Image Processing Reference
In-Depth Information
3.9.2 P
RODUCT AND
S
UM OF
E
IGENVALUES
The product and sum of eigenvalues of any matrix are equal to the determinant and
the trace of that matrix, respectively. We
first show that the product of eigenvalues is
equal to the determinant of the matrix. Let A be an n
n matrix with characteristic
polynomial P(l),
(l)
, then
P
(l) ¼jl
I
A
j¼(l l
1
)(l l
2
) (l l
n
)
(
3
:
113
)
Set
l ¼
0, then
n
j
A
j¼(l
1
)(l
2
) (l
n
) ¼ (
1
)
l
1
l
2
l
n
(
3
:
114
)
or
n
n
(
1
)
j
A
j¼(
1
)
l
1
l
2
l
n
(
3
:
115
)
Hence, we have
l
1
l
2
l
n
¼j
A
j
(
3
:
116
)
Now we show that the sum of eigenvalues is equal to the trace of the matrix:
2
4
3
5
l
a
11
a
12
a
1n
a
21
l
a
22
a
2n
P
(l) ¼jl
I
A
j¼
.
.
.
.
(
3
:
117
)
.
.
a
n1
a
n2
l
a
nn
By expanding the above determinant along the
first column, we have
X
n
i
¼
2
(
i
þ
1
a
i1
j
M
i1
j
P
(l) ¼ (l
a
11
)j
M
11
j
1
)
(
3
:
118
)
where M
ij
is the determinant of matrix obtained by deleting the ith row and the j
th
column of A. Expanding determinant of M
11
, we have
X
n
i
¼
3
(
j
M
11
j¼(l
a
22
)j
M
11
j
i
þ
1
a
i1
j
M
i1
j
1
)
(
3
:
119
)
Continuing these expansions, we will have
P
(l) ¼ (l
a
11
)(l
a
22
) (l
a
nn
) þ
P
0
(l)
(
3
:
120
)
where P
0
(l)
is a polynomial of degree n
2. Therefore, the leading coef
cient of P(l)
(l)
is one and the second leading coef
cient is a
11
þ
a
22
þþ
a
nn
. Also from Equation
3.113, the second leading coef
cient of P(l)
(l)
is
l
1
þ l
2
þþl
n
. Therefore,
l
1
þ l
2
þþl
n
¼
a
11
þ
a
22
þþ
a
nn
¼ Trace(
A
)
(
3
:
121
)
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