%No.  For example, $3= |-3| \neq -1 | 3 | = - 3$.

\end{document}
