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\begin{center}\large\bf
Linear Algebra 2270-2 \\
\normalsize\rm
Due in Week 15
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The fourteenth  week completes chapter 7, which is the end of the course.

\begin{description}

\item[{\bf Section 7.3}.] Exercises  1, 5, 9

\item[{\bf Section 7.4}.] Exercises  3, 7, 13, 15, 17, 23

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%\subsection*{Problem Notes}
%
%Issues for Strang's problems will be communicated here. For help, please send email, call 581-6879, or visit JWB 113.
%
%
%\subsection*{Some Answers}
%
%
%\bigskip
%
%{\bf 7.1}. Exercises 3, 7, 10 have textbook answers.
%
%{\bf 7.1-1}. With $w = 0$ linearity gives $T(v + 0)= T(v)+T(0)$. Thus $T(0)=0$. With $c= -1$
%linearity gives $T(-\vec{0})=-T(\vec{0})$. This is a second proof that $T(0)=0$.
%
%{\bf 7.1-8}. (a) The range of $T (v_1,v_2)= (v_1-v_2,0)$ is the line of vectors $(c,0)$. The nullspace
%is the line of vectors $(c,c)$. (b) $T(v_1,v_2,v_3)=(v_1,v_2)$ has Range $\Rr^2$, kernel
%$\{(0, 0, v_3)\}$ (c) $T(0)=0$ has Range $\{0\}$, kernel $\Rr^2$ (d)  $T(v_1,v_2)=(v_1,v_1)$
%has Range = multiples of $(1,1)$, kernel = multiples of $(1,-1)$.
%
%{\bf 7.1-11}. For multiplication $T(v)= Av$: $V = \Rr^n$, $W = \Rr^m$; the outputs fill the column space;
%$v$ is in the kernel if $Av = 0$.
%
%
%\bigskip
%
%{\bf 7.2}. Exercises 5, 14, 20  have textbook answers.
%
%{\bf 7.2-1}. For $Sv = d^2v/dx^2$,
%$v_1, v_2, v_3, v_4 = 1, x, x^2, x^3$,
%$Sv_1 = Sv_2 = 0$, $Sv_3 = 2v_1$, $Sv_4 = 6v_2$. The matrix for $S$ is
%$B=\left(\begin{array}{rrrrrr}
% 0& 0& 2& 0 \\
% 0& 0& 0& 6 \\
% 0& 0& 0& 0 \\
% 0& 0& 0& 0
% \end{array}\right)$.
%
%{\bf 7.2-2}. $Sv = d^2v/dx^2 = 0$ for linear functions $v(x)=a + bx$. All $(a, b, 0, 0)$ are in the
%nullspace of the second derivative matrix $B$.
%
%{\bf 7.2-10}. The matrix for $T$ is $A =
%\threebythree
%{1& 0& 0}
%{1& 1& 0}
%{1& 1& 1}$.
%For the output
%$\colvector{1\\ 0\\ 0}$
%choose input $v =\colvector{1\\ -1\\ 0}= A^{-1}\colvector{1\\ 0\\ 0}$.
%This means: For the output $w_1$ choose the input $v_1- v_2$.
%
%{\bf 7.2-21}. Basis $w$ to basis $v$:
%$\threebythree
%{0.0 & 1&   0}
%{0.5 & 0&  -0.5}
%{0.5 &-1&   0.5}
%$.
%Basis $v$ to basis $w$: inverse matrix =
%$\threebythree
%{1&  1& 1}
%{1&  0& 0}
%{1& -1& 1}
%$. The key
%idea: The matrix multiplies the coordinates in the $v$ basis to give the coordinates in the
%$w$ basis.
%
%{\bf 7.2-26}. Start from $A = LU$. Row 2 of $A$ is $\ell_{21}$(row 1 of $U$)+ $\ell_{22}$ (row 2 of $U$). The change
%of basis matrix is always invertible, because basis goes to basis.

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