MATH 425b TAKEHOME FINAL EXAM SOLUTIONS
SPRING 2008
Prof. Alexander
(1)(a) Let
γ
(
t
) =
p
k
+
t
(
p
0
k

p
k
)
,t
∈
[0
,
1], which traces a line from
p
k
to
p
0
k
. Let
σ
t
=
[
p
0
···
p
k

1
γ
(
t
)] be the corresponding simplex. Since these points are on opposite sides on
H
, there must be a value
t
0
where
γ
(
t
0
)
∈
H
. The mapping
σ
0
is given by
σ
0
(
x
) =
p
0
+
k
X
m
=1
x
m
(
p
m

p
0
)
,
so
∂
(
σ
0
)
i
∂x
j
=
∂
∂x
j
k
X
m
=1
x
m
(
p
j

p
0
)
i
= (
p
j

p
0
)
i
,
which means the
j
th column of the derivative matrix
σ
0
0
(
x
) is
p
j

p
0
, for all
x
. For
σ
t
the
derivative matrix is the same except the last column is
γ
(
t
)

p
0
instead of
p
k

p
0
. Let
H
0
be the translate of
H
which passes through the origin; since
p
0
,...,p
k

1
,γ
(
t
0
) all lie in
H
, the
k
vectors
p
1

p
0
,...,p
k

1

p
0
,γ
(
t
0
)

p
0
all lie in the subspace
H
0
. Since
H
0
is
(
k

1)dimensional, these
k
vectors (the columns of
σ
0
t
0
(
x
)) must be linearly dependent.
This means
σ
0
t
0
(
x
) has determinant 0, that is,
J
(
t
0
)(
x
) = 0, for all
x
. Now each entry of the
last column of
σ
0
t
(
x
) has the form
a
+
bt
with
a,b
∈
R
, so expanding the determinant around
this last column shows that
J
(
t
) also has the form
c
+
dt
. Since
σ
0
is invertible, we have
J
(0)
6
= 0, while
J
(
t
0
) = 0, so we must have
d
6
= 0. Since
J
(
t
0
) = 0 this means
J
(0) and
J
(1)
must have opposite sign.