Image Processing Reference
In-Depth Information
orthogonal functions that eliminate redundancies in small rectangular image patches
have been constructed as one way to compress image data so that we can store and
transport images efficiently (e.g., JPEG images).
Orthogonal sets are often used as bases; that is, they are used to synthesize as well
as to analyze vector spaces. To illustrate this, we go back to our ordinary space of E 3
and ask ourselves the following: How do we find the coordinates of a point x if we
have three orthogonal vectors e 1 , e 2 , e 3 that we use as a basis? We do this by the so-
called projection operation, which measures the “orthogonal shadow” of the vector
representing the point (which starts at the origin and ends, at the point itself) on the
three basis vectors. More formally, a projection is an operator that equals to identity
if it is repeated more than once. Paraphrased, the resulting vector of a projection does
not change by further projections. The point is projected orthogonally to define the
tips of the three vectors that lie on the basis vectors, which when added vectorially
end up with a result that is identical to the coordinate vector of the point, the one that
goes from the origin to the point.
To teach projection to a computer we should have a more precise mathematical
procedure of making projections than the term “orthogonal shadows”. To do that we
work backwards. We write the vector x as if we knew its coordinates or coefficients,
c 1 , c 2 and c 2 in the basis
{
e i }
, that is
x =
i
c i e i
(3.30)
Now we take the scalar product of the left- as well as the right-hand side of this
equation with one of the bases, e.g., e 1 .
x , e 1 = [
i
c i e i ] , e 1 = c 1 e 1 , e 1 + c 2 e 2 , e 1 + c 3 e 3 , e 1
(3.31)
Since our basis vectors are orthogonal, the scalar products between different bases
will vanish
e 1 , e 2
=
e 1 , e 3
=
e 2 , e 3
=0.Sowehave
x , e 1
= c 1
e 1 , e 1
=
c 1 =
x , e 1
/
e 1 , e 1
(3.32)
But
, which is the square of the length of e 1 , can be computed regardless
of x since e 1 is known. In other words, by using the scalar product operation, we
have achieved projecting x on e 1 since we could find c 1 , which was the unknown
coordinate scaling factor. If we could do it for e 1 , we should be able to do it for the
remaining two coefficients as well. This is done without effort by just replacing the
index 1 with the desired index, that is:
e 1 , e 1
c i =
x , e i
/
e i , e i
(3.33)
Eq. (3.33) is the reason for why the scalar product is often referred to as the projec-
tion operator because, except for the denominator, it is just a scalar product of any
member x of the space, with the known basis vectors
{ e i } i . The denominator does
not spoil the equivalence of the scalar product to projection since the denominator
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