Download HOMEWORK 1 SOLUTIONS Levandosky, Linear Algebra 1.2 (a

HOMEWORK 1 SOLUTIONS
Levandosky, Linear Algebra
1.2 (a) Using the properties of vector addition and scalar multiplication, we
find

   

 
−2
1
−1
1
 3  − 2 1  =  3  −  2 
2
1
2
2


1 − (−2)
= 3−2 
2−2
 
3
=  1 .
0
(b) Similarly, we find


 
  
    
4
0
1
12
0
3
3 0  + 2 1  − 3 1  =  0  +  2  −  3 
−1
4
0
−3
8
0


12 + 0 − 3
= 0+2−3 
−3 + 8 + 0


9
=  −1  .
5
1.3 (a) Substituting and computing as above, we find



 

3
2
1
v − 2w + x =  −2  − 2  1  +  4 
3
−1
−5


3 − 2(2) + 1
=  −2 − 2(1) + 4 
3 − 2(−1) − 5
 
0
=  0 .
0
1
2
HOMEWORK 1 SOLUTIONS
(b) Similarly, we find

 
 

3
2
1
v + w + x =  −2  +  1  +  4 
3
−1
−5


3+2+1
=  −2 + 1 + 4 
3−1−5


6
=  3 .
−3
1.8 (a)
-2u+v
-2u
v
u
(b)
2v
v
u
3u+2v
3u
HOMEWORK 1 SOLUTIONS
3
(c)
v-u
v
-u
u
(d)
v
-v
u
u-v
2.1 (a) Observe that
¸
·
¸
¸
·
·
1 1
2 2
1
2
1
=−
+
= − v + w.
0
2
1
3
3
3
3
(b) Observe that
·
¸
·
¸
·
¸
2 1
1 2
2
1
0
−
= v − w.
=
1
3 2
3 1
3
3
(c) Observe that
x=
·
x1
x2
¸
·
= x1
1
0
¸
·
+ x2
0
1
¸
.
(d) By part (c), we have
·
¸
·
¸
· ¸
x1
1
0
x=
= x1
+ x2
.
x2
0
1
Subtituting in our results from parts (a) and (b) then gives
2
2
1
1
x = x1 (− v + w) + x2 ( v − w)
3
3
3
3
−x1
2x2
2x1
x2
=(
+
)v + (
− )w.
3
3
3
3
4
HOMEWORK 1 SOLUTIONS
3.2 The given vectors are linearly dependent (as are any three vectors in R2 , as
we’ll soon see). There are many possible ways to express one of the vectors
as a linear combination of the other two. One such expression is:
· ¸
·
¸
· ¸
3 3
1 1
2
=
−
.
1
4 2
4 2
3.7 Suppose there exist constants a, b, c ∈ R such that
a(u + v) + b(u + w) + c(v + w) = 0.
We must show that a, b, and c are all equal to 0 (since this is the very
definition of linear independence!). Note that we can rewrite the above
equation as
(a + b)u + (a + c)v + (b + c)w = 0.
Since {u, v, w} is a linearly independent set by hypothesis, the above equation implies a + b = 0, a + c = 0, and b + c = 0. The only solution to these
three equations is a = b = c = 0.
(Here’s a quick proof:
Subtract the second equation from the first to obtain b − c = 0. Now add
this equation to the third to obtain 2b = 0, and hence b = 0. The first
equation then gives a = −b = 0, and the third equation gives c = −b = 0.)
3.8 The set {u − v, v − w, u − w} is not a linearly independent set. Indeed,
observe that
(u − v) + (v − w) − (u − w) = u − v + v − w − u + w = 0.
3.10 True. For suppose S 0 ⊂ S is a subset of vectors which are linearly dependent. For simplicity, let us assume S 0 = {v1 , . . . , vm } with 1 ≤ m ≤ k (we
could always relabel the vectors to ensure this). Then there exist constants
c1 , . . . , cm ∈ R, at least one of which is nonzero, such that
c1 v1 + . . . cm vm = 0.
Then note that
c1 v1 + . . . cm vm + 0vm+1 + . . . + 0vk = 0,
and we still have that at least one of the coefficients is nonzero (since one of
the ci is nonzero). This implies that the set S is linearly dependent, which
contradicts our hypothesis.
So, if S is a set of linearly independent vectors, then every subset of S
must also be linearly independent.
· ¸ · ¸
1
1
3.11 False. Here is a very silly example. Consider the set S = {
,
}⊂
0
· 0 ¸
1
R2 . This set is clearly linearly dependent, but the subset S 0 = {
} is
0
trivially linearly independent.