CO 350 Linear Optimization (Spring 2009)
Solution to Assignment 1
[10]
1.
(Exercise 2 in page 22 of the course notes.)
The dimension of
R
3
coincides with the number of vectors we have (3). Therefore,
{
A
1
, A
2
, A
3
}
is
a basis for
R
3
iff it is linearly independent. Let
α
1
, α
2
, α
3
∈
R
. Then
α
1
A
1
+
α
2
A
2
+
α
3
A
3
= 0
implies
α
1
+
α
2
=
α
3
,
2
α
2
=
α
3
,
α
1
+
α
2
=

α
3
.
First and the last equations imply
α
3
= 0, then the second equation implies
α
2
= 0, finally the first or
the last equation gives
α
1
= 0. Therefore,
{
A
1
, A
2
, A
3
}
is a basis for
R
3
.
A
4
=
3
4
A
1

1
4
A
2

1
2
A
3
,
A
5
=

1
2
A
1
+
1
2
A
2
,
A
6
=
1
4
A
1
+
1
4
A
2
+
1
2
A
3
.
[10]
2.
(Exercise 3 in page 22 of the course notes.)
Suppose to the contrary that
{
A
1
, A
2
, . . . , A
k
, A
k
+1
}
is linearly dependent.
Then there are real
numbers
α
1
, α
2
, . . . , α
k
, α
k
+1
, at least one of which is nonzero, such that
α
1
A
1
+
α
2
A
2
+
· · ·
+
α
k
A
k
+
α
k
+1
A
k
+1
= 0
.
(1)
Case 1:
α
k
+1
= 0.
Then
α
1
A
1
+
α
2
A
2
+
· · ·
+
α
k
A
k
= 0 and at least one of the coefficients,
α
i
, is nonzero.
But this
contradicts the linear independence of the set
{
A
1
, A
2
, . . . , A
k
}
.
Case 2:
α
k
+1
6
= 0.
In this case, we can solve (1) for
A
k
+1
:
A
k
+1
=
k
X
i
=1

α
i
α
k
+1
A
i
.
Now this contradicts our assumption that
A
k
+1
is not in the subspace generated by the vectors
A
1
, A
2
, . . . , A
k
.
Hence in both possible cases, we encountered a contradiction and the proof is complete.
[10]
3.
(Exercise 4 in page 22 of the course notes.)
Consider any subset
S
of linearly independent vectors in
R
m
. If
S
is not a basis, then

S

=
k < m
.
Therefore, there exists a vector
A
k
+1
∈
R
m
such that
A
k
+1
is not in the subspace spanned by vectors in
S
. Now, applying the statement proved in the solution of the previous exercise, we see that
S
∪{
A
k
+1
}
is linearly independent. We can apply this argument (
m

k
) times to reach a basis. Therefore, given
a linearly independent subset of vectors (say
k
of them) in
R
m
, we can always find
m

k
vectors to
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 Fall '07
 S.Furino,B.Guenin
 Linear Algebra, Aij xj

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