Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 49
Two.II.1.25 (a) Assume that the set {⃗u, ⃗v, ⃗w} is linearly independent, so that any relationship d 0 ⃗u+
d 1 ⃗v + d 2 ⃗w = ⃗0 leads to the conclusion that d 0 = 0, d 1 = 0, and d 2 = 0.
Consider the relationship c 1 (⃗u) + c 2 (⃗u + ⃗v) + c 3 (⃗u + ⃗v + ⃗w) = ⃗0. Rewrite it to get (c 1 + c 2 +
c 3 )⃗u + (c 2 + c 3 )⃗v + (c 3 ) ⃗w = ⃗0. Taking d 0 to be c 1 + c 2 + c 3 , taking d 1 to be c 2 + c 3 , and taking d 2
to be c 3 we have this system.
c 1 + c 2 + c 3 = 0
c 2 + c 3 = 0
c 3 = 0
Conclusion: the c’s are all zero, and so the set is linearly independent.
(b) The second set is dependent.
1 · (⃗u − ⃗v) + 1 · (⃗v − ⃗w) + 1 · ( ⃗w − ⃗u) = ⃗0
Beyond that, the two statements are unrelated in the sense that any of the first set could be either
independent or dependent. For instance, in R 3 , we can have that the first is independent while the
second is not
⎛
{⃗u, ⃗v, ⃗w} = { ⎝ 1 ⎞ ⎛
0⎠ , ⎝ 0 ⎞ ⎛
1⎠ , ⎝ 0 ⎞
⎛
0⎠} {⃗u − ⃗v, ⃗v − ⃗w, ⃗w − ⃗u} = { ⎝ 1 ⎞ ⎛
−1⎠ , ⎝ 0 ⎞ ⎛ ⎞
−1
1 ⎠ , ⎝ 0 ⎠}
0 0 1
0 −1 1
or that both are dependent.
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
{⃗u, ⃗v, ⃗w} = { ⎝ 1 0⎠ , ⎝ 0 1
0 0
⎠ ,
⎝ 0 1
1
⎠} {⃗u − ⃗v, ⃗v − ⃗w, ⃗w − ⃗u} = {
⎛
⎝ 1 −1
0
⎞
⎠ ,
⎛
⎝ 0 ⎞ ⎛
0 ⎠ , ⎝ −1 ⎞
1 ⎠}
−1 1
Two.II.1.26 (a) A singleton set {⃗v} is linearly independent if and only if ⃗v ≠ ⃗0. For the ‘if’ direction,
with ⃗v ≠ ⃗0, we can apply Lemma 1.4 by considering the relationship c · ⃗v = ⃗0 and noting that the
only solution is the trivial one: c = 0. For the ‘only if’ direction, just recall that Example 1.11 shows
that {⃗0} is linearly dependent, and so if the set {⃗v} is linearly independent then ⃗v ≠ ⃗0.
(Remark. Another answer is to say that this is the special case of Lemma 1.16 where S = ∅.)
(b) A set with two elements is linearly independent if and only if neither member is a multiple of the
other (note that if one is the zero vector then it is a multiple of the other, so this case is covered).
This is an equivalent statement: a set is linearly dependent if and only if one element is a multiple
of the other.
The proof is easy. A set {⃗v 1 , ⃗v 2 } is linearly dependent if and only if there is a relationship
c 1 ⃗v 1 + c 2 ⃗v 2 = ⃗0 with either c 1 ≠ 0 or c 2 ≠ 0 (or both). That holds if and only if ⃗v 1 = (−c 2 /c 1 )⃗v 2 or
⃗v 2 = (−c 1 /c 2 )⃗v 1 (or both).
Two.II.1.27
This set is linearly dependent set because it contains the zero vector.
Two.II.1.28 The ‘if’ half is given by Lemma 1.14. The converse (the ‘only if’ statement) does not
hold. An example is to consider the vector space R 2 and these vectors.
( ( (
1 0 1
⃗x = , ⃗y = , ⃗z =
0)
1)
1)
Two.II.1.29
(a) The linear system arising from
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 −1 0
c 1
⎝1⎠ + c 2
⎝ 2 ⎠ = ⎝0⎠
0 0 0
has the unique solution c 1 = 0 and c 2 = 0.
(b) The linear system arising from
⎛
c 1
⎝ 1 ⎞ ⎛
1⎠ + c 2
⎝ −1
⎞ ⎛
2 ⎠ = ⎝ 3 ⎞
2⎠
0 0 0
has the unique solution c 1 = 8/3 and c 2 = −1/3.
(c) Suppose that S is linearly independent. Suppose that we have both ⃗v = c 1 ⃗s 1 + · · · + c n ⃗s n and
⃗v = d 1
⃗t 1 + · · · + d m
⃗t m (where the vectors are members of S). Now,
can be rewritten in this way.
c 1 ⃗s 1 + · · · + c n ⃗s n = ⃗v = d 1
⃗t 1 + · · · + d m
⃗t m
c 1 ⃗s 1 + · · · + c n ⃗s n − d 1
⃗t 1 − · · · − d m
⃗t m = ⃗0