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\section*{Strang: Chapter 7}
\begin{description}

\item[{\bf Section 7.1}.] Exercises  1, 3, 7, 8, 10, 11

\item[{\bf Section 7.2}.] Exercises  1, 2, 5, 10, 14, 20, 21, 26

\item[{\bf Section 7.3}.] Recommended singular value decomposition problems: Exercises  4, 7, 21, 23

\end{description}


\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.

\bigskip
7.3: Problems 4, 7, 23 have answers in Strang's book.

{\bf 7.3-21}. Column times row multiplication gives $A = U\Sigma V^T = \sum \sigma_i \vec{u}_i \vec{v}_i^T$ and also
$A^+ = V\Sigma^+ U^T = \sum (1/\sigma_i)\vec{v}_i\vec{u}_i^T$. Multiplying $A^+ A$ and using orthogonality
of each $\vec{u}_i$ to all 
other
$\vec{u}_j$
gives
the
projection
matrix
$A^+ A=\sum (1) \vec{v}_i\vec{v}_i^T$.
Similarly
$A A^+=\sum (1)\vec{u}_i\vec{u}_i^T$ from $VV^T=I$.

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