The orthogonal projection of a vector
onto a given subspace
is the vector
that is closest to
.
Before explaining orthogonal projections, we are going to revise some important concepts.
Let
be a vector space. Remember that
two vectors
and
belonging to
are orthogonal when their inner
product is
zero:
Let
be a subspace of
.
The orthogonal complement
of
,
denoted by
,
is the unique subspace satisfying
The two subspaces
and
are complementary
subspaces, which means
that
where
denotes a direct sum. By the
properties of direct sums, any vector
can be uniquely written
as
where
and
.
We can now define orthogonal projections.
Definition
Let
be a linear space. Let
be a subspace of
and
its orthogonal complement. Let
with its unique
decomposition
in
which
and
.
Then, the vector
is called the orthogonal projection of
onto
and it is denoted by
.
Thus, the orthogonal projection is a special case of the so-called
oblique projection, which is
defined as above, but without the requirement that the complementary subspace
of
be an orthogonal complement.
Example
Let
be the space of
column vectors.
Define
Its
orthogonal complement
is
as
we can easily verify by checking that the vector spanning
is orthogonal to the two vectors spanning
.
Now, consider the
vector
Then,
The distance between two vectors is measured by the norm of their difference.
It turns out that
is the vector of
that is closest to
.
Proposition
Let
be a
finite-dimensional
vector space. Let
be a subspace of
.
Then,
for
any
.
Sincewhere
,
the vector
belongs to
and, as a consequence, is orthogonal to any vector belonging to
,
including the vector
.
Therefore,
where
in step
we have used Pythagoras' theorem.
By taking the square root of both sides, we obtain the stated result.
Suppose that
is the space of
complex vectors
and
is a subspace of
.
By the results demonstrated in the lecture on
projection matrices (that are
valid for oblique projections and, hence, for the special case of orthogonal
projections), there exists a projection matrix
such
that
for
any
.
The projection matrix
iswhere:
is any matrix whose columns form a basis for
;
is any matrix whose columns form a basis for
.
In the case of orthogonal projections, the formula above becomes simpler.
Proposition
Let
be the space of complex
vectors. Let
be a subspace of
.
Let
be a matrix whose columns form a basis for
.
Denote by
the conjugate transpose of
.
Then, the matrix
is the projection matrix such that
for
any
.
We choose the columns of
in such a way that they form an
orthonormal basis for
.
As a consequence, as explained in the lecture on unitary matrices (see the
section on non-square matrices with
orthonormal columns), we
have
where
denotes the conjugate transpose of
.
Moreover, since the columns of
are orthogonal to the columns of
,
we
have
and
The
columns of
are linearly independent since they form a basis. Hence,
for any
,
which implies that
for any
.
Thus,
is full-rank (hence invertible). We use these results to derive the following
equality:
which
implies, by the definition of
inverse matrix,
that
Thus,
When we confine our attention to real vectors, conjugate transposition becomes
simple transposition and the formula for the projection matrix
becomeswhich
might be familiar to those of us that have previously dealt with
linear
regressions and the
OLS
estimator.
When the columns of the matrix
are orthonormal, we have a further simplification:
and
Denote by
the columns of
.
Then, for any
,
we
have
which
is the formula for projections on orthonormal sets that we have already
encountered in the lectures on the
Gram-Schmidt process and
on the QR decomposition.
Please cite as:
Taboga, Marco (2021). "Orthogonal projection", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/orthogonal-projection.
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