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only lower or upper triangular elements havetobeconsidered) in the six unknown
components of
L
.Bysolving these equations in a certain order, it is possible to have
only one unknown in each equation.
Consider the lower triangular portion of each matrixinEq. (2.16) (the upper
triangular portionwoulddo as well).Byequating the elements in the first column,
starting with the first row and proceeding downward, wecan compute
L
11
,
L
21
, and
L
31
in thatorder:
A
11
L
11
A
11
=
L
11
=
A
21
=
L
11
L
21
L
21
=
A
21
/
L
11
A
31
=
L
11
L
31
L
31
=
A
31
/
L
11
The second column,starting with second row, yields
L
22
and
L
32
:
A
22
−
L
21
+
L
22
L
21
A
22
=
L
22
=
A
32
=
L
21
L
31
+
L
22
L
32
L
32
=
(
A
32
−
L
21
L
31
)
/
L
22
Finally the third column,third row gives us
L
33
:
A
33
L
31
+
L
32
+
L
33
L
31
−
L
32
A
33
=
L
33
=
−
n
matrix. Weobservethat a typical
element in the lower triangular portion of
LL
T
is of the form
Wecan now extrapolate the results foran
n
×
j
(
LL
T
)
i j
=
L
i
1
L
j
1
+
L
i
2
L
j
2
+···+
L
i j
L
jj
=
L
ik
L
jk
,
i
≥
j
k
=
1
Equating thisterm to the corresponding elementof
A
yields
j
A
i j
=
L
ik
L
jk
,
i
=
j
,
j
+
1
,...,
n
,
j
=
1
,
2
,...,
n
(2.17)
k
=
1
The rangeofindices shown limits the elements to the lower triangular part.For the
first column (
j
=
1), weobtain fromEq. (2.17)
L
11
=
A
11
L
i
1
=
A
i
1
/
L
11
,
i
=
2
,
3
,...,
n
(2.18)
Proceeding to other columns, weobservethat the unknown in Eq. (2.17) is
L
i j
(the
other elements of
L
appearing in the equation have already been computed). Taking
the term containing
L
i j
outside the summationinEq. (2.17), weobtain
j
−
1
A
i j
=
L
ik
L
jk
+
L
i j
L
jj
k
=
1
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