UC Berkeley
Department of Electrical Engineering and Computer Science
EECS 227A
Nonlinear and Convex Optimization
Solutions 5
Fall 2009
Reading:
Boyd and Vandenberghe, Chapter 5
Solution 5.1
Note that we have
∇
f
(
x
) =
1
for any
x
∈
R
n
. Since for any feasible point
∇
h
(
x
) = 2
x
6
= 0,
any local minimum will be regular. From the Lagrange multiplier theorem, there exists
λ
*
such that
∇
f
(
x
*
) +
λ
*
∇
h
(
x
*
) = 0
,
or equivalently
1
+ 2
λ
*
x
*
= 0. Using the requirement that
h
(
x
*
) =
k
x
*
k
2

1 = 0, this yields
the two possibilities
(
x
*
i
,λ
*
i
) =
±
(
1
√
n
,

√
n
2
)
.
By using the second order optimality conditions, we have
y
0
(
∇
2
f
(
x
*
) +
λ
*
∇
2
h
(
x
*
)
)
y
≥
0
for all
y
∈
V
(
x
*
) : =
{
y

2(
x
*
)
T
y
= 0
}
. Since
∇
2
f
(
x
*
) = 0 and
∇
2
h
(
x
*
) = 2
I
, the above
condition is equivalent to 2
λ
*
y
T
Iy
≥
0 for all
y
∈
V
(
x
*
), which can hold only for nonnegative
λ
*
. Hence the only point satisfying the ﬁrst and second order conditions for a local minimum
is
x
*
=

1
√
n
1
. Since the constraint set is compact and
f
is continuous, an optimal point
x
*
exists, and the above point must be the optimum.
One geometric interpretation of the Lagrange multiplier conditions is that the gradient
direction
∇
f
(
x
*
) =
1
is parallel with the gradient
∇
h
(
x
*
) = 2
x
*
associated with the constraint
h
(
x
) =
k
x
k
2

1 = 0, which deﬁnes the boundary of a circle.
Solution 5.2
(a) Note that the constraint set for
k
th
problem is
C
k
=
{
x
∈
R
n
:
k
x
k
2
= 1
,e
T
i
x
= 0
,i
= 1
,...,k
}
satisﬁes
C
1
⊇
C
2
⊇
...
⊇
C
n
. Since all these problems have the same objective, we
must have the optimal value satisﬁes
λ
1
≤
λ
2
...
≤
λ
n
(b) Suppose
α
1
,...,α
n
satisﬁes
∑
n
i
=1
α
i
e
i
= 0. By multiplying both sides with
e
T
k
, note
that
e
T
k
e
i
= 0
,
∀
i
6
=
k
and
e
T
k
e
k
= 1, we have