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\newcommand{\docid}{Exam 2} \input localmacs


\renewcommand{\today}{November 22, 2000}

\begin{document}
\doctitle


\begin{problem}
  (4 points each.) Provide the precise definition of each of the
  following concept:
\begin{subproblem}
   The \emph{basis} of a vector space $V$.
\end{subproblem}

\vspace{1.8in}

%$B$ is a basis for $V$ iff the span of $B$ is $V$ and $S$ is linearly
%independent.

\begin{subproblem}
  A \emph{finite-dimensional} vector space $V$.
\end{subproblem}

\vspace{1.8in}
%A vector space that has a finite basis.


\begin{subproblem}
  The \emph{eigenvalues} of a matrix $A$.
\end{subproblem}

\vspace{1.8in}
%$\lambda$ is an eigenvalue if there is a non-zero vector $\vec{x}$
%such that $A\vec{x} = \lambda \vec{x}$.

\begin{subproblem}
  Two matrices $A$ and $B$ are \emph{similar}.
\end{subproblem}

%$A$ and $B$ are similar iff there is a invertible matrix $P$ such that
%$A = P B P^{-1}$.

\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. (2 points each
  for correct answer; 2 points each for correct reason.)
  \begin{subproblem}
    $A$ is invertible iff $\lambda = 0$ is not a eigenvalue of $A$.
  \end{subproblem}
  \vspace{1.8in}
  
  %TRUE.  $\lambda$ is an eigenvalue of $A$ iff the nullspace of $A -
%  \lambda$ is $\mathbf{0}$ (the vector space with just the zero
%  vector).  So 0 is an eigenvalue of $A$ iff nullspace of $A$ is not
%  $\mathbf{0}$ iff $A$ is not invertible.

  \begin{subproblem}
    Let $A$ be a $m \times n$ matrix. Then the column space of $A$
    plus the nullspace of $A^{T}$ is $n$.
  \end{subproblem}


  \vspace{1.8in}
  %FALSE.  The rank theorem tells us that the rank of $A$ plus the
%  dimension of the nullspace of $A$ is $n$.  So the rank of $A^T$ plus
%  the dimension of the nullspace of $A$ is $m$. The rank of $A^T$,
%  rank of $A$, dimension of the column space of $A$ and the rowspace of
%  $A$ are the same.

  \begin{subproblem}
    The dimension of $\mathbb{P}^3$ (all polynomials of degree 3 or
    less) is 3.
  \end{subproblem}
  \vspace{1.8in}
 
 % FALSE.  $\{1,t,t^2,t^3 \}$ is a basis for $\mathbb{P}^3$ hence
%  $\mathbb{P}^3$ has dimension $4$.

  \begin{subproblem}
    All diagonalizable matrices are invertible.
  \end{subproblem}

\vspace{1.8in}
  %FALSE.  $A 
%  \begin{bmatrix}
%    1 & 0 \\
%    0 & 0 
%  \end{bmatrix}$ is diagonalizable not invertible.
\end{problem}

\newpage

\begin{problem} (20 points.)
  Let $A =
   \begin{bmatrix}
     1 &  0 & 2 & 1 \\
     0 &  1 & 0 & 1 \\
     1 & 1  & 2 & 2 \\
   \end{bmatrix}$.
   \begin{subproblem}
     Row reduce $A$.
   \end{subproblem}

 \begin{subproblem}
   Find a basis for the column space of $A$.  What the dimension of the
   column space $A$?
   \end{subproblem}
 
  \begin{subproblem}
    Find a basis for the row space of $A$.  What the dimension of the
    row space $A$?
   \end{subproblem}
 
  \begin{subproblem}
    Find a basis for the null space of $A$.  What the dimension of the
    null space $A$?
   \end{subproblem}
\end{problem}

%$A$ row reduces to $
%\begin{bmatrix}
%  1 &  0 & 2 & 1 \\
%  0 &  1 & 0 & 1 \\
%  0 & 0  & 0 & 0 \\
%\end{bmatrix}$.  The first 
%two columns of $A$ form a basis for the column space of $A$ which has
%dimension $2$.  The first two rows of $A$ form a basis for the row
%space of $A$ which has dimension $2$.  $(-2,0,1,0)$ and $(-1,-1,0,1)$
%form a basis for the null space of $A$ which also has dimension $2$.

\newpage


\begin{problem} (20 points.)
  Let $A =
   \begin{bmatrix}
     2 &  d & 1 \\
     0 &  2 & 2 \\
     0 & 0 & 3 \\
   \end{bmatrix}$.

   \begin{subproblem}
     For what values of $d$ is $A$ diagonalizable?
   \end{subproblem}
   \begin{subproblem}
     For any such value, diagonalize $A$ (do not find $P^{-1}$).
   \end{subproblem}
\end{problem}
 
%The goal is to diagonalize $A$.  The eigenvalues are $2$ and $3$.
%This is only possible if eigenspace corresponding the $\lambda =2$ has
%dimension $2$.  If $\lambda = 2$, $A - \lambda I =
%\begin{bmatrix}
%  0 &  d & 1 \\
%     0 &  0 & 2 \\
%     0 & 0 & 3 \\
%\end{bmatrix}$.  So the eigenspace corresponding the $\lambda =2$ has
%dimension $2$ iff $d = 0$.  Lets assume $d=0$.  Then a basis for
%eigenspace corresponding the $\lambda =2$ is $(1,0,0), (0,1,0)$.  If
%$\lambda = 1$ (and $d=0$), $A - \lambda I =
%\begin{bmatrix}
%  -1 &  0 & 1 \\
%     0 &  -1 & 2 \\
%     0 & 0 & 0 \\
%\end{bmatrix}$ which row reduces to
%$\begin{bmatrix}
%  1 &  0 & -1 \\
%     0 &  1 & -2 \\
%     0 & 0 & 0 \\
%\end{bmatrix}$.  So  a basis for
%eigenspace corresponding the $\lambda =3$ is $(1,2,1)$.  Hence $P =
%\begin{bmatrix}
%  1 & 0 & 1 \\
%  0 & 1 & 2 \\
%  0 & 0 & 1 \\
%\end{bmatrix}$ and $D = 
%\begin{bmatrix}
%  2 & 0 & 0 \\
%  0 & 2 & 0 \\
%  0 & 0 & 3 
%\end{bmatrix}$. \\ 

\newpage

\begin{problem}  (15 points.)
  Consider the vector space $V = \mathbf{M}_2$ of $2 \times 2$
  matrices. Let
  \begin{equation*}
    \mathcal{B} = \left \{ 
  E_{1,1} = \begin{bmatrix}
    1 & 0 \\
    0 & 0
  \end{bmatrix},
  E_{1,2} = \begin{bmatrix}
    0 & 1 \\
    0 & 0
  \end{bmatrix},
    E_{2,1} = \begin{bmatrix}
    0 & 0 \\
    1 & 0
  \end{bmatrix},
  E_{2,2} = \begin{bmatrix}
    0 & 0 \\
    0 & 1
  \end{bmatrix}
\right\}.
\end{equation*}
  
  From the sample exam (problem \#12) we know $\mathcal{B}$ is a basis
  for $V$.

  \begin{subproblem}
    Let $T: V \rightarrow V$ be given by the rule $T(M) = M - M^T$.
    Show that $T$ is a linear transformation.
  \end{subproblem}
  \begin{subproblem}
    Compute the matrix $N$ of $T$ relative of the basis $\mathcal{B}$.
    (As in problem \#12 $N$ is a $4 \times 4$ matrix, not a $2 \times
    2$ matrix.)
  \end{subproblem}
\end{problem}

%(a) $T(cM_1 + dM_2) = (cM_1 + dM_2) - (cM_1 + dM_2)^T = (cM_1 + dM_2)
%- c(M_1)^T + d(M_2)^T = c(M_1 - (M_1)^T) + d (M_2 - (M_2)^T) = c T
%(M_1) + d T(M_2)$. So $T$ is a linear transformation.

%(b) $T(E_{1,1}) = \vec{0}$, $T(E_{1,2})=E_{1,2} - E_{2,1}$,
%$T(E_{2,1})=E_{2,1} - E_{1,2}$ and $T(E_{2,2}) = \vec{0}$. $$N =
%\begin{bmatrix}
%  0 & 0 & 0 & 0 \\
%  0 & 1 & 1 & 0 \\
%  0 & -1 & -1 & 0 \\
%  0 & 0 & 0 & 0
%\end{bmatrix}.$$

\newpage

\begin{problem} (15 points.)
  Let $A$ be a $5 \times 8$ matrix whose rank is 5. Show that
  $A\vec{x} = \vec{b}$ is consistent and has infinitely many solutions
  for all $\vec{b}$ in $\mathbb{R}^5$.
\end{problem}

%The column space of $A$ has dimension 5.  Since the column space of
%$A$ is a subspace of $\mathbf{R}^5$ and the dimension of
%$\mathbf{R}^5$ is 5, the column space of $A$ is $\mathbf{R}^5$.  Hence
%for every $\vec{b}$ in $\mathbb{R}^5$ is consistent.  The nullspace of
%$A$ has dimension $8-5= 3$.  Thus the equation $A\vec{x} = \vec{0}$
%has infinitely solutions.  Therefore there are infinitely many
%solutions for all equations $A\vec{x} = \vec{b}$.

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