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Chapter 04.07
LU Decomposition
After reading this chapter, you should be able to:
1.
identify when LU decomposition is numerically more efficient than Gaussian
elimination,
2.
decompose a nonsingular matrix into LU, and
3.
show how LU decomposition is used to find the inverse of a matrix.
I hear about LU decomposition used as a method to solve a set of simultaneous linear
equations.
What is it?
We already studied two numerical methods of finding the solution to simultaneous linear
equations – Naïve Gauss elimination and Gaussian elimination with partial pivoting.
Then,
why do we need to learn another method?
To appreciate why LU decomposition could be a
better choice than the Gauss elimination techniques in some cases, let us discuss first what
LU decomposition is about.
For a nonsingular matrix
[
]
A
on which one can successfully conduct the Naïve Gauss
elimination forward elimination steps, one can always write it as
[
]
[
] [
]
U
L
A
=
where
[
]
L
= Lower triangular matrix
[
]
U
= Upper triangular matrix
Then if one is solving a set of equations
]
[
]
][
[
C
X
A
=
,
then
[
] [
] [
]
[
]
C
X
U
L
=
as
[
] [
]
(
29
U
L
A
]
[
=
Multiplying both sides by
[
]
1

L
,
[
]
[
] [
] [
]
[
]
[
]
C
L
X
U
L
L
1
1


=
[
] [
] [
]
X
U
I
=
[
]
[
]
C
L
1

as
[
]
[
]
(
29
]
[
1
I
L
L
=

[
] [
]
[
]
[
]
C
L
X
U
1

=
as
[
] [
]
(
29
]
[
U
U
I
=
Let
[
]
[
]
[
]
Z
C
L
=

1
04.07.1
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Chapter 04.07
then
[
] [
]
[
]
C
Z
L
=
(1)
and
[
] [
]
[
]
Z
X
U
=
(2)
So we can solve Equation (1) first for
]
[
Z
by using forward substitution and then use
Equation (2) to calculate the solution vector
[
]
X
by back substitution.
This is all exciting but LU decomposition looks more complicated than Gaussian
elimination.
Do we use LU decomposition because it is computationally more
efficient than Gaussian elimination to solve a set of n equations given by [
A
]
[
X
]=[
C
]?
For a square matrix
]
[
A
of
n
n
×
size, the computational time
1
DE
CT

to decompose the
]
[
A
matrix to
]
][
[
U
L
form is given by
DE
CT

=

+
3
20
4
3
8
2
3
n
n
n
T
,
where
T =
clock cycle time
2
.
The computational time
FS
CT

to solve by forward substitution
[
] [
]
[
]
C
Z
L
=
is given by
FS
CT

=
(
29
n
n
T
4
4
2

The computational time
BS
CT

to solve by back substitution
[
] [
]
[
]
Z
X
U
=
is given by
BS
CT

=
(
29
n
n
T
12
4
2
+
So, the total computational time to solve a set of equations by LU decomposition is
LU
CT

=
DE
CT

+
FS
CT

+
BS
CT

=

+
3
20
4
3
8
2
3
n
n
n
T
+
(
29
n
n
T
4
4
2

+
(
29
n
n
T
12
4
2
+
=
+
+
3
4
12
3
8
2
3
n
n
n
T
Now let us look at the computational time taken by Gaussian elimination.
The computational
time
FE
CT

for the forward elimination part,
FE
CT

=

+
3
32
8
3
8
2
3
n
n
n
T
,
and the computational time
BS
CT

for the back substitution part is
1
The time is calculated by first separately calculating the number of additions, subtractions,
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 Spring '07
 Gallivan

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