2 TETRACHROMATIC
METAMERISM
Let w, x, y and z be four linearly independent, N-
vectors of samples of the spectral responses, at a set
of wavelengths λ
1
,...λ
N
, of four photodetectors; also,
let the (corresponding samples of the) light spectrum
be given by s. Denote the ”colour” response of the
photodetector set by c = [c
w
,c
x
,c
y
,c
z
]. Assume then
that the response to a light beam falling on four such,
nearly placed, photodetectors is given by
c
T
=
w
x
y
z
s
T
=: Ms
T
or
c
w
c
x
c
y
c
z
=
w
1
... w
N
x
1
... x
N
y
1
... y
N
z
1
... z
N
s
1
.
..
s
N
This provides a linear transformation R
N
→ R
4
,
s 7→ r, that reduces the dimensionality from N to 4. M
has full rank and its entries are nonnegative and, typi-
cally, positive. Thus, for a nonnegative s, c is nonneg-
ative: c ∈ R
4+
. The kernel K of this transformation
is given by the set of vectors k for which
Mk =
w
x
y
z
k =
0
0
0
0
Thus, K, the set of the metameric blacks, is the
space of vectors orthogonal to (each element of) the
subspace L := span{w,x,y,z} = {aM : a ∈ R
4
},
which is isomorphic to R
4
. Also, any two spectra
1
s and
2
s such that
1
s −
2
s ∈ L
⊥
= K, produce
the same colour response c = [c
w
,c
x
,c
y
,c
z
]. L has
dimension 4 and L
⊥
has dimension N − 4; also,
MM
T
: R
4
→ R
4
is invertible. K contains ”spec-
tra” (we might call them virtual spectra) that are
neither nonnegative nor nonpositive
5
. The cosets
s + K := {s + k : s ∈ R
N
,k ∈ K} provide a partition
of R
N
. In a decomposition s = f + k, f ∈ L, k ∈ K,
which is unique, f is called a fundamental metamer
and k is called a metameric black. The spectra in the
coset f + K are said to be metameric and are mapped
by M to the same colour point c ∈ R
4
; only the
nonnegative spectra in such coset are realizable, the
remaining are merely virtual.
5
The spectra in R
N
that are nonnegative are those in the
wedge or ”2
N
-tant” [+, +, ... +]:= R
N+
2.1 A Basis for K
Calling the colour point [0, 0, 0, 0] black, then K
is the set of spectra that ”evoke” the colour black;
call them metameric blacks. Since the components
of M are nonnegative, the only nonnegative spec-
trum that is a metameric black is the 0 spectrum; all
other metameric black spectra include both positive
and negative components.
Cohen’s method (Cohen and Kappauf, 1982),
based on CIE data, consists of finding f as f =
[M
T
(MM
T
)
−1
M]s and then writting k = s − f.
We derive a basis for K of narrow-band spectra in
a 4-step process where 4 triangular, sparse matrices of
row vectors of local support are derived. In the first
matrix
1
A you have a basis for the orthogonal comple-
ment of span{w}, in the second one
2
A, a basis for the
orthogonal complement of span{w, x}, then, in
3
A, a
basis for span{w,x,y}
⊥
and finally, in
4
A, a basis for
span{w,x,y,z}
⊥
. We assume that the components of
w, x, y and z, are positive so that the matrix
1
A below
is computable and also that each of the matrices
2
A,
3
A and
4
A, as defined below, are computable.
Let
1
A be the N × (N − 1) matrix with i
th
row of
the form [0, ..., 0, 1, −w
i
/w
i+1
,0,...,0]; thus,
1
M has a
diagonal of 1’s. Clearly, each row of
1
A is orthogonal
to w and, since linearly independent, they provide a
basis for span{w}
⊥
.
Let each row of
2
A result from linearly combining
each pair of consecutive rows of from
1
A. In this way,
each row is still orthogonal span{w}
⊥
and, by using
appropriate weights in the combination, you can make
it also orthogonal to span{x}
⊥
. In fact, let the i
th
row
of the N × (N − 2) matrix
2
A be given by
[0,...,0,1,m
i,i+1
+ β
i
m
i+1,i+1
,β
i
m
i+1,1+2
,0,...,0]
where the m’s are the components of
1
A, and
β
i
= −
x
i
+m
i,i+1
x
i+1
m
i+1,i+1
x
i+1
+m
i+1,i+2
x
i+2
;
again, the diagonal of
2
A is a diagonal of 1’s. Like-
wise, by making sure a certain linear combination of
each two consecutive rows in
2
A is orthogonal to y,
you get the N × (N − 3)-matrix
3
A with i
th
row of the
form
[0,...,0,1,m
i,i+1
+ βm
i+1,i+1
,m
i,i+2
+βm
i+1,i+2
,βm
i+1,i+3
,0,...,0]
where the m’s are now the components of
2
A and
β
i
= −
y
i
+y
i+1
m
i,i+1
+y
i+2
m
i,i+2
y
i+1
m
i+1,i+1
+y
i+2
m
i+1,i+2
+y
i+3
m
i+1,i+3
.
The diagonal of
3
A is a diagonal of ones. Finally, a
linear combination of each two consecutive rows in
3
A that is orthogonal to z, provides the N × (N − 4)-
matrix
4
A with i
th
row
[0,...,0,1,m
i,i+1
+ βm
i+1,i+1
,m
i,i+2
+βm
i+1,i+2
,m
i,i+3
+ βm
i+1,i+3
,βm
i+1,1+4
,0,
...,0],
where the m’s are now the components of M
3
and
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