Solutions to Homework 2.
Math 110, Fall 2006.
Prob 1.4.4.
(a) Yes, since the linear system
a
+
b
=
1
2
a
+ 3
b
=
0

a
=

3
a

b
=
5
has a solution
a
= 3,
b
=

2.
(b) No, the corresponding linear system has no solution.
(c) Yes, the corresponding linear system has a solution
a
= 4,
b
=

3.
(d) Yes, the corresponding linear system has a solution
a
=

2,
b
= 5.
(e) No, the corresponding linear system has no solution.
(f) No, the corresponding linear system has no solution.
Prob 1.4.15.
Since any set
H
is contained in its linear span, we have
S
1
∩
S
2
⊆
S
1
⊆
span(
S
1
),
S
1
∩
S
2
⊆
S
2
⊆
span(
S
2
), hence
S
1
∩
S
2
⊆
span(
S
1
)
∩
span(
S
2
)
.
(1)
The righthand side of (1) is a subspace, hence taking the span of the lefthand side, we still remain within
that subspace, i.e.,
span(
S
1
∩
S
2
)
⊆
span(
S
1
)
∩
span(
S
2
)
.
An example of equality is provided by identical sets
S
1
=
S
2
; an example of inequality by two vectors which
are multiples of each other, say
S
1
=
{
(1
,
0)
}
,
S
2
=
{
(2
,
0)
}
. Then
S
1
∩
S
2
=
∅
, hence span(
S
1
∩
S
2
) =
{
(0
,
0)
}
,
while span(
S
1
)
∩
span(
S
2
) = span(
S
1
) =
{
(
t,
0) :
t
∈
IR
}
.
Prob 1.5.1.
(a) False: consider a dependent set
{
v,
0
}
where
v
6
= 0;
v
is a not a multiple of 0. (b) True:
directly from the deﬁnition. (c) False, also directly from the deﬁnition. (d) False, take the same example as
in (a). (e) True (shown in class). (f) True, from the deﬁnition.
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 Spring '08
 VALDIMARSSON
 Math, Linear Algebra, Vector Space, basis, Berlin UBahn, PROB

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