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if and only if all of its coefﬁcients are 0; therefore, X is linearly
independent.
(ii) Deﬁne Vn = 1, x , x 2 , . . . , x n . Prove that 1, x , x 2 , . . . , x n is a
basis of Vn , and conclude that dim(Vn ) = n + 1.
Solution. Absent.
4.9 It is shown in analytic geometry that if
slopes m 1 and m 2 , respectively, then
only if m 1 m 2 = −1. If
i 1
1 and
and 2
2 are nonvertical lines with
are perpendicular if and = {αvi + u i : α ∈ R}, for i = 1, 2, prove that m 1 m 2 = −1 if and only if the dot product v1 · v2 =
0.
Solution. The lines 1 and 2 are perpendicular if and only if v1 and v2 are
perpendicular if and only if v1 · v2 = 0. If v1 = (a , b), then m 1 = b/a (the
slope of the line joining the origin O with (a , b)), and if v2 = (c, d ), then
m 2 = d /c. Hence,
m 1 m 2 = −1 ⇔ −1 = (b/a )(d /c)
⇔ b/a = −c/d
⇔ ac = −bd
⇔ ac + bd = 0
⇔ v1 · v2 = 0
⇔ 1 and 2 are perpendicular.
4.10 (i) A line in space passing through a point u is deﬁned as
{u + αw : α ∈ R} ⊆ R3 ,
where w is a ﬁxed nonzero vector. Show that every line through u
is a coset of a onedimensional subspace of R3 .
Solution. If the origin (0, 0, 0) lies on a line , then u = 0 and
= {αw : α ∈ R}, where w is some ﬁxed nonzero vector. In
vector space notation, = w , a onedimensional subspace of
R3 . (The converse is true, and it is easy to prove.) (ii) A plane in space passing through a point u is deﬁned as the subset
{v ∈ R3 : (v − u ) · n = 0} ⊆ R3 ,
where n = 0 is a ﬁxed normal vector and (v − u ) · n is a dot product. Prove that a plane through u is a coset of a twodimensional
subspace of R3 . ...
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 Fall '11
 KeithCornell

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