Homework SourseLinear Algebra Determine Dimensions of Vector Spaces of the
Linear Algebra, Determine Dimensions of Vector Spaces of the following questions:
1.
2.
3.
4.
V p(z)EP2(z) {dzp(0)=0} . Consider the vector space The dimension of Vis two. of polynomials Describe your isomorphism Iso by giving the elements of V satisfying the following Isofao.0 where, olas1 1 2)-10,1 where 3, satisfying 1,1 21 a3,2 SUBMITSolution
Ans-
Consider the element
(5
s
+5
;
7
t
)
2h
5
ih
7
i
. Since
is an isomorphism, it must
b e onto, so there is an integer
m
such that
(
m
) = (5
s
+ 5
;
7
t
)
:
However,
(
m
) = (5
ms;
7
mt
)
;
so we need to nd
m
so that
5
s
+ 5 = 5
ms
and
7
t
= 7
mt
. Thus, by setting
comp onents equal and canceling 5 and 7,
s
+ 1 =
ms
and
t
=
mt:
If
t
6
= 0
, then this forces
m
= 1
, but then we get
s
+ 1 = 1
s
=
s
, which is not
p ossible. Thus,
t
= 0
, but then for any integer
n
,
(
n
) = (5
ns;
7
nt
) = (5
ns;
0)
;
so
(
Z
) =
h
5
s
if
0
g6
=
h
5
ih
7
i
. In other words, for any integer
m
, the second
comp onent of
(
m
)
must b e zero. For example, there is no integer
m
for which
(
m
) = (5
;
7)
:
Therefore,
is not onto, contradicting the assumption that it was an isomor-
phism. Therefore,
Z
is not isomorphic to
h
5
ih
7
i
, so by Theorem 9.6,
Z
is not
equal to
h
5
ih
7
i
.
7 Chapter 9, Problem 44 (not graded, but take
a lo ok)
By Theorem 9.4, page 187, we have
D
13
=Z
(
D
13
)
Inn
(
D
13
)
;
which is pretty close to what we want. In order to prove that
D
13
itself is
isomorphic to Inn
(
D
13
)
,we need to do the following:
1. Prove that
Z
(
D
13
) =
f
R
0
g
, where
R
0
is the identity element of
D
13
(it is
a trivial rotation by a multiple of 360 degrees).
2. Prove that
D
13
=
f
R
0
g
D
13
. This can b e done by either dening an
isomorphism from
D
13
to
D
13
=
f
R
0
g
, or by dening a homomorphism from
D
13
to
D
13
which is onto and has kernel equal to
f
R
0
g
. I will present b o
