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AS4sol/MATH1111/YKL/1011
THE UNIVERSITY OF HONG KONG
DEPARTMENT OF MATHEMATICS
MATH1111: Linear Algebra
Assignment 4 Suggested Solution
1.
(a) No. As
b
=
1
2
a
+
1
2
c
,
L
(
b
) =
1
2
L
(
a
) +
1
2
L
(
c
) =
1
2
b
+
1
2
a
= (
3
2
5
2
)
T
if
L
is a linear
transformation.
(b)
i.
T
(
a
) =
±
10
12
²
=
13
2
b
+ (
±
1)
c
,
T
(
b
) =
±
±
4
±
4
²
=
±
2
b
.
The matrix representation is
±
13
2
±
2
±
1
0
²
.
ii. The standard matrix representation is
±
±
8 6
±
9 7
²
.
So the eigenvalues are
±
2 and 1, with eigenvectors
±
1
1
²
and
±
2
3
²
.
Set
G
=
³±
1
1
²
;
±
2
3
²´
.
2.
(a)
i. As
A
is row equivalent to
0
@
1 1 2
0 1 1
0 0 0
1
A
, we see that (1 0 1)
T
;
(1 1 3)
T
form
a basis for
c
(
A
).
Also,
N
(
A
) = Span((
±
1
±
1 1)
T
).
Direct checking shows that
0
@
1
0
1
1
A
,
0
@
1
1
3
1
A
,
0
@
±
1
±
1
1
1
A
are linearly independent.
So
c
(
A
) +
N
(
A
) =
R
3
.
It is also easy to see that
c
(
A
)
\
N
(
A
) =
f
0
g
. (One way is to apply the dimension
formula: dim
c
(
A
) = 2, dim
N
(
A
) = 1
)
dim(
c
(
A
)
\
N
(
A
)) = 0.)
Hence
c
(
A
)
²
N
(
A
) =
R
3
.
ii. No. There are matrices
B
such that
R
3
6
=
N
(
B
) +
c
(
B
).
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This note was uploaded on 05/04/2011 for the course MATH 1111 taught by Professor Dr,li during the Spring '10 term at HKU.
 Spring '10
 Dr,Li

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