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\def\bmatrix#1{\left[\matrix{#1}\right]}
\def\fr#1#2{{#1\over #2}}
{\bf Section 3.1}
\noindent{\bf 6.}
An orthonormal basis would be
$$\{\fr1{\sqrt6}\bmatrix{1\cr1\cr-2}\}.$$
This was found by inspection starting with the second vector, which is
orthogonal to $(1,-1,k),$ where $k$ is any number and then choosing $k$
such that the dot product with the first vector was zero. Then, it was
normalized. It is the complete orthogonal space because a space
orthogonal to a 2-dimensional space in $R^3$ has dimension 1.
\noindent{\bf 12.}
A basis would be
$$\{\bmatrix{-2\cr-2\cr 1}\}.$$
This was found by a similar method to 6, and because it is also in
$R^3,$ is proved complete by the same means.
\noindent{\bf 19.}
{\it (a)}
The trivial space $\{0\}$ is orthogonal to itself, but $\{0\}^\perp =
R^n$ is not orthogonal to itself.
{\it (b)}
Orthogonality is reflexive, so if $V=Z,$ this statement is false if $V$
is nontrivial.
\noindent{\bf 32.}
$A$ and $B$ have the same column spaces and left null spaces, the column
space having basis
$$\{\bmatrix{1\cr 3}\}$$
and the left null space having orthogonal basis
$$\{\bmatrix{-3\cr 1}\}.$$
However, they have different row and null spaces, $A$ having row space
with basis
$$\{\bmatrix{1\cr 2}\},$$
and null space with basis
$$\{\bmatrix{-2\cr 1}\}.$$
and $B$ having row space with basis
$$\{\bmatrix{1\cr 0}\},$$
and null space with basis
$$\{\bmatrix{0\cr 1}\}.$$
{\bf Section 3.2}
\noindent{\bf 12.}
Orthonormal basis is
$$Q = \bmatrix{-2/\sqrt{5}\cr 1/\sqrt{5}},$$
which is on the line.
$$P = QQ^T = \fr1{5}\bmatrix{4&-2\cr-2&1}.$$
\noindent{\bf 14.}
The point $(x,y,t) = (1,-2,1)$ is a basis for this subspace.
By the same method as in 12,
$$P = \fr16\bmatrix{1\cr -2\cr 1}\bmatrix{1&-2&1} =
\fr16\bmatrix{1&-2&1\cr -2&4&-2\cr 1&-2&1}.$$
\noindent{\bf 17(a)} % only (a)
The projection of $b$ onto the subspace through $a$ is $a^Tb/||b|| * a,$
giving
$$(5/9)\bmatrix{1\cr1\cr1}$$
\noindent{\bf 24.}
$a_1$ is a normal, so $b$ projected onto $a_1$ is $(1,0).$
$b$ projected onto $a_2$ is $3a_2/\sqrt5.$
The sum is $(1+3/\sqrt5, 6/\sqrt5).$
\bye
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