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MTH 235 : Linear Algebra
Sample problems  Midterm 1
1.
(a) Argue that
Z
/
5
Z
is a ﬁeld by using addition and multiplication tables.
(b) Argue that
Z
/
4
Z
is
not
a ﬁeld.
Solution.
(a) The addition and multiplication is inherited from that of the inte
gers
Z
, so it satisﬁes most of the axioms of a ﬁeld. We need to show existence of
additive/multiplicative identity and additive/multiplicative inverses.
+
0 1 2 3 4
0
0 1 2 3 4
1
1 2 3 4 0
2
2 3 4 0 1
3
3 4 0 1 2
4
4 0 1 2 3
·
0 1 2 3 4
0
0 0 0 0 0
1
0 1 2 3 4
2
0 2 4 1 3
3
0 3 1 4 2
4
0 4 3 2 1
The additive identity is 0, and

1 = 4,

2 = 3,

3 = 2,

4 = 1 so every element has
an additive inverse. The multiplicative identity is 1 and 2

1
= 3, 3

1
= 2, 4

1
= 4
so that every element has a multiplicative inverse.
(b) The tables for
Z
/
4
Z
are:
+
0 1 2 3
0
0 1 2 3
1
1 2 3 0
2
2 3 0 1
3
3 0 1 2
·
0 1 2 3
0
0 0 0 0
1
0 1 2 3
2
0 2 0 2
3
0 3 2 1
While
Z
/
4
Z
has all its additive inverses (

1 = 3,

2 = 2,

3 = 1) it does not have
all of its multiplicative inverses since there is no element
a
∈
Z
/
4
Z
so that 2
·
a
= 1.
1
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View Full Document 2. Let
V
be the vector space of functions
f
:
R
→
R
. Show that
W
⊂
V
is a subspace,
where
(a)
W
=
{
f

f
(1) = 0
}
(b)
W
=
{
f

f
(3) =
f
(1)
}
(c)
W
=
{
f

f
(

x
) =

f
(
x
)
}
Solution.
Need only to show that
W
is nonempty and if
α,β
∈
W
and
c
∈
R
then
cα
+
β
∈
W
.
(a) The constant function
f
(
x
) = 0 is in
W
. Now suppose that
f,g
∈
W
, i.e., that
f
(1) = 0 =
g
(1). We want to show that
cf
+
g
∈
W
, i.e., that (
cf
+
g
)(1) = 0.
(
cf
+
g
)(1) =
cf
(1) +
g
(1) =
c
·
0 + 0 = 0
so
W
is a subspace.
(b) Once again, any constant function is in
W
. Now assume that
f,g
∈
W
, i.e., that
f
(1) =
f
(3) and
g
(1) =
g
(3):
(
cf
+
g
)(1) =
cf
(1) +
g
(1) =
cf
(3) +
g
(3) = (
cf
+
g
)(3)
so
W
is a subspace.
(c) The function
f
(
x
) = sin(
x
) is in this set. Suppose that
f,g
∈
W
, i.e., that
f
(

x
) =

f
(
x
) and
g
(

x
) =

g
(
x
):
(
cf
+
g
)(

x
) =
cf
(

x
) +
g
(

x
) =

cf
(
x
)

g
(
x
) =

(
cf
+
g
)(
x
)
so
W
is a subspace.
2
3. Prove that
W
0
=
{
(
a
1
,a
2
,...,a
n
)
∈
F
n

a
1
+
a
2
+
···
+
a
n
= 0
}
is a subspace of
F
n
,
but
W
1
=
{
(
a
1
,a
2
,...,a
n
)
∈
F
n

a
1
+
a
2
+
···
+
a
n
= 1
}
is not.
Solution.
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This note was uploaded on 11/14/2010 for the course PHYCS 498 taught by Professor Aa during the Spring '10 term at University of Illinois, Urbana Champaign.
 Spring '10
 aa

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