1
Math 110 Homework 1
Partial Solutions
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1.2.13 This is not a vector space; it fails (at least) VS4. The zero element
here is the pair (0,1). With the “addition” deﬁned here, the pair (1,0)
has no additive inverse. In other words, there is no (
a
1
, a
2
) such that
(
a
1
, a
2
) + (1
,
0) = (0
,
1).
1.2.21 This is very straightforward; I will sketch the details. The zero of
Z
is the pair (0
V
,
0
W
). All of the axioms can be veriﬁed by applying the
analogous axioms for
V
and
W
. As a cultural aside, this object
Z
is
called the
direct sum
of
V
and
W
, denoted
Z
=
V
⊕
W
.
1.3.10
W
1
can be checked to be a subspace. First, 0 = (0
,
0
, . . . ,
0)
∈
W
1
since
0 + 0 +
···
+ 0 = 0. Second, if
x
= (
a
1
, . . . , a
n
) and
y
= (
b
1
, . . . , b
n
)
are in
W
1
, then
x
+
y
= (
a
1
+
b
1
, . . . , a
n
+
b
n
)
∈
W
1
since (
a
1
+
b
1
) +
···
+ (
a
n
+
b
n
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 Winter '07
 Liu
 Linear Algebra, Algebra, Addition, Vector Space, ej, scalar multiplication, 0w, additive inverse

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