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Lecture 10
[Interlude
:
Det
0
A
(
H
) = (Det
A
) Tr(
A

1
H
) if
A
∈
M
n
(
R
) is invertible and
H
∈
M
n
(
R
).
Proof.
Write
A
=
±
a
1
...
a
n
²
with columns
a
1
,...,a
n
∈
R
n
. In particular,
A
=
±
e
1
...
e
n
²
for the standard basis
e
1
,...,e
n
of
R
n
. Since Det
A
is multilinear in
a
1
,...,a
n
,
Det is diﬀerentiable and
Det
0
A
(
H
) =
n
X
j
=1
Det
±
a
1
...
h
j
...
a
n
²
.
Special Case
:
A
=
I
.
Det
0
I
(
H
) =
n
X
j
=1
Det
±
e
1
...
h
j
...
e
n
²
=
n
X
j
=1
h
jj
= Tr
H.
Any invertible
A
: Note that if
U
∈
M
n
(
R
) then Det
A
·
Det(
A

1
U
) = Det
U
.
That is, if we deﬁne
λ
:
M
n
(
R
)
→
M
n
(
R
) :
U
7→
A

1
U
then
Det = (Det
A
)
·
Det
◦
λ.
1
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View Full Document Apply the chain rule: as
λ
is linear, so
Det
0
A
(
H
) = (Det
A
)
·
Det
0
λ
(
H
)
(
λ
0
A
(
H
))
= (Det
A
)
·
Det
0
I
(
λ
(
H
))
= (Det
A
)
·
Tr(
A

1
H
)
Remark:
In general, if
A
∈
M
n
(
R
) then Det
0
A
(
H
) = Tr(
˜
AH
) where
˜
A
is the
‘adjugate’ of A (the matrix whose (
i,j
) entry is (

1)
i
+
j
A
ji
and
A
ji
is the
(
j,i
) cofactor of
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This note was uploaded on 07/08/2011 for the course MAT 6932 taught by Professor Staff during the Summer '10 term at University of Florida.
 Summer '10
 Staff
 Algebra

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