Biomedical Engineering Reference
In-Depth Information
•
S
min
{
}
The minimum and maximum eigenvalues of a matrix
A
are denoted
A
and
S
max
{
}
A
.
•
The eigenvectors corresponding to the minimum and maximum eigenvalues of a
matrix
A
are denoted
ˑ
min
{
A
}
and
ˑ
max
{
A
}
.
•
The minimum and maximum generalized eigenvalues of a matrix
A
with a metric
B
are denoted
S
min
{
A
,
B
}
and
S
max
{
A
,
B
}
, and the corresponding eigenvectors are
denoted
ˑ
min
{
A
,
B
}
and
ˑ
max
{
A
,
B
}
.
Here, if the matrix
B
is nonsingular, the following relationships hold:
B
−
1
A
S
max
{
A
,
B
}=
S
max
{
}
,
B
−
1
A
ˑ
max
{
A
,
B
}=
ˑ
max
{
}
,
B
−
1
A
S
min
{
,
}=
S
min
{
}
,
A
B
B
−
1
A
ˑ
min
{
A
,
B
}=
ˑ
min
{
}
.
Using
x
to denote a column vector with its dimension commensurate with the size
of the matrices, this appendix shows that
x
T
Ax
x
T
Bx
=
S
max
{
max
x
A
,
B
}
,
(C.96)
and
x
T
Ax
x
T
Bx
=
ˑ
max
{
argmax
x
A
,
B
}
.
(C.97)
x
T
Ax
x
T
Bx
Since the value of the ratio
is not affected by the norm of
x
,weset
the norm of
x
so as to satisfy the relationship
x
T
Bx
(
)/(
)
=
1. Then, the maximization
problem in Eq. (
C.96
) is rewritten as
x
T
Ax
subject to
x
T
Bx
max
x
=
1
.
(C.98)
We change this constrainedmaximization problem to an unconstrainedmaximization
problem by introducing the Lagrange multiplier
ʺ
. We define the Lagrangian
L
(
x
,ʺ)
such that
x
T
Ax
x
T
Bx
L
(
x
,ʺ)
=
−
ʺ(
−
1
).
(C.99)
L
(
,ʺ)
The maximization in Eq. (
C.98
) is equivalent to maximizing
x
with no con-
straints.
To obtain the maximum of
L
(
x
,ʺ)
, we calculate the derivatives
∂
L
(
x
,ʺ)
=
2
(
Ax
−
ʺ
Bx
),
(C.100)
∂
x
∂
L
(
,ʺ)
∂ʺ
x
x
T
Bx
=−
(
−
1
).
(C.101)
