
% extra credit ch6 2250, S2015
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\begin{document}
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 {\bf Name~}\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\hfill
 {Class Time~}\underline{~~~~~~~~}\newline


\begin{center}
\bf Math 2250 Extra Credit Problems \\
\bf Chapter 6 \\
{\small\bf S2015}
\end{center}

{\bf Submitted work}. Please submit one stapled package with this sheet
on top. Kindly check-mark the problems submitted and
label the paper \fbox{\bf Extra Credit}. Label each solved problem with
its corresponding problem number, e.g., \fbox{Xc10.3-20}.


\Prob{Xc6.1-12}{(Eigenpairs of a $2\times 2$)}
 Let $A=\twobytwo{9 & -10}{2 & 0}$. Find the eigenpairs of $A$. Then report
 eigenpair
packages $P$ and $D$ such that $AP=PD$.


\Prob{Xc6.1-20}{(Eigenpairs of a $3\times 3$)} Let
$A=\threebythree{5 & -6 & 3}{6 & -7 & 3}{6 & -6 & 2}$. Find the
eigenpairs of $A$. Then report eigenpair packages $P$ and $D$ such
that $AP=PD$.
% A1:=matrix([[9,-10],[2,0]]);eigenvects(A1);
% [4, 1, {vector([2, 1])}], [5, 1, {vector([5/2, 1])}]

\Prob{Xc6.1-32}{(Complex eigenpairs of a $2\times 2$)} Let
$A=\twobytwo{0 & -6}{24 & 0}$. Find the eigenpairs of $A$. Then
report eigenpair packages $P$ and $D$ such that $AP=PD$.
% A:=matrix([[0,-6],[24,0]]);eigenvects(A);

\Prob{Xc6.1-36}{(Eigenvalues of band matrices)}
 Find the eigenvalues of the matrix $A$ below without the aid of
 computers.
$$
 A = \left(\begin{array}{rrrrrr}
 1 & 2 & 0 & 0 & 0 & 0 \\
 2 & 1 & 2 & 0 & 0 & 0 \\
 0 & 2 & 1 & 2 & 0 & 0 \\
 0 & 0 & 2 & 1 & 2 & 0 \\
 0 & 0 & 0 & 2 & 1 & 2 \\
 0 & 0 & 0 & 0 & 0 & 1
 \end{array}
 \right)
$$
%B:=matrix([
%  [1 , 2 , 0 , 0 , 0 , 0],
%  [2 , 1 , 2 , 0 , 0 , 0],
%  [0 , 2 , 1 , 2 , 0 , 0] ,
%  [0 , 0 , 2 , 1 , 2 , 0] ,
%  [0 , 0 , 0 , 2 , 1 , 2] ,
%  [0 , 0 , 0 , 0 , 0 , 1]]);
% eigenvals(B);
% -1, 3, 1+2*sqrt(3), 1-2*sqrt(3), 1, 1

\Prob{Xc6.2-6}{(Eigenpair packages of a $3\times 3$)}
 Let $A=\threebythree{2 & -2 & 1}{2 & -2 & 1}{2 & -2 & 1}$. Find the eigenpairs of $A$. Then report
 eigenpair
packages $P$ and $D$ such that $AP=PD$. Check the answer by hand,
expanding both products $AP$ and $PD$, finally showing equality.
% G:=matrix([[2 , -2 ,1],[2 , -2 , 1],[2 , -2 , 1]]);eigenvects(G);
% [1, 1, {vector([1, 1, 1])}], [0, 2, {vector([1, 0, -2]), vector([0, 1, 2])}]

\Prob{Xc6.2-18}{(Fourier's model for a $3\times 3$)}
 Assume Fourier's model for a certain matrix $A$:
$$
 A\left( c_1\colvector{1\\ 0 \\ -2}+c_2\colvector{1\\ 1\\
 0}+c_3\colvector{0\\ 0\\ 1}\right) =
 3c_1\colvector{1\\ 0 \\ -2}+c_2\colvector{1\\ 1\\
 0}+c_3\colvector{0\\ 0\\ 1}.
$$
Find $A$ explicitly from $AP=PD$. Check your answer by finding the
eigenpairs of $A$.
% C:=matrix([[3 , -2 ,0],[0 , 1 , 0],[-4 , 4 , 1]]);eigenvects(C);
% [3, 1, {vector([1, 0, -2])}], [1, 2, {vector([1, 1, 0]), vector([0, 0, 1])}]

\Prob{Xc6.2-28}{(Eigenpairs and diagonalization of a $4\times 4$)}
 Determine the eigenpairs of $A$ below. If diagonalizable, then
 report eigenpair packages $P$ and $D$ such that $AP=PD$.
$$
 A = \left(\begin{array}{rrrrrr}
 1 & 2 & 0 & 0 \\
 2 & 1 & 2 & 0 \\
 0 & 2 & 1 & 2 \\
 0 & 0 & 0 & 13
 \end{array}
 \right)
$$
%F:=matrix([
%  [1 , 2 , 0 , 0],
%  [2 , 1 , 2 , 0],
%  [0 , 2 , 1 , 2] ,
%  [0 , 0 , 0 , 3]]);
% eigenvals(F);
% [1, 1, {vector([-1, 0, 1, 0])}], [1+2*sqrt(2), 1, {vector([1, sqrt(2), 1, 0])}],
% [1-2*sqrt(2), 1, {vector([1, -sqrt(2), 1, 0])}],[3, 1, {vector([1, 1, 0, -1])}]

 \vfill {\bf End of extra credit problems chapter 6.}

\end{document}
