C&O 350 Linear Optimization – Fall 2009 – Solutions to Assignment 1
Question #
Max. marks
Part marks
1
6
2
2
2
2
8
2
2
2
2
3
10
4
5
5
12
3
3
3
3
Total
41
1. (Linear Algebra Review)
6 marks = [2+2+2]
Solution:
(a)
S
is a linearly independent set if and only if for all scalars
c
1
, . . . , c
n
, the equation
∑
n
i
=1
c
i
v
(
i
)
= 0
implies that
c
1
=
c
2
=
· · ·
=
c
n
= 0
.
(b) Since the set is linearly dependent, there exist scalars
c
1
, . . . , c
n
that are not all zeros such that
∑
n
i
=1
c
i
v
(
i
)
= 0.
Pick
k
∈ {
1
, . . . , n
}
such that
c
k
6
= 0.
Then, we have

c
k
v
k
=
∑
i
∈{
1
,...,n
}
,i
6
=
k
c
i
v
(
i
)
.
Thus,
v
k
can be written as

∑
i
∈{
1
,...,n
}
,i
6
=
k
c
i
c
k
v
(
i
)
, a linear combinations of other vectors in
S
.
(c) The given set is
not
a subspace of
R
2
. Observe that both the vectors
v
= [

1
,
1]
T
and
w
= [1
,
1]
T
are in the given set, but
v
+
w
= [0
,
1]
T
is not in the set.
2. (Linear Algebra Review)
[8 marks = 2+2+2+2]
Solution:
(a) TRUE. rank(
A
)
< n
implies that the column vectors of
A
are linearly dependent. Therefore, the zero
vector can be written as a nontrivial linear combination of the column vectors of
A
. Hence, there exist
scalars
c
1
, . . . , c
n
that are not all zeros such that
Ac
= 0, where
c
denotes the column vector (
c
1
, . . . , c
n
)
T
.
(b) FALSE. Notice that a system
Ax
=
b
may have a unique solution if rank(
A
) =
n
. Therefore, we can
construct a counter example that has rank(
A
) =
n
and
n < m
.
One such example is:
A
=
1
3
and
b
=
1
3
. Then the unique solution to
Ax
=
b
is
x
= 1.
(c) FALSE. In general, if we start with a feasible system of linear inequalities, and then we change the
righthandside, then the resulting system may be infeasible.
A counter example:
A
=
1

1
, b
=
2
0
and
b
0
=
0

2
.
Although
Ax
≤
b
has a solution (any
x
∈
[0
,
2] would do),
Ax
≤
b
0
implies that
x
≤
0 and

x
≤ 
2, which is obviously infeasible.
(d) TRUE. Since
A
is square and
Ax
=
b
has a unique solution, we know that
A
must be nonsingular,
and hence so is
A
T
. Thus, (
A
T
)

1
exists. Then
A
T
x
=
b
0
⇐⇒
(
A
T
)

1
(
A
T
x
) = (
A
T
)

1
(
b
0
)
⇐⇒
x
=
(
A
T
)

1
b
0
, and so
x
= (
A
T
)

1
b
0
is the unique solution to
A
T
x
=
b
0
.
3. (LP Problem formulation)
1
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[10 marks]
3 for correct definition of decision variables
2 for objective function (1 for correct data, 1 for “min”)
4 for the constraints
1 for nonnegativity constraints
Solution:
Decision Variables:
For
i
= 1
,
2
,
3
,
4
,
5, we define the decision variables
x
i
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 Spring '09
 cheriyan
 Linear Algebra, Algebra, Scalar, Optimization, Debt, objective function, Constraint

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