Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 53
(c) It is linearly dependent if and only if either vector is a multiple of the other. That is, it is not
independent iff
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎝ a d⎠ = r · ⎝ b e⎠
g h
or
⎝ b e⎠ = s · ⎝ a d⎠
h g
(or both) for some scalars r and s. Eliminating r and s in order to restate this condition only in
terms of the given letters a, b, d, e, g, h, we have that it is not independent — it is dependent — iff
ae − bd = ah − gb = dh − ge.
(d) Dependence or independence is a function of the indices, so there is indeed a formula (although
at first glance a person might think the formula involves cases: “if the first component of the first
vector is zero then . . . ”, this guess turns out not to be correct).
Two.II.1.39 Recall that two vectors from R n are perpendicular if and only if their dot product is
zero.
(a) Assume that ⃗v and ⃗w are perpendicular nonzero vectors in R n , with n > 1. With the linear
relationship c⃗v + d ⃗w = ⃗0, apply ⃗v to both sides to conclude that c · ‖⃗v‖ 2 + d · 0 = 0. Because ⃗v ≠ ⃗0
we have that c = 0. A similar application of ⃗w shows that d = 0.
(b) Two vectors in R 1 are perpendicular if and only if at least one of them is zero.
We define R 0 to be a trivial space, and so both ⃗v and ⃗w are the zero vector.
(c) The right generalization is to look at a set {⃗v 1 , . . . , ⃗v n } ⊆ R k of vectors that are mutually
orthogonal (also called pairwise perpendicular): if i ≠ j then ⃗v i is perpendicular to ⃗v j . Mimicing the
proof of the first item above shows that such a set of nonzero vectors is linearly independent.
Two.II.1.40 (a) This check is routine.
(b) The summation is infinite (has infinitely many summands). The definition of linear combination
involves only finite sums.
(c) No nontrivial finite sum of members of {g, f 0 , f 1 , . . .} adds to the zero object: assume that
c 0 · (1/(1 − x)) + c 1 · 1 + · · · + c n · x n = 0
(any finite sum uses a highest power, here n). Multiply both sides by 1 − x to conclude that
each coefficient is zero, because a polynomial describes the zero function only when it is the zero
polynomial.
Two.II.1.41 It is both ‘if’ and ‘only if’.
Let T be a subset of the subspace S of the vector space V . The assertion that any linear relationship
c 1
⃗t 1 + · · · + c n
⃗t n = ⃗0 among members of T must be the trivial relationship c 1 = 0, . . . , c n = 0 is a
statement that holds in S if and only if it holds in V , because the subspace S inherits its addition and
scalar multiplication operations from V .
Subsection Two.III.1: Basis
Two.III.1.16 By Theorem 1.12, each is a basis if and only if each vector in the space can be given in
a unique way as a linear combination of the given vectors.
(a) Yes this is a basis. The relation
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 3 0 x
c 1
⎝2⎠ + c 2
⎝2⎠ + c 3
⎝0⎠ = ⎝y⎠
3 1 1 z
gives
⎛
⎝ 1 3 0 x
⎞
2 2 0 y
3 1 1 z
⎠ −2ρ1+ρ2 −2ρ 2+ρ 3
−→ −→
−3ρ 1 +ρ 3
⎛
⎝ 1 3 0 x
0 −4 0 −2x + y
0 0 1 x − 2y + z
which has the unique solution c 3 = x − 2y + z, c 2 = x/2 − y/4, and c 1 = −x/2 + 3y/4.
(b) This is not a basis. Setting it up as in the prior item
⎛
c 1
⎝ 1 ⎞ ⎛
2⎠ + c 2
⎝ 3 ⎞ ⎛
2⎠ = ⎝ x ⎞
y⎠
3 1 z
⎞
⎠