\documentclass[time,mathptm]{article}
\newcommand{\docdate}{October 19, 2001}
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\newcommand{\doctype}{Exam }
\newcommand{\professor}{Professors Cholak and Polini}
\newcommand{\coursenum}{Math 221}
\newcommand{\courseblurb}{Linear Algebra}

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      -- \courseblurb \\ \doc}
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\begin{document}
\doctitle



\begin{problem}  (20 points -- 4 each)
  TRUE/FALSE.  Determine whether the followings statements are true or
  false.  (Comment: no reason needed.)

   \begin{subproblem}
     Let $T : \mathbb{R}^n \to \mathbb{R}^m$ be a linear
     transformation.  Then there is a $n \times m$ matrix $A$ such
     $T(\vec{x})= A \vec{x}$.
  \end{subproblem}
  \vspace{1.25in}
%  FALSE.  $A$ must be $m \times n$.

  \begin{subproblem}
    Let $T : \mathbb{R}^m \to \mathbb{R}^n$ be a linear transformation. Let
    $S = \{\vec{v}_1, \ldots, \vec{v}_n\}$ be a set of linearly
    independent vectors in $\mathbb{R}^m$. Then the set of vectors
    $\{T(\vec{v}_1), \ldots, T(\vec{v}_n)\}$ is also linearly
    independent.
  \end{subproblem}
\vspace{1.25in}
 % FALSE. Consider $T : \mathbb{R}^m \to \mathbb{R}^n$ by $T(\vec{x}) = {\vec
%    0}$.


  \begin{subproblem}
    Let $A$ and $B$ be two invertible matrices. Then $ABA^{-1}$ is
    invertible.
  \end{subproblem}
\vspace{1.25in}
%  TRUE.  $(AB^{-1}A^{-1})(ABA^{-1}) = I$.

  \begin{subproblem}
    If $BA = I$ then both $A$ and $B$ are invertible.
  \end{subproblem}
\vspace{1.25in}
 % FALSE. $A = \begin{bmatrix} 2 & -1 \\ -1 & 1 \\ 0 & 0
%\end{bmatrix}$ and   $ B =
%\begin{bmatrix}
%  1 & 1 & 1 \\ 1 & 2 & 3 \end{bmatrix}$ then $BA =I_2$.

  \begin{subproblem}
    Let $T({\vec x})$ be a linear transformation.  $T$ is one to
    one if and only if the kernel of $T$ is $\{\vec{0}\}$.
  \end{subproblem}

\vspace{1.3in}
 % TRUE.  Assume that $T(\vec{x}) = A\vec{x}$.  Then kernel of $A$ is
%  $\{\vec{0}\}$ iff there are no free variables iff $A \vec{x} =
%  \vec{b}$ has at most one solution for all $\vec{b}$ iff $T$ is one
%  to one.


\begin{subproblem}
If $A$ is a $3 \times 4$ matrix of rank $2$ then the
dimension of the kernel of $A$ is $1$.
\end{subproblem}
\vspace{1.3in}

%FALSE. By the rank theorem the dimension of the kernel is the
%number of columns (4) minus the rank (2), so $2$.


\begin{subproblem}
If $A$ and $B$ are square matrices then $(A+B)^2= A^2 +
  2AB + B^2$
\end{subproblem}
\vspace{1.3in}

%FALSE. You need to have $AB$=$BA$ for this to be true
%which is not usually the case.
%Take
%$A= \begin{bmatrix}
%   1 & 1 \\
%   1 & 0
%\end{bmatrix}$ and $B= \begin{bmatrix}
%   1 & 1 \\
%   0 & 1
%\end{bmatrix}$.
 

\begin{subproblem}
  Let $A$ be an $m \times n$ matrix, if the rank of $A$ is $m$ then
  the linear transformation $A \vec{x}$ is onto.
\end{subproblem}

%TRUE.  If the rank of $A$ is $m$ then $A\vec{x} = \vec{b}$ is
%consistent for all $\vec{b}$ which means that $A \vec{x}$ is onto.
\vspace{1.25in}
\begin{subproblem}
  Let $V=Span\{\vec{v_1}, \vec{v_2}, \vec{v_3}, \vec{v_4} \}$ be a
  subspace of ${\nr}^4$ of dimension $3$. Then $\{ \vec{v_1},
  \vec{v_2}, \vec{v_3}, \vec{v_4} \}$ form a basis of $V$.
\end{subproblem}

\vspace{1.3in}

%FALSE. A basis of $V$ has only 3 vectors in it.


\begin{subproblem}
  Let $V$ be a subspace of ${\nr}^n$. A basis of $V$ is a set of
  linearly independent vectors in $V$ that is as large as possible.
\end{subproblem}

\vspace{1.3in}

%TRUE.  The dimension of $V$ is $m$ which less than or equal to $n$.
%There are at most $m$ linearly independent vectors in $V$ and $m$
%linearly independent vectors in $V$ forms a basis of $V$.

\newpage
\end{problem}


\begin{problem}(15 points)

  \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 $2 \begin{bmatrix}
%      1 \\ 1
%    \end{bmatrix} +  \begin{bmatrix}
%      -1 \\ 2
%    \end{bmatrix} = \begin{bmatrix}
%      1 \\ 4
%    \end{bmatrix}$.
  
  \vspace{4in}
 \begin{subproblem}
    Let $T : \mathbb{R}^2 \to \mathbb{R}^2$ be a linear transformation. If
    $T( \begin{bmatrix} 1 \\ 1
    \end{bmatrix}) = (
    \begin{bmatrix}
      2 \\ 3
    \end{bmatrix})
    , T(
   \begin{bmatrix}
     -1 \\ 2
  \end{bmatrix}) =
  \begin{bmatrix}
  4 \\ 5
\end{bmatrix}
$ then $T(
\begin{bmatrix}
  1 \\ 4
\end{bmatrix}) = ?$.
  \end{subproblem}

 % $T(\begin{bmatrix} 1 \\ 4
%    \end{bmatrix}) = T(2\begin{bmatrix} 1 \\ 1
%    \end{bmatrix}+  \begin{bmatrix}
%      -1 \\ 2
%  \end{bmatrix}) = 2T(\begin{bmatrix} 1 \\ 1
%    \end{bmatrix}) + T(\begin{bmatrix}
%     -1 \\ 2
%  \end{bmatrix}) = 2\begin{bmatrix}
%      2 \\ 3
%    \end{bmatrix}+\begin{bmatrix}
%      4 \\ 5
%    \end{bmatrix}=
%  \begin{bmatrix}
%      4 \\ 6
%    \end{bmatrix}+\begin{bmatrix}
%      4 \\ 5
%    \end{bmatrix}=\begin{bmatrix}
%      8 \\ 11 \end{bmatrix}$, since $T$ is a linear transformation.
 
 \end{problem}


\newpage

\begin{problem}(30 points)

  Let $A=
  \begin{bmatrix} 1&-1&2&1&-1\\-2&2&-4&-1&3\\3&-2&6&3&1\\4&-4&8&4&-4
  \end{bmatrix}$, $\vec{b_1} =
    \begin{bmatrix}-2\\8\\4\\-8\end{bmatrix}$ and $\vec{b_2} =
    \begin{bmatrix}0\\2\\4\\0\end{bmatrix}$
    

   \begin{subproblem}
     Row reduce the matrix $[A | \vec{b}_1 | \vec{b}_2]$ into reduced
     row echelon form (this is not that hard).
    \end{subproblem}

  \begin{subproblem}
    Find a basis for $\ker(A)$. What is the dimension of $\ker(A)$?
  \end{subproblem}

  \begin{subproblem}
    Find a basis for the Image$(A)$. What is the dimension of
    Image$(A)$?
  \end{subproblem}

  \begin{subproblem}
    The vectors $\vec{b}_1$ and $\vec{b}_2$ are the $\text{Image}(A)$.
    Why?
  \end{subproblem}

  \begin{subproblem}
    Find all solutions of $A\vec{x} = \vec{b_1}$.
  \end{subproblem}

    \begin{subproblem}
    Find all solutions of $A\vec{x} = \vec{b_2}$.
    \end{subproblem}

    \begin{subproblem}
     Use your answers to describe \emph{all} possible matrices $C$ with the
    property that $AC =
    \begin{bmatrix}
      -2 & 0 \\
      8 & 2 \\
      4 & 4 \\
      -8 & 0
    \end{bmatrix}$.
     \end{subproblem}



  
%     The matrix $[A | \vec{ b}_1 | \vec{b}_2]$ reduces to $B =
%\begin{bmatrix} 1 & 0 & 2& 0 & 2 & 4 & 2\\ 0 & 1 & 0& 0 & 4 & 10 & 4 \\
%  0 & 0 & 0& 1 & 1 & 4 & 2 \\ 0 & 0 & 0 & 0&0 & 0 & 0
%  \end{bmatrix}$.
  
%  Hence a basis for the kernel are the two vectors $ \vec{x}_1 =
%  \begin{bmatrix} -2 \\ 0 \\ 1 \\ 0 \\ 0
%  \end{bmatrix}$ and $\vec{x}_2= \begin{bmatrix} -2 \\ -4 \\0 \\ -1 \\ 1
%  \end{bmatrix}$.  $\ker(A)$ has dimension 2.
  
%  The pivot columns of rref$(A)$ are the first, second and fourth.  So
%  the first, second and fourth columns of $A$ are a basis for the
%  Image($A$). Image($A$) has dimension 3.
  
%  In rref$([A | \vec{b}_1 | \vec{b}_2 ])$ the last two columns are a
%  linear combination of the pivot columns and hence $\vec{b}_1$ and
%  $\vec{b}_2$ are linear combinations of the columns of $A$.
  
%  A solution for $A \vec{x} = \vec{b_1}$ is $\vec{x}_p =
%  \begin{bmatrix}
%    4 \\ 10 \\0 \\ 4 \\ 0
%  \end{bmatrix}$ and all the solutions are in the form
%  $\vec{x}_p + \vec{x}$, where $\vec{x}$ is in the kernel of $A$. Or a
%  solution looks like $\vec{x}_p + c_1 \vec{x}_1 + c_2 \vec{x}_2$ for
%  $c_1$ and $c_2$ scalars. Let $H_1=\vec{x}_p + c_1 \vec{x}_1 + c_2
%  \vec{x}_2$.
  
%  A solution for $A \vec{x} = \vec{b_2}$ is $\vec{x}_q =
%  \begin{bmatrix}
%    2 \\ 4 \\0 \\ 2 \\ 0
%  \end{bmatrix}$ and all the solutions are in the form
%  $\vec{x}_q + \vec{x}$, where $\vec{x}$ is in the kernel of $A$. Or a
%  solution looks like $\vec{x}_q + c_1 \vec{x}_1 + c_2 \vec{x}_2$ for
%  $c_1$ and $c_2$ scalars. Let $H_2=\vec{x}_q + c_1 \vec{x}_1 + c_2
%  \vec{x}_2$.
  
%  Then $C = [ \vec{u}_1 \ \vec{u}_2]$, where $\vec{u}_i \in H_i$.

\end{problem}

\newpage ~ \newpage

\begin{problem} (15 points)
\begin{subproblem}
  Write down the standard matrix $A$ for the linear transformation $T:
  {\nr}^3 \longrightarrow {\nr}^3$ which is a projection on the $y-z$
  plane followed by a counterclockwise rotation of an angle of
  $\frac{\pi}{3}$ (60 degrees) in the $y-z$ plane (i.e., find $A$ such
  that $T(\vec{x}) = A \vec{x}$).
\end{subproblem}
\vspace{4.25in}
%$A = [T(\vec{e}_1) ~ T(\vec{e}_2) ~ T(\vec{e}_3)]$.  $\vec{e}_1$ is
%projected to $\vec{0}$ which is rotated to $\vec{0} = 
%\begin{bmatrix}
%  0 \\ 0 \\ 0
%\end{bmatrix}
%$.  $\vec{e}_2$ is projected to itself $\vec{e}_2$ which is then
%rotated to $\begin{bmatrix} 0 \\ 1/2 \\ \sqrt{3}/2
%\end{bmatrix}
%$.   $\vec{e}_3$ is projected to itself $\vec{e}_3$ which is then
%rotated to $\begin{bmatrix} 0 \\ -\sqrt{3}/2  \\ 1/2
%\end{bmatrix}
%$.  So $A= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1/2 & -\sqrt{3}/2 \\
%  0 & \sqrt{3}/2 & 1/2
% \end{bmatrix}$

\begin{subproblem}
  What is the image of $T$ (describe this geometrically)? What is the
  dimension of the image of $T$?  What is the rank of $T$?  (Hint: you
  should need to do any computation or row reduction to answer this
  question or any of the following questions.)
\end{subproblem}
\vspace{1in}
%Clearly the image of $T$ is the $y-z$ plane which has dimension $2$.
%So $A$ must have rank 2.

 \begin{subproblem}
   What is the dimension of the kernel of $T$? What is the image of
   $T$ (describe this geometrically)?
  \end{subproblem}
  \vspace{1in}
 % By the rank-nullity theorem the dimension of the kernel of $T$ must
%  be 1.  The kernel is the $x$-axis which has dimension 1.

\begin{subproblem}
Is $T$ 1-1? Why or why not? Is $T$ onto? Why or why not?
\end{subproblem}
\vspace{1in}
%$T$ is not 1-1 because the kernel of $T$ is not the zero vector. $T$
%is not onto because the image of $T$ is not all ${\nr}^3$.
\end{problem} 

\newpage

\begin{problem} (20 points -- 4 each)
  Provide the precise definition of each of the following concept:

\begin{subproblem}
  A vector $\vec{x}$ is a \emph{linear combination} of the vectors
  $\vec{v}_1, \ldots, \vec{v}_n$.
\end{subproblem}
\vspace{1.35in}
%If there are scalars $c_i$'s such that $ \vec{x} = c_1\vec{v}_1+
%\ldots + c_n\vec{v}_n$.

\begin{subproblem}
  The \emph{span} of a set of vectors $\{\vec{v}_1, \ldots, {\vec v}_n\}$.
\end{subproblem}
\vspace{1.35in}
%All linear combinations of the vectors $\{\vec{v}_1, \ldots, \vec{v}_n
%\}$.

\begin{subproblem}
  A set of vectors $\{\vec{v}_1, ..., \vec{v}_n\}$ being
  \emph{linearly independent}.
\end{subproblem}
\vspace{1.35in}
%%%%%No vector $\vec{v}_j$ is a linear combination of the others.

%\begin{subproblem}
%  A \emph{linear transformation} $T$.
%\end{subproblem}

%$T(\vec{x})$ is a linear transformation if there is a matrix $A$ such
%that $T(\vec{x})= A \vec{x}$.

\begin{subproblem}
  A function $T:X \rightarrow Y$ is \emph{onto}.
\end{subproblem}
\vspace{1.35in}
%If for all $y \in Y$, there is a $x \in X$ such that $T(x) = y$.

\begin{subproblem}
  $A$ is an \emph{invertible matrix}.
\end{subproblem}
\vspace{1.35in}
%There is a matrix $A^{-1}$ such that $AA^{-1} = I = A^{-1}A$.  This
%implies that $A$ must be square.

\end{problem}


\end{document}

