4
Solutions of Exercise 1.4
#1, 2(b) (c) (e) (f), 3(a) (b) (c) 4(a), 6, 10, 12, 13, 14, 15
1. (a) True. (b) False. The span of
∅
is
{
0
}
. (c) True. We need to prove this. Let
{
W
i
}
be the collection of all subspaces of
V
which contain
S
. Then
∩
i
W
i
conatins
S
and
being an intersection of subspaces it is again a subspace. Claim:
∩
i
W
i
is the smallest
subspace of
V
which contains
S
. In fact, if
W
is a subspace which contain
S
then
W
=
W
i
for some
i
. Hence
W
⊇∩
i
W
i
, which says
∩
i
W
i
is the smallest subspace of
V
which contain
S
. Therefore span(
S
)=
∩
i
W
i
. (d) False. The constant must be nonzero.
(e) True. (f) False.
2.
3. (a)Yes. (b)No. (c) No.
4. (a) The ±rst polynomial is a linear combination of the last two if and only if we can
±nd
a,b
such that
x
3

3
x
+5=
a
(
x
3
+2
x
2

x
+1)+
b
(
x
3
+3
x
2

1). This is equivalent
to the existence of the solution of the simultaneous linear equations
a
+
b
=1
2
a
b
=0

a
=

3
a

b
=5
Now we see that the equation has solution
a
=3
,b
=

2.
6. For any (
a,b,c
)
∈
F
3
we need to ±nd
x,y,z
∈
F
for which (
x
(1
,
1
,
0) +
y
(1
,
0
,
1) +
z
(0
,
1
,
1). Therefore we need to solve
x
+
y
=
a
x
+
z
=
b
y
+
z
=
c
The solution to this equation is
x
=
b

c
2
,y
=
a

b
+
c
2
,z
=

a
+
b
2
.
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 Spring '08
 GUREVITCH
 Math, Linear Algebra, Vector Space, Elementary algebra

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