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4.1 Semi-Convergence for SIRT Methods 35
By using (4.1) we can then write
k−1
(I − λB) j =
j=0
=
k−1 T T
V V − λV FV j
k−1 T
= V (I − λF)V j
j=0
j=0
k−1 j T
V (I − λF) V ⎛
k−1
= V ⎝ (I − λF) j
⎞
⎠ V T
j=0
= V EkV T ,
where the i’th diagonal element of Ek is
k−1
(1 − λσ 2 i ) j = 1 + (1 − λσ 2 i ) + (1 − λσ 2 i ) 2 + . . . + (1 − λσ 2 i ) k−1
j=0
j=0
= 1 − (1 − λσ2 i )k
1 − (1 − λσ2 i ) = 1 − (1 − λσ2 i )k
λσ2 i
where the formula for geometric series is used to obtain the last result. The
matrix Ek then has the following form:
Ek = diag
1 − (1 − λσ 2 1 )k
λσ 2 1
Assuming that x0 = 0 we can then write x k as
, . . . , 1 − (1 − λσ2 p )k
λσ2
, 0, . . .,0 .
p
x k = V (λEk)V T c = V (λEk)V T A T Mb (4.2)
= V (λEk)V T V Σ T U T M 1
2( ¯b + δb)
p 2
= 1 − (1 − λσi ) k uT i
i=1
,
M 1
2 ( ¯ b + δb)
where ui and vi are the columns of U and V respectively and ϕ k i = 1−(1−λσ2 i )k
for i = 1, 2, . . ., p are the filter factors [20, p. 138].
The minimum-norm solution to the weighted least squares problem with the
noise-free right-hand side ¯x = argminAx − ¯ bM can, using SVD, be written as
where
σi
¯x = V EΣ T U T M 1
2¯b, (4.3)
1
E = diag
σ2 ,
1
1
σ2 , . . . ,
2
1
σ2
, 0, . . .,0 .
p
vi,