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combination of the blue rows of S is zero:
α r , s c nr + s , J = 0
(10.7)
r∈ Γ
s∈ Λ r
Here 0 is a column vector of zeros, one per blue row. Again, J is the set of columns of C in
the submatrix S .
Column j of the n
γ ) n columns
of S and is bad otherwise. Let G be the indices of the good columns in B and let g =
×
n matrix B is good if it is associated with at least ( 1
|
G
|
.
Then there are g
( 1
γ ) n good columns and b
γn bad columns in B ( g + b = n )
because, if not, g
( 1
γ ) n
1 and the number of columns altogether in S is at most
γ ) n , which is an increasing function of g whose value is less than n 2
γ 2 n 2
gn + b ( 1
when g
1, which is less than the number of columns of S .
Since B has at least g =
( 1
γ ) n
|
G
|≥
( 1
γ ) n good columns and B is γ -nice, any set of up to
rows are linearly independent. In particular, the rows of B indexed by Λ r are linearly
independent. This implies that
γn
α r , s b s , G
= 0
s∈ Λ r
where 0 is a zero column with
|
Λ r |
rows. Thus, there must be a column index t
G such
that
α r , s b s , t
= 0
(10.8)
s∈ Λ r
Let K =
{
j
|
nj + t
J
}
be the columns of S corresponding to the good column of B
with index t . It follows that
|
K
|≥
( 1
γ ) n
.
Let u i = c i , J∩K , the intersection of the i th row of S with columns whose indices are in
K . Similarly, let v i be the intersection of the i th row of A with columns in K . It follows
from the definition of C that u ni + j = b j , t v i .From( 10.7 )wehavethat
α r , s c nr + s , J∩K = 0
r∈ Γ
s∈ Λ r
v r = 0
α r , s b s , t
r
Γ
s
Λ r
However, the rows
|
Γ
|
rows v r constitute a
γn
×|
K
|
submatrix of the γ -nice matrix A
where
|
K
|≥
( 1
γ ) n
. Since its rows are linearly independent, each of the coefficients
s∈ Λ r α r , s b s , t must be zero, contradicting the statement of ( 10.8 ). It follows that C =
A
B is γ 2 -ok.
10.13 Applications of the Borodin-Cook Method
In this section we illustrate the Borodin-Cook method of Section 10.11 by applying it to a
variety of representative problems.
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