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Problem Set 3  Solutions
J. Scholtz
October 24, 2006
Question 1: 1.3/31
a) Cosets and Subspaces
We want to show that
v
+
W
is a subspace if and only if
v
∈
W
.
(
⇐
) Suppose that
v
+
W
is a subspace.
v
+
W
must contain 0. Then there exists
u
∈
W
such that
v
+
u
= 0, hence
W
contains

v
, and sincd it is a subspace
itself then
W
contains also
v
.
(
⇒
) If
v
∈
W
, then the set of form
{
v
+
w,w
∈
W
}
=
W
, since that is closed
under addition. Therefore
v
+
W
=
W
which we know is a subspace.
b) Equivalent representations
Suppose that
v
1
+
W
=
v
2
+
W
, then this is equivalent to the statement that
for every
w
1
∈
W
, there exists
w
2
∈
W
such that
v
1
+
w
1
=
v
2
+
w
2
, which is
equivalent with the statement
v
1

v
2
=
w
2

w
1
, but this is equivalent with the
statement that
v
1

v
2
∈
W
since
w
2

w
1
∈
W
since
W
is a suspace. Since all
these statements were equivalent then the if and only condition is satisﬁed.
c) Are these operations well deﬁned?
If
v
1
+
W
=
v
0
1
+
W
and
v
2
+
W
=
v
0
2
+
W
then
v
1

v
0
1
∈
W
and
v
2

v
0
2
∈
W
.
Therefore (
v
1
+
v
2
)

(
v
0
1
+
v
0
2
)
∈
W
, hence (
v
1
+
v
2
)+
W
= (
v
0
1
+
v
0
2
)+
W
, which
in turn means that (
v
1
+
W
) + (
v
2
+
W
) = (
v
0
1
+
W
) + (
v
0
2
+
W
).
For scalar multiplication: if
v
1
+
W
=
v
0
1
+
W
then
v
1

v
0
1
∈
W
, hence
a
(
v
1

v
0
1
)
∈
W
therefore
av
1

av
0
1
∈
W
which in turn implies that
av
1
+
W
=
av
0
1
+
W
.
But this is equivalent to the statement:
a
(
v
1
+
W
) =
a
(
v
0
1
+
W
).
d) They form a vector space!
We need to check all the axioms:
VS1(Commutativity): (
v
+
W
) + (
u
+
W
) = (
v
+
u
) +
W
= (
u
+
v
) +
W
=
(
u
+
W
) + (
v
+
W
), where we used the fact that
V
is a vectorspace.
VS2(Associativity): Use the same trick knowing that
V
is a vectorspace.
VS3(Existence of 0): The coset 0 +
W
does the job since (0 +
W
) + (
a
+
W
) =
(
a
+ 0) +
W
=
a
+
W
.
1
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View Full Document VS4(Inverses): Use the inverses from
V
: (

v
+
W
) + (
v
+
W
) = (
v

v
) +
W
=
0 +
W
.
VS5(Identity): Take the identity from
V
: 1(
v
+
W
) = 1
v
+
W
=
v
+
W
.
The last three are inherited due to the nature of deﬁnition of addition and scalar
multiplication in
V/W
.
Question 2: 1/6/33
a)
Let
V
=
W
1
⊕
W
2
, that means that
W
1
∩
W
2
= 0. Now suppose that
v
∈
β
1
∩
β
2
,
this would mean that
v
∈
W
2
and
v
∈
W
1
, which is a contradiction.
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This note was uploaded on 07/03/2011 for the course MATH 110 taught by Professor Gurevitch during the Spring '08 term at University of California, Berkeley.
 Spring '08
 GUREVITCH
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