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