AS1sol/MATH1111/YKL/1011
THE UNIVERSITY OF HONG KONG
DEPARTMENT OF MATHEMATICS
MATH1111: Linear Algebra
Assignment 1 Suggested Solution
Remark
:
You may ±nd that the solutions (especially the proofs) are rather lengthy. It is because we hope
to explain every step detailedly. After reading each answer, you should dig out the central ideas and work out
the answer again by your words.
1.
(a) Let
x
= (
x y
)
T
2
R
2
. To show that
x
is a linear combination of
x
1
and
x
2
, we need
to prove that
x
=
c
1
x
1
+
c
2
x
2
is solvable for some scalars
c
1
;c
2
, i.e.
±
x
=
c
1
c
+
c
2
a
y
=
c
1
d
+
c
2
b
has a solution (for
c
1
;c
2
). The augmented matrix is
²
c a
x
d b
y
³
:
As
x
2
6
= 0
, at least one of
c
or
d
is nonzero.
Case 1.
c
6
= 0. Then the row operations
1
c
R
1
and
±
dR
1
+
R
2
gives
²
1
a
c
x
c
0
b
±
da
c
y
±
dx
c
³
:
Note that
b
±
da
c
6
= 0, for otherwise,
x
2
= (
a b
)
T
=
a
c
(
c d
)
T
=
a
c
x
1
(contradicting
to the condition
x
1
6
=
±x
2
for any
±
2
R
). Thus the operation (
b
±
da
c
)
±
1
R
2
further
transforms the augmented matrix into
²
1
a
c
x
c
0 1
(
b
±
da
c
)
±
1
(
y
±
dx
c
)
³
:
From it we solve for
c
1
and
c
2
.
Case 2.
d
6
= 0. The proof is similar and the details are left to you.
(b) In case
x
1
is a scalar multiple of
x
2
, there are (plenty of) vectors that cannot be
written as a linear combination of
x
1
and
x
2
. Below is an example.
Choose
u;v
2
R
such that
av
6
=
bu
, which can be done because at least one of
a;b
is
not equal to 0. Take
x
= (
a
+
u b
+
v
)
T
. We prove the following.
Claim:
x
is not a linear combination of
x
1
and
x
2
.
1
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View Full DocumentProof. Suppose not. i.e.
x
=
±x
1
+
²x
2
where
±;²
2
R
. As
x
1
=
cx
2
for some
constant
c
, we have
x
= (
c
+ 1)
x
2
. This implies
a
+
u
= (
c
+ 1)
a
and
b
+
v
= (
c
+ 1)
b:
Or equivalently
ca
=
u
and
cb
=
v
. This implies
cav
=
cbu
; as
av
6
=
bu
, we conclude
c
= 0 and thus
x
1
=
cx
2
= 0
. Contradiction arises.
2. As
A
is nonsingular, we get by multiplying
A
±
1
to both sides of
A
3
=
A
. We have
A
2
=
I
.
[If then, you answer
A
=
±
I
. Well this question is signi±cant to you because
you are
WRONG
. Life is not so simple in matrices.]
To see what
A
can satisfy the equation, we carry out a bruteforce calculation. Let
A
=
±
a b
c d
²
:
Then
A
2
=
±
a
2
+
bc
b
(
a
+
d
)
c
(
a
+
d
)
d
2
+
bc
²
. From
A
2
=
I
, we conclude
a
2
+
bc
= 1
;
b
(
a
+
d
) = 0 =
c
(
a
+
d
)
;
d
2
+
bc
= 1
:
Case 1.
b
= 0. Then
a
2
= 1 =
d
2
, so
a;d
can be
±
1. We have
b
a
d
a
+
d
c
A
0
1
1
2
0
±
1 0
0 1
²
0
1
²
1
0
any
±
1
0
c
²
1
²
0
²
1
1
0
any
±
²
1 0
c
1
²
0
²
1
²
1
²
2
0
±
²
1
0
0
²
1
²
Case 2.
c
= 0. We consider
A
T
. Note that
A
2
=
I
implies (
A
T
)
2
=
I
. Applying Case 1,
we get
A
=
±
±
1 0
0 1
²
;
or
A
=
±
±
1
c
0
²
1
²
where
c
can be any real number.
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 Spring '10
 Dr,Li

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