Lecture 11
Interlude:
Theorem 1.
Det
0
A
(
H
) =
Tr
(
e
AH
)
Proof.
It is enough to see both sides agree when
H
is a standard basis matrix
E
ij
.
Left Hand Side:
Det
0
A
(
E
ij
)
=
d
dt
Det
(
A
+
tE
ij
)

t
=0
=
d
dt
(
DetA
+ (

1)
i
+
j
tA
ij
)

t
=0
(where
A
ij
is the minor of A)
=
(

1)
i
+
j
A
ij
.
Right Hand Side:
(
BE
ij
)
pq
=
∑
n
r
=1
B
pr
(
E
ij
)
rq
=
∑
n
r
=1
B
pr
δ
ir
δ
jq
=
B
pi
δ
jq
∴
Tr
(
BE
ij
)
=
∑
n
p
=1
B
pi
δ
jp
=
B
ji
.
In particular,
Tr
(
e
AE
ij
)
=
e
A
ji
= (

1)
i
+
j
A
ij
.
Lemma 1.
Let
F
:
U
→
R
where
U
⊂
X
and
X
is a normed space. Suppose
F
is differentiable at
a
∈
U
and has a local maximum value at
a
then
F
0
a
(
h
) =
0
for all
h
. (Exercise)
1
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Proof.
Suppose BWOC there exists
x
∈
U
such that
F
0
a
(
x
)
6
= 0.
WLOG we can take
F
0
a
(
x
)
>
0 (if
F
0
a
(
x
)
<
0 look at

x
).
Since
a
is a local maximum, there exists
δ
1
such that
k
y

a
k
< δ
1
⇒
f
(
y
)
≤
f
(
a
)
.
Now let
F
0
a
(
x
) =
s >
0.
Let
ε >
0 such that
ε <
s
k
x
k
. Now since
F
is differentiable at
a
, there exists
δ
2
>
0 such that if
k
h
k
< δ
2
, then

F
(
a
+
h
)

F
(
a
)

F
0
a
(
h
)

< ε
k
h
k
.
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 Summer '10
 Staff
 Logic, Algebra, HMS H3, All wheel drive vehicles

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