Homework 3 Foundations of Computational Math 1 Fall
2010
The solutions will be posted on Wednesday, 9/22/09
Problem 3.1
Suppose
A
∈
R
n
×
n
is a nonsymmetric nonsingular diagonally dominant matrix with the
following nonzero pattern (shown for
n
= 6)
* * * * * *
* *
0 0 0 0
*
0
*
0 0 0
*
0 0
*
0 0
*
0 0 0
*
0
*
0 0 0 0
*
It is known that a diagonally dominant (row or column dominant) matrix has an
LU
factor
ization and it can be computed stably without pivoting.
3.1.a
. Describe an algorithm that solves
Ax
=
b
in a stable manner as eﬃciently as
possible.
3.1.b
. Given that the number of operations in the algorithm is of the form
Cn
k
+
O
(
n
k

1
), where
C
is a constant independent of
n
and
k >
0, what are
C
and
k
?
Solution:
Since the matrix is assumed to be diagonally dominant, no pivoting is required for stabil
ity and existence as was discussed in class and the text. Suppose we simply apply Gaussian
elimination without pivoting. After one step of
LU
factorization the
n

1
×
n

1 matrix to
be factored on steps 2 to n is dense and
2
3
n
3
+
O
(
n
2
) computations are needed to complete
the factorization. That is, the linear combination of the ﬁrst row with each row 2
≤
i
≤
6
replaces all of the 0 elements with nonzero elements:
M

1
1
A
=
* * * * * *
0
* * * * *
0
* * * * *
0
* * * * *
0
* * * * *
0
* * * * *
These new nonzeros must be eliminated on subsequent steps.
Since the matrix is diagonally dominant any permuted
PAP
T
, i.e., using the same gen
eral permutation on the left and right, keeps diagonal elements on the diagonal but changes
1
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View Full Documentthe ordering. Oﬀdiagonal elements stay in their original rows and columns but occupy dif
ferent positions. Therefore,
PAP
T
is also diagonally dominant and can be factored without
pivoting.
Now consider step one of the
LU
factorization where we interchange rows and columns
to choose the pivot element. We do not need to do so for stability or existence as long as we
choose a pivot element from the diagonal elements of the matrix, i.e.,
P
i
=
Q
i
.
Choose
P
1
=
(
e
n
e
2
... e
n

1
e
1
)
=
Q
1
i.e., interchange rows
n
and 1 and columns
n
and 1 thereby choosing
α
nn
as the pivot for
step one. For example, with
n
= 6 the matrix structure for the permuted matrix is
P
1
AP
T
1
=
*
0 0 0 0
*
0
*
0 0 0
*
0 0
*
0 0
*
0 0 0
*
0
*
0 0 0 0
* *
* * * * * *
So
M

1
1
only needs to combine row 1 with row 6 (in general with row
n
) to produce a ﬁrst
column of 0 below the diagonal, i.e.,
M

1
1
P
1
AP
T
1
=
*
0 0 0 0
*
0
*
0 0 0
*
0 0
*
0 0
*
0 0 0
*
0
*
0 0 0 0
* *
0
* * * * *
No further pivoting is needed and each step combines row
i
with row
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 Spring '11
 Gallivan
 Determinant, Matrices, lower triangular

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