\def\bmatrix#1{\left[\matrix{#1}\right]}
\def\fr#1#2{{#1\over #2}}
\def\proj{\mathop{\rm proj}}

    {\bf Section 3.3}

\noindent{\bf 6.}

$$A = \bmatrix{1&1\cr 1&-1\cr -2&4}.$$
$$A^TA = \bmatrix{6&-8\cr -8&18}.$$
$$P = A(A^TA)^{-1}A^T = \bmatrix{1&1\cr 1&-1\cr -2&4}
\fr1{22}\bmatrix{9&4\cr4&3}\bmatrix{1&1&-2\cr1&-1&4}
= \fr1{22}\bmatrix{20&6&2\cr 6&4&-6\cr2&-6&20}.$$

$$p = Pb = \fr1{22}\bmatrix{20&6&2\cr
6&4&-6\cr2&-6&20}\bmatrix{1\cr2\cr7} =
\fr1{22}\bmatrix{46\cr-28\cr130}$$
With $b = p + q,$ and $p$ known, $q = b-p,$ and $q$ is in the left null
space of $A$ by the definition of orthogonality with the column space
which contains $p.$

\noindent{\bf 7.}

$$A = \bmatrix{1&1\cr0&1\cr1&-1}$$
$$A^TA = \bmatrix{2&0\cr0&3}$$
$$A(A^TA)^{-1}A^T =
\bmatrix{1&1\cr0&1\cr1&-1}\bmatrix{1/2&0\cr0&1/3}\bmatrix{1&0&1\cr1&1&-1}.$$

\noindent{\bf 12.}

{\it (a)}

$$\left\{\pmatrix{1\cr-1\cr0\cr0}, \pmatrix{0\cr1\cr0\cr-1}\right\}$$
is a basis for the orthogonal complement.

{\it (b)}

If $v_1$ and $v_2$ are the two given basis vectors for ${\bf V},$ they
are orthogonal and $v_1/\sqrt3$ and $v_2$ form an orthonormal basis, so
$$Q = \fr1{\sqrt3}\bmatrix{1&0\cr1&0\cr0&\sqrt3\cr1&0}$$
$$P = QQ^T =
\bmatrix{1/3&1/3&0&1/3\cr1/3&1/3&0&1/3\cr0&0&1&0\cr1/3&1/3&0&1/3}$$

{\it (c)}

This is the zero vector because an orthogonal vector $b$ has no parallel
component to the plane, and any parallel part in ${\bf V}$ would give a
longer distance.

    {\bf Section 3.4}

\noindent{\bf 16.}

$$A = \bmatrix{1&1\cr2&3\cr2&1} = \bmatrix{1/3&0\cr
2/3&1/\sqrt2\cr2/3&-1/\sqrt2}\bmatrix{3&3\cr0&\sqrt2} = QR.$$

Generally, $A$ is an $m\times n$ matrix, $Q$ is an $m\times r$ (where
$r$ is the rank of $A$) matrix, and $R$ is an $r\times r$ matrix.

\noindent{\bf 17.}

$$Pb = QQ^Tb = \bmatrix{1/9&2/9&2/9\cr 2/9&17/18&-1/18\cr
2/9&-1/18&17/18}\bmatrix{1\cr1\cr1} = \bmatrix{5/9\cr10/9\cr10/9}$$

    {\bf Section 3.5}

\noindent{\bf 12.}

$$y = F_8c.$$
$$y' = F_4c'.$$
$$(y')' = F_2(c')'.$$
$$((y')')' = F_1((c')')' = 1.$$
$$((y')')'' = F_1((c')')'' = 1.$$
$$(y')' = \bmatrix{((y')')' + w_2^0((y')')''\cr ((y')')' - w_2^0((y')')''} = \bmatrix{2\cr0}.$$
$$((y')'')' = F_1((c')'')' = 1.$$
$$((y')'')'' = F_1((c')'')'' = 1.$$
$$(y')'' = \bmatrix{2\cr 0}.$$
$$y' = \bmatrix{4\cr0\cr0\cr0}$$
$$y'' = F_4c'' = \underline 0.$$
$$y = \bmatrix{4\cr0\cr0\cr0\cr4\cr0\cr0\cr0}.$$

The same computation with $(0,1,0,1,0,1,0,1)$ has $y'$ and $y''$ switch
values, giving
$$y = \bmatrix{4\cr0\cr0\cr0\cr-4\cr0\cr0\cr0}.$$

\bye