Cones II
Contents
9.5. Cones II#
9.5.1. Dual Cones#
Dual cones are defined for finite dimensional inner product spaces.
Dual cones technically belong to the dual space
Recall that the
dual space
(Dual cone)
Let
Let
is called the dual cone of
In the Euclidean space
Geometric interpretation
For a vector
, the set is a halfspace passing through origin. is the normal vector of the halfspace along (in the direction of) the halfspace.If
belongs to the dual cone of , then for every , we have .Thus, the set
is contained in the halfspace .In particular, if
is a cone, then it will also touch the boundary of the half space as contains the origin.
9.5.1.1. Properties#
Dual cone is a cone.
Proof. Let
Thus, for some
Thus, for every
Dual cone is convex.
Proof. Let
Then for an arbitrary
Thus,
We note that dual cone is a convex cone even if the original set
(Containment reversal in dual cone)
Let
The dual cone of the subset contains the dual cone of the superset.
Proof. Let
Thus,
(Closedness)
A dual cone is a closed set.
Proof. The dual cone of a set
Fix a
The set
We can now see that
Thus,
(Interior of dual cone)
The interior of the dual cone
Proof. Let
Let
Since
Now, let
for all
Hence,
(Non-empty interior implies pointed dual cone)
If
Proof. Let
Thus,
(Dual cone of a subspace)
The dual cone of a subspace
More precisely,
Proof. Let
Let us now assume that there is a vector
Then, there exists
Thus,
Thus,
9.5.1.2. Self Dual Cones#
(Self dual cone)
A cone
By equality, we mean that the dual cone
(Nonnegative orthant)
The non-negative orthant
Let
Now, for some
Now consider the vector
Thus,
(Positive semidefinite cone)
The positive semi-definite cone
Let
Choose an arbitrary
Express
Since
Then,
But since
Now, suppose
Thus,
This completes the proof that
9.5.2. Polar Cones#
(Polar cone)
Let
Let
We note that polar cones are just the negative of dual cones. Thus, they exhibit similar properties as dual cones.
(Polar cone of a ray)
Let
Let
.Then for every
, we have .Equivalently,
since .Hence
Also note that
is closed and convex.We shall show later in polar cone theorem that
Hence the polar cone of the set
is
.
9.5.2.1. Properties#
Polar cone is a cone.
Proof. Let
Thus, for some
Thus, for every
Polar cone is convex.
Proof. Let
Then for an arbitrary
Thus,
We note that polar cone is a convex cone even if the original set
(Containment reversal in polar cone)
Let
The polar cone of the subset contains the polar cone of the superset.
Proof. Let
Thus,
(Closedness)
A polar cone is a closed set.
Proof. The polar cone of a set
Fix a
The set
We can now see that
Thus,
(Interior of polar cone)
The interior of the polar cone
Proof. Let
Let
Since
Now, let
for all
Hence,
(Non-empty interior implies pointed polar cone)
If
Proof. Let
Thus,
(Polar cone of a subspace)
The polar cone of a subspace
More precisely,
Proof. Let
Let us now assume that there is a vector
Then, there exists
Thus,
Thus,
(Polar cone of a null space)
Let
Recall from linear algebra that
Hence by Theorem 9.61,
We can verify this result easily.
Let
.Then there exists
such that .For every
, we have .Hence
Hence
.
(Polar cone and closure)
For any nonempty set
Proof. We first show that
We have
.Hence by Property 9.24,
We now show that
Let
.Then for every
, we have .Let
.There exists a sequence
of such that .But
for every .Hence, taking the limit, we have
.Hence for every
, we have .Hence
.Hence
.
(Polar cone and convex hull)
For any nonempty set
Proof. We first show that
We have
.Hence by Property 9.24,
We now show that
Let
.Then for every
, we have .Let
.Then there exist
and with such thatThen
But
since for every .Hence
.Hence for every
, we have .Hence
.Hence
.
(Polar cone and conic hull)
For any nonempty set
Proof. We first show that
We have
.Hence by Property 9.24,
We now show that
Let
.Then for every
, we have .Let
.Then there exist
and such thatThen
But
since for every .Hence
.Hence for every
, we have .Hence
.Hence
.
(Polar cone theorem)
For any nonempty cone
In particular, if
Proof. First we assume that
Pick any
.By definition, we have
for every .Hence
.Hence
.Now choose any
.Since
is nonempty, closed and convex, hence by projection theorem (Theorem 10.7), there exists a unique projection of on , denoted by that satisfiesSince
is a cone, hence .Since
is a cone and , hence .By putting
, we getBy putting
, we getTogether, we have
Putting this back into the projection inequality, we get
Hence
.Since
, hence .We also have
.Adding these two, we get
This means that
It follows that
.Hence
.Hence
.
We have so far shown that if
Now consider the case where
Then
is a closed convex cone.By previous argument
But
Hence
9.5.3. Normal Cones#
(Normal vector)
Let
(Normal vector)
Let
Let
Then,
Note that
(Normal cone)
The set of all vectors normal to a set
We customarily define
A normal cone is always a convex cone.
Proof. Let
For any
Thus,
Assume
But then for any
Thus,
Now, let
since sum of two nonpositive quantities is nonpositive.
Thus,
Combining these two observations,
A normal cone is closed.
Specifically, if
Proof. For some fixed
Note that
Let
Then, for every
, .Thus, for every
, .Thus,
.Thus,
.
Going in the opposite direction:
Let
.Then, for every
, .Thus, for every
, .Thus,
is a normal vector to at .Thus,
.Thus,
.
Combining, we get:
Now, since
Since each half space is convex and intersection of
convex sets is convex, hence, as a bonus, this proof also
shows that
(Normal cone of entire space)
Let
Let
.Then we must have
This is equivalent to
The only vector that satisfies this inequality is
.Hence
.
(Normal cone of unit ball)
Proof. The unit ball at origin is given by:
Consider
Therefore, for any