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\section*{Strang: Chapter 4}
\begin{description}

\item[{\bf Section 4.1}.] Exercises 5, 9, 11, 12, 16, 17, 19, 20, 21, 26

\item[{\bf Section 4.2}.] Exercises 1, 2, 3, 11, 12, 17, 21, 27, 31

\item[{\bf Section 4.3}.] Exercises 1, 6, 12, 17, 18, 21

\item[{\bf Problem week9-1}.] Define a function $T$ from $\Rr^n$ to $\Rr^m$ by the matrix multiply formula $T(\vec{x})=A\vec{x}$. Prove that for all vectors $\vec{u}, \vec{v}$ and all constants $c$, (a) $T(\vec{u}+\vec{v})=T(\vec{u})+T(\vec{v})$, (b) $T(c\vec{u})=cT(\vec{u})$. {\small\em Definition: $T$ is called a {\bf linear transformation} if $T$ maps $\Rr^n$ into $\Rr^m$ and satisfies (a) and (b).}

\item[{\bf Problem week9-2}.]  Let $T$ be a linear transformation from $\Rr^n$ into $\Rr^n$ that satisfies $\|T(\vec{x})\|=\|\vec{x}\|$ for all $\vec{x}$. Prove that the $n\times n$ matrix $A$ of $T$ is orthogonal, that is, $A^TA=I$, which means the columns of $A$ are {\bf orthonormal}:
     $$\col(A,i)\cdot\col(A,j)=0\quad\mbox{for}\quad i\ne j,\quad\mbox{and}\quad \col(A,i)\cdot\col(A,i)=1.$$

\item[{\bf Problem week9-3}.] Let $T$ be a linear transformation given by $n\times n$ orthogonal matrix $A$. Then $\|T(\vec{x})\|=\|\vec{x}\|$ holds. Construct an example of such a matrix $A$ for dimension $n=3$, which corresponds to holding the $z$-axis fixed and rotating the $xy$-plane $45$ degrees counter-clockwise. Draw a 3D-figure which shows the action of $T$ on the unit cube $S=\{ (x,y,z) ~:~ 0\le x \le 1, 0\le y\le 1, 0\le z \le 1\}$.

\item[{\bf Section 4.4}.] Exercises 3, 6, 10, 13, 15, 18, 30, 32\\
{\bf 4.4-18}: It helps to find explicitly $Q$ and $R$, which can be quickly checked in {\tt Maple}.

\end{description}


\subsection*{Some Answers}



{\bf 4.1}. Exercises 9, 12, 16, 21 have textbook answers.

{\bf 4.1-5}. The problem is about the equation $\mbox{nullspace}= \mbox{rowspace}^\perp$, valid for both matrix $A$ and $A^T$. The {\bf Fundamental Theorem of Linear Algebra}, Part II, can be summarized as the text
\begin{quote} {\em The nullspace is perpendicular to the rowspace}. \end{quote}
The text is justified from equality of vectors: $Ax=0$ is equivalent to scalar equations $\row(A,1)\cdot x=0,\ldots,\row(A,m)\cdot x=0$,
which says that $x$ is perpendicular to all rows of $A$, hence perpendicular to the rowspace of $A$.

(a) Apply the equation to $A^T$. Then nullspace$(A^T)$ $\perp$ rowspace$(A^T)$, equivalent to nullspace$(A^T)$ $\perp$ colspace$(A)$, implies any solution $y$ to $A^Ty=0$ is perpendicular to any $Ax$.
Since $b=Ax$, then $y\perp b$ or $y^Tb=0$. \newline
(b) If $A^Ty=(1,1,1)$ has a solution, then $y$ is in rowspace$(A)$. Then nullspace($A$) $\perp$ rowspace($A$) implies $y\cdot x=0$ for all $x$ in the nullspace of $A$.

{\bf 4.1-9}. Answers: colspace, perpendicular. See problem 4.1-5 above for nullspace $\perp$ rowspace, applied here for matrix $A^T$.
\newline
 {\bf Prove $A^TAx=0$ implies $Ax=0$}.\newline
 Because $y=Ax$ is a linear combination of the columns of $A$, then $y$ is in colspace$(A)$ = rowspace$(A^T)$. If $A^TAx=0$, then $A^Ty=0$, which implies $y$ is in nullspace$(A^T)$. Use nullspace($A^T$) $\perp$ rowspace($A^T$). Then nullspace$(A)$ and rowspace$(A^T)$ meet only in the vector $y=0$, which says $y=Ax=0$.\newline
 {\bf Prove $Ax=0$ implies $A^TAx=0$}.\newline
 First, assume $Ax=0$. Multiply by $A^T$ to get $A^TAx=A^T0$. The right side is the zero vector, which gives $A^TAx=0$.\newline
 See also problem 4.2-27, which repeats this same argument.


{\bf 4.1-11}.

For $A$: The nullspace is spanned by $(-2,1)$, the row space is spanned by $(1,2)$. The
column space is the line through $(1,3)$ and $N(A^T)$ is the line through $(3,-1)$. In each case,


For $B$: The nullspace of B is is a line spanned by $(0,1)$, the row space is a line spanned by
$(1,0)$. The column space and left nullspace are the same as for A. As in (a), the line pairs are perpendicular.

{\bf 4.1-17}. If $S$ is the subspace of $\Rr^3$ containing only the zero vector, then $S^\perp$ is $\Rr^3$. If $S$ is
spanned by $(1,1,1)$, then $S^\perp$ is the plane spanned by any two independent vectors perpendicular to $(1,1,1)$. For example, the vectors $(1,-1,0)$  and $(1,0,-1)$. If $S$ is
spanned by $(2,0,0)$ and $(0,0,3)$, then $S^\perp$ is the line spanned by $(0,1,0)$, computed as the cross product of the two vectors, then scaled to be a unit vector.

{\bf 4.1-26}. $A=\threebythree{2&2&-1}{-1&2&2}{2&-1&2}$

This example shows a matrix with perpendicular columns. Then
$A^TA= 9I$ is diagonal: $(A^TA)_{ij}=$ (column $i$ of $A$) $\cdot$ (column $j$ of $A$).
When the columns are unit vectors, then $A^TA=I$.

\bigskip

{\bf 4.2}. Exercises 1, 3, 11, 21, 31 have textbook answers.

{\bf 4.2-2}.

(a) The projection of $b=(\cos\theta,\sin\theta)$ onto $a=(1,0)$ is $p=(\cos\theta,0)$.

(b) The projection of $b=(1,1)$ onto $a=(1,-1)$ is $p=(0,0)$ since $a^Tb= 0$.

{\bf 4.2-12}.  $P_1 =\threebythree{1&0&0}{0&1&0}{0&0&0}$
= projection matrix onto the column space of A (the $xy$ plane)

$P_2 =\threebythree{0.5&0.5&0}{0.5&0.5&0}{0.0&0.0&1}$
=
Projection matrix onto the second column space.
Certainly $(P_2)^2= P_2$.


{\bf 4.2-17}. If $P^2= P$ then $(I-P)^2=(I-P)(I-P)=I^2-PI-IP+P^2=I-P$. When
$P$ projects onto the column space, then $I-P$ projects onto the left nullspace.

{\bf 4.2-27}. If $A^TAx= 0$ then $Ax$ is a vector in the nullspace of $A^T$. But $Ax$ is a vector in the column space
of $A$. To be in both of those perpendicular spaces, $Ax$ must be zero. So $A$ and $A^TA$
have the same nullspace.


\bigskip

{\bf 4.3}. Exercises 1, 18, 21 have textbook answers.

{\bf 4.3-6}. $a=(1,1,1,1)$ and $b=(0,8,8,20)$ give $\dd\hat{x}=\frac{a^Tb}{a^Ta}= 9$ and the projection is
$\dd\hat{x}a= p = (9,9,9,9)$. Then $\dd e^Ta= (-9,-1,-1,11)^T(1,1,1,1)= 0$ and $\dd \|e\|=\sqrt{204}$.

{\bf 4.3-12}.

(a) $a=(1,\ldots,1)$ has $a^Ta= m$, $a^Tb =b_1+\cdots+b_m$. Therefore $\dd\hat{x}= a^Tb/m$ is the
mean of the b's

(b) $\dd e= b -\hat{x}a$, $b=(1,2,b)$, $\|e\|=\sum_{i=1}^m (b_1 - \hat{x})^2$ = variance

(c) $p=(3,3,3)$, $e=(-2,-1,3)$, $\dd p^Te= 0$.
$ P =\frac13 \threebythree{1&1&1}{1&1&1}{1&1&1}$.

{\bf 4.3-17}. $\left(\begin{array}{rrr} 1 & -1 \\ 1 & 1 \\ 1 & 2 \end{array}\right)
\colvector{C \\ D} = \colvector{7 \\ 7 \\ 21}$. The solution $\vec{x} = \colvector{9 \\ 4}$ comes from
$\twobytwo{3&2}{2&6}\colvector{C\\ D} = \colvector{35\\ 42}$.

\bigskip


{\bf Problem week9-1}. Write out both sides of identities (a) and (b), replacing $T(\vec{w})$ by matrix product $A\vec{w}$ for various choices of $\vec{w}$. Then compare sides to finish the proof.

{\bf Problem week9-2}. Equation $\|T(\vec{x})\|=\|\vec{x}\|$ means lengths are preserved by $T$. It also means $\|A\vec{x}\|=\|\vec{x}\|$, which applied to $\vec{x}=\col(I,k)$ means $\col(A,k)$ has length equal to $\col(I,k)$ (=1).
Write $\|\vec{w}\|^2=\vec{w}\cdot\vec{w}=\vec{w}^T\vec{w}$ (the latter a matrix product). Then write out the equation
$\|A\vec{x}\|^2=\|\vec{x}\|^2$, to see what you get, for various choices of unit vectors $\vec{x}$.

{\bf Problem week9-3}. The equations for such a transformation can be written as plane rotation equations in $x$, $y$ plus the
identity in $z$. They might look like $x'=x\cos\theta -y\sin\theta$, $y'=$ similar, $z'=z$. Choose $\theta$ then test it by seeing what happens to $x=1$, $y=0$, $z=0$, the answer for which is a rotation of vector $(1,0,0)$. The answer for $A$ is obtained by writing the scalar equations as a matrix equation $(x',y',z')^T=A(x,y,z)^T$.

\bigskip


{\bf 4.4}. Exercises 3, 6, 13, 15, 18, 30, 32 have textbook answers.

{\bf 4.4-10}.
(a) If $q_1, q_2, q_3$ are orthonormal then the dot product of $q_1$ with $c_1q_1+c_2q_2+c_3q_3=0$ gives $c_1=0$.
Similarly $c_2=c_3=0$. Independent $q$'s. (b) $Qx=0$ implies $x=0$ implies $x=0$.

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