In this example, the simplex method moves to an
adjacent vertex
.
We must show that
A
k
+1
satisfies the properties of a working set.
Result
If
A
k
is nonsingular, then
A
k
+1
is nonsingular.
Proof: At the end of iteration
k
, row
s
of
A
k
is replaced by
a
T
t
.
A
k
+1
=
a
T
w
1
a
T
w
2
.
.
.
a
T
t
.
.
.
a
T
w
n
←
row
s
UCSD Center for Computational Mathematics
Slide 5/46, Friday, February 13th, 2009
The vectors
a
w
1
,
a
w
2
, . . . ,
a
w
s

1
,
a
w
s
+1
, . . . ,
a
w
n
are linearly
independent.
⇒
If
A
k
+1
is singular, then row
a
t
must be dependent on
a
w
1
,
a
w
2
, . . . ,
a
w
s

1
,
a
w
s
+1
, . . . ,
a
w
n
, i.e.,
a
t
=
X
i
6
=
s
a
w
i
y
i
for some
{
y
i
}
not all zero
Multiplying by
p
T
k
gives
a
T
t
p
k
=
X
i
6
=
s
(
a
T
w
i
p
k
)
y
i
UCSD Center for Computational Mathematics
Slide 6/46, Friday, February 13th, 2009
From the previous slide:
a
T
t
p
k
=
X
i
6
=
s
(
a
T
w
i
p
k
)
y
i
But
p
k
is defined so that
A
k
p
k
=
e
s
, i.e.,
a
T
w
i
p
k
= 0
for all
i
6
=
s
a
T
w
s
p
k
= 1
⇒
a
T
t
p
k
= 0.
UCSD Center for Computational Mathematics
Slide 7/46, Friday, February 13th, 2009
If
a
T
t
p
k
= 0 then
a
T
t
x
≥
b
t
is not decreasing along
p
k
⇒
a
T
t
x
≥
b
t
cannot be a blocking constraint
⇒
a contradiction.
⇒
A
k
+1
is nonsingular.
UCSD Center for Computational Mathematics
Slide 8/46, Friday, February 13th, 2009