Page 30 - Computational Colour Science Using MATLAB

LINEAR AND NON-LINEAR TRANSFORMS 17
For a given standard matrix A it is trivial to compute w given x. However, it is
not trivial to compute x given w. This situation corresponds, of course, to the
problem of solving simultaneous equations and in the example given by
Equation (2.3) there are four variables (the entries of the column matrix x) and
three equations. When the number of equations is less than the number of
variables we say that the system is under-determined (a system is said be over-
determined if the number of equations is greater than the number of variables).
A common method for solving for x in Equation (2.3) is to multiply both sides
1
of the equation by the inverse of the standard matrix which we denote A .
However, the inverse of a non-square matrix is not defined and therefore
numerical methods must be employed to compute the pseudoinverse matrix
+
denoted by A .
Recall (Section 2.2) that the product of a matrix and its inverse yields the
identity matrix and therefore we can write
1 1 1
A w ¼ A Ax or x ¼ A w: ð2:4Þ
A solution for x is therefore possible if we can compute the inverse (or
pseudoinverse) of the standard matrix A.
Linear algebra can be used to find mappings between one set of data and
another and this is sometimes called function approximation. Now, the problem
is to find the standard matrix A given examples of data from each of the sets.
Imagine, for example, that we have a set of n camera RGB response values and
we wish to find a linear transform between the corresponding n known XYZ
tristimulus values. Formally, if we define T as the 36n matrix of tristimulus
values and C as the 36n matrix of camera values, then we seek a transformation
3
3
of type T: < !< or explicitly in this case T: C ! T. Thus, we need to find the
coefficients a 11 to a for the following three equations:
33
X ¼ a 11 R þ a 12 G þ a 13 B,
Y ¼ a 21 R þ a 22 G þ a 23 B,
Z ¼ a 31 R þ a 32 G þ a 33 B.

In matrix algebra we need to find the 3 3 standard matrix A where

T ¼ AC. ð2:5Þ
Note that if we considered each row of A separately for the first row we can write
that

X ¼ a 11 R þ a 12 G þ a 13 B,
where X is the X tristimulus value, RGB are the camera values, and a , a and
12
11
a 13 are the coefficients that form the first row of A. If we now consider the n
known samples we can write