MATH 223, Linear Algebra
Winter 2008
Solutions to Assignment 2
C
stands for the set of complex numbers, and
R
for the set of real numbers.
1. Here are some subsets of the complex vector space
C
3
. In each case, decide
whether the given set
S
is a subspace of
C
3
; justify your answers.
(a)
S
=
{
a

b
3
ib
(2 +
i
)
a

2
b
:
a,b
any complex numbers
}
.
Solution: This is a subspace.
First
~
0
∈
S
since we can take
a
=
b
= 0.
Next, suppose that
~v
1
and
~v
2
are in
S
. Then
~v
1
=
a
1

b
1
3
ib
1
(2 +
i
)
a
1

2
b
2
for some complex numbers
a
1
and
b
1
, and
~v
2
=
a
2

b
2
3
ib
2
(2 +
i
)
a
2

2
b
2
for some complex
a
2
,
b
2
. So
~v
1
+
~v
2
=
a
1

b
1
+
a
2

b
2
3
ib
1
+ 3
ib
2
(2 +
i
)
a
1

2
b
1
+ (2 +
i
)
a
2

2
b
2
=
(
a
1
+
a
2
)

(
b
1
+
b
2
)
3
i
(
b
1
+
b
2
)
(2 +
i
)(
a
1
+
a
2
)

2(
b
1
+
b
2
)
, which is of the right form to be
an element of
S
. Hence
S
is closed under addition.
Finally, suppose
~v
∈
S
and
α
is a (complex) scalar. Then
~v
=
a

b
3
ib
(2 +
i
)
a

2
b
for some complex numbers
a
and
b
, so
α~v
=
(
αa
)

(
αb
)
3
i
(
αb
)
(2 +
i
)(
αa
)

2(
αb
)
, again an element of
S
. Thus
S
is closed
under scalar multiplication.
(
S
is in fact
Span
{
1
0
2 +
i
,

1
3
i

2
}
.
)
(b)
S
=
{
a

b
3
ib
+ 2
(2 +
i
)
a

2
b
:
a,b
any complex numbers
}
.
Solution: This is not a subspace. In fact,
~
0
/
∈
S
. If it were (note the
use of the subjunctive) we would need
a

b
= 3
ib
+2 = (2+
i
)
a

2
b
= 0
1
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View Full Documentfor some complex
a
and
b
. The ﬁrst two equations would tell us
a
=
b
=
2
3
i
, but then (2 +
i
)
a

2
b
6
= 0. So it’s impossible for
S
to
have the zero vector. (In fact, it is not closed under either addition or
scalar multiplication, either. But you don’t need to show that now.)
(c)
S
=
{
a

b
3
ib
(2 +
i
)
a

2
b
:
a,b
any
real
numbers
}
.
Solution: This is not a subspace. (Although it does have the zero
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 Fall '07
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 Linear Algebra, Algebra, Real Numbers, Vector Space, Complex Numbers, Sets, Complex number, w1 w2

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