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Given a point and an affine subspace
, the projection of
onto
is the point
such that,
One can obviously define other notions of projection but the above is probably the most commonly used in geometry. If then the projected point
is the unique point in
such that the vector
is orthogonal to
. Also is well known that projecting any segment to an affine subspace can only shrink its length. The proofs of these facts are easy to see. But in fact these facts are just corollaries of the following two results.
Given any nonempty closed convex set
|
and,
Let
|
Applying these results to affine spaces (which are nonempty closed convex sets) yields the results mentioned earlier. This projection that maps to
is known as the metric projection. The proofs of these facts are in order.
Proof: First we show that the minimum distance to is indeed obtained. This is easy and the details are as follows. Since
is nonempty there is some point
. Let
denote
. Clearly a closest point to
can only lie in
where
denotes a closed ball of radius
with center
. But
is a closed and bounded set and is therefore compact. It is also nonempty. The function
defined for
as
is a continuous function on a compact set and attains its minimum at some point
.
Next we prove the point where
attains minimum, is unique. If
is another point at which
then in the triangle
, which has two equal sides, the perpendicular from
onto segment
is clearly shorter than the length
. However the base of this perpendicular lies on the segment itself. By convexity of
the entire segment is in
and thus we have found a point in
closer to
than
which is a contradiction.
Next we prove that the metric projection is a contraction.
Proof: Let the points be projected to
and
respectively and assume
. Consider the hyperplanes
through
that are orthogonal to the vector
. See figure below.
The main observation is that must lie on the other side of
as
. If
lies on the same side of
as
then one can always find a point on the segment
closer to
than
which would contradict that
is the closest point to
in
. Similarly
cannot lie on same side of
as
. Thus
is at least the distance between
and
which is
.
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