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Winter, 2010
Assignment 4, due
in class
Monday, February 8, 2010
1. Find a basis for each of the row space, column space and null space of the
following matrix
A
over the complex numbers. What is its rank? Express
each row of
A
as a linear combination of the rows in your basis for the
row space; express each of the columns of
A
as a linear combination of the
vectors in your basis for the column space.
A
=
1
2
i
2 + 4
i
1 + 8
i
4 + 14
i
1

i
3 + 2
i
7 +
i
12 + 7
i
23 + 9
i
i
0

2

2 +
i

3 + 2
i
4

1 +
i
10
i
1 + 11
i
4 + 22
i
.
Solution: We perform
R
2
7→
R
2
+ (

1 +
i
)
R
1
, then
R
3
7→
R
3

iR
1
,
then
R
4
7→
R
4

4
R
1
, then
R
1
7→
R
1

2
iR
2
, then
R
3
7→
R
3

2
R
2
, then
R
4
7→
R
4
+ (1 + 7
i
)
R
2
, then
R
1
7→
R
1
+ (

2

4
i
)
R
3
, then
R
2
7→
R
2
+ (

5 +
i
)
R
3
and ﬁnally,
R
4
7→
R
4

2
iR
3
to produce the
RREF matrix
1 0
2
i
1 + 2
i
0
0 1 1

i
3
0
0 0
0
0
1
0 0
0
0
0
.
A basis for the row space is
{
~v
1
,~v
2
,~v
3
}
=
{
(
1 0 2
i
1 + 2
i
0
)
,
(
0 1 1

i
3 0
)
,
(
0 0 0 0 1
)
}
.
A basis for the column space is
{
C
1
,C
2
,C
5
}
=
1
1

i
i
4
,
2
i
3 + 2
i
0

1 +
i
,
4 + 14
i
23 + 9
i

3 + 2
i
4 + 22
i
.
The rank
r
(
A
) is 3. A basis for the null space is

2
i

1 +
i
1
0
0
,

1

2
i

3
0
1
0
.
If the original rows are relabelled
R
1
,...,R
4
, we have
R
1
= 1
~v
1
+ 2
i~v
2
+ (4 + 14
i
)
~v
3
,
R
2
= (1

i
)
~v
1
+ (3 + 2
i
)
~v
2
+ (23 + 9
i
)
~v
3
,
R
3
=
i~v
1
+ 0
~v
2
+ (

3 + 2
i
)
~v
3
and
R
4
= 4
~v
1
+ (

1 +
i
)
~v
2
+ (4 + 22
i
)
~v
3
. If
the original columns are labelled
C
1
,...,C
5
, then of course
C
1
= 1
C
1
+ 0
C
2
+ 0
C
3
and so on, but also
C
3
= 2
iC
1
+ (1

i
)
C
2
and
C
4
= (1 + 2
i
)
C
1
+ 3
C
2
.
2. Let
V
=
Z
4
2
and
W
1
=
Span
1
0
1
0
,
0
1
0
1
and
W
2
=
Span
0
1
1
0
,
1
0
0
1
and be subspaces of
V
. Find a basis for
W
1
+
W
2
and one for
W
1
∩
W
2
.
1
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This note was uploaded on 02/10/2010 for the course MATH math 223 taught by Professor Loveys during the Winter '10 term at McGill.
 Winter '10
 Loveys
 Linear Algebra, Algebra, Complex Numbers

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