20 Using Spectral Form of Mathematical Description …
297
It is easy to see that ΛE i 3 = 1 i 3 1. Similarly, E
T
i 1
Λ = 1 i 1 1
T . Further,
(V E i 2 )1 = (V 1)E i 2 = E E i 2 = E i 2 ,
1
T
(V E i 2 ) = [(V E i 2 )
T 1]
T
= [(V E i 2 )1]
T
= [(V 1)E i 2 ]
T
= [E E i 2 ]
T
= E
T
i 2
.
Consequently,
E
T
i 1
P
−1
(V E i 2 )ΛE i 3 = 1 i 3 E
T
i 1
P
−1 E i 2 = C i 1 i 2 C i 3 ,
E
T
i 1
Λ(V E i 2 )P
−1 E i 3 = 1 i 1 E
T
i 2
P
−1 E i 3 = C i 1 C i 2 i 3 ,
E
T
i 1
Λ(V E i 2 )ΛE i 3 = 1 i 1 1
T
(V E i 2 )1 i 3 1 = 1 i 1 1 i 2 1 i 3 = C i 1 C i 2 C i 3 ,
i.e.,
C i 3 i 2 i 1 = C i 1 i 2 i 3 − C i 1 i 2 C i 3 − C i 1 C i 2 i 3 + C i 1 C i 2 C i 3 .
Summing up by i 1 , i 2 , i 3 the products of random values ζ
( j 1 )
i 1
ζ
( j 2 )
i 2
ζ
( j 3 )
i 3
and both the
left-hand side and the right-hand side of the above relation for expansion coefficients,
we can write that
I
∗( j 1 j 2 j 3 )
h
= I
∗( j 3 j 2 j 1 )
h
− I
∗( j 2 j 1 )
h
I
∗( j 3 )
h
− I
∗( j 3 j 2 )
h
I
∗( j 1 )
h
+ I
∗( j 3 )
h
I
∗( j 2 )
h
I
∗( j 1 )
h
.
Next, consider the multiplicity k = 4:
C i 4 i 3 i 2 i 1 = E
T
i 4
P
−1
(V E i 3 )P
−1
(V E i 2 )P
−1 E i 1 ,
C i 1 i 2 i 3 i 4 = E
T
i 1
P
−1
(V E i 2 )P
−1
(V E i 3 )P
−1 E i 4 .
Also using properties of the matrix multiplication and transpose, we have
[E
T
i 4
P
−1
(V E i 3 )P
−1
(V E i 2 )P
−1 E i 1 ]
T
= E
T
i 1
[P
−1
]
T
(V E i 2 )
T
[P
−1
]
T
(V E i 3 )
T
[P
−1
]
T E i 4
= E
T
i 1
(Λ − P
−1
)(V E i 2 )(Λ − P
−1
)(V E i 3 )(Λ − P
−1
)E i 4
= −E
T
i 1
P
−1
(V E i 2 )P
−1
(V E i 3 )P
−1 E i 4 + E
T
i 1
P
−1
(V E i 2 )P
−1
(V E i 3 )ΛE i 4
+ E
T
i 1
P
−1
(V E i 2 )Λ(V E i 3 )P
−1 E i 4 + E
T
i 1
Λ(V E i 2 )P
−1
(V E i 3 )P
−1 E i 4
− E
T
i 1
P
−1
(V E i 2 )Λ(V E i 3 )ΛE i 4 − E
T
i 1
Λ(V E i 2 )P
−1
(V E i 3 )ΛE i 4
− E
T
i 1
Λ(V E i 2 )Λ(V E i 3 )P
−1 E i 4 + E
T
i 1
Λ(V E i 2 )Λ(V E i 3 )ΛE i 4 .
Applying same properties as well as in the case k = 3, we obtain
E
T
i 1
P
−1
(V E i 2 )P
−1
(V E i 3 )ΛE i 4 = 1 i 4 E
T
i 1
P
−1
(V E i 2 )P
−1 E i 3 = C i 1 i 2 i 3 C i 4 ,
E
T
i 1
P
−1
(V E i 2 )Λ(V E i 3 )P
−1 E i 4 = E
T
i 1
P
−1
(V E i 2 )1 · 1
T
(V E i 3 )P
−1 E i 4
297
It is easy to see that ΛE i 3 = 1 i 3 1. Similarly, E
T
i 1
Λ = 1 i 1 1
T . Further,
(V E i 2 )1 = (V 1)E i 2 = E E i 2 = E i 2 ,
1
T
(V E i 2 ) = [(V E i 2 )
T 1]
T
= [(V E i 2 )1]
T
= [(V 1)E i 2 ]
T
= [E E i 2 ]
T
= E
T
i 2
.
Consequently,
E
T
i 1
P
−1
(V E i 2 )ΛE i 3 = 1 i 3 E
T
i 1
P
−1 E i 2 = C i 1 i 2 C i 3 ,
E
T
i 1
Λ(V E i 2 )P
−1 E i 3 = 1 i 1 E
T
i 2
P
−1 E i 3 = C i 1 C i 2 i 3 ,
E
T
i 1
Λ(V E i 2 )ΛE i 3 = 1 i 1 1
T
(V E i 2 )1 i 3 1 = 1 i 1 1 i 2 1 i 3 = C i 1 C i 2 C i 3 ,
i.e.,
C i 3 i 2 i 1 = C i 1 i 2 i 3 − C i 1 i 2 C i 3 − C i 1 C i 2 i 3 + C i 1 C i 2 C i 3 .
Summing up by i 1 , i 2 , i 3 the products of random values ζ
( j 1 )
i 1
ζ
( j 2 )
i 2
ζ
( j 3 )
i 3
and both the
left-hand side and the right-hand side of the above relation for expansion coefficients,
we can write that
I
∗( j 1 j 2 j 3 )
h
= I
∗( j 3 j 2 j 1 )
h
− I
∗( j 2 j 1 )
h
I
∗( j 3 )
h
− I
∗( j 3 j 2 )
h
I
∗( j 1 )
h
+ I
∗( j 3 )
h
I
∗( j 2 )
h
I
∗( j 1 )
h
.
Next, consider the multiplicity k = 4:
C i 4 i 3 i 2 i 1 = E
T
i 4
P
−1
(V E i 3 )P
−1
(V E i 2 )P
−1 E i 1 ,
C i 1 i 2 i 3 i 4 = E
T
i 1
P
−1
(V E i 2 )P
−1
(V E i 3 )P
−1 E i 4 .
Also using properties of the matrix multiplication and transpose, we have
[E
T
i 4
P
−1
(V E i 3 )P
−1
(V E i 2 )P
−1 E i 1 ]
T
= E
T
i 1
[P
−1
]
T
(V E i 2 )
T
[P
−1
]
T
(V E i 3 )
T
[P
−1
]
T E i 4
= E
T
i 1
(Λ − P
−1
)(V E i 2 )(Λ − P
−1
)(V E i 3 )(Λ − P
−1
)E i 4
= −E
T
i 1
P
−1
(V E i 2 )P
−1
(V E i 3 )P
−1 E i 4 + E
T
i 1
P
−1
(V E i 2 )P
−1
(V E i 3 )ΛE i 4
+ E
T
i 1
P
−1
(V E i 2 )Λ(V E i 3 )P
−1 E i 4 + E
T
i 1
Λ(V E i 2 )P
−1
(V E i 3 )P
−1 E i 4
− E
T
i 1
P
−1
(V E i 2 )Λ(V E i 3 )ΛE i 4 − E
T
i 1
Λ(V E i 2 )P
−1
(V E i 3 )ΛE i 4
− E
T
i 1
Λ(V E i 2 )Λ(V E i 3 )P
−1 E i 4 + E
T
i 1
Λ(V E i 2 )Λ(V E i 3 )ΛE i 4 .
Applying same properties as well as in the case k = 3, we obtain
E
T
i 1
P
−1
(V E i 2 )P
−1
(V E i 3 )ΛE i 4 = 1 i 4 E
T
i 1
P
−1
(V E i 2 )P
−1 E i 3 = C i 1 i 2 i 3 C i 4 ,
E
T
i 1
P
−1
(V E i 2 )Λ(V E i 3 )P
−1 E i 4 = E
T
i 1
P
−1
(V E i 2 )1 · 1
T
(V E i 3 )P
−1 E i 4