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replaced by
Λ
:(
Φ
T
d
e
Φ
) in (1.47). The only issue
here is to find a suitable matrix
Λ
.
Similar to Equation (1.24), we consider the
following increment vector
following quadratic equation which is emerged
from Equation (1.43)
T
T
f
∆
x
−
∆
λ
f
∆
x
=
det
11
12
0
(1.54)
T
T
f
∆
x
f
∆
x
−
∆
λ
12
22
(
)
−
T
∆
x
=
D
=
Λ Φ Φ
:
d
Γ
v
,
e
=
1
,
…
,
N
e
e
e
e
(1.48)
We are interested in finding
Λ
i
∗
such that
Equation (1.54) results in two positive eigenval-
ues ∆
λ
1
> 0 and ∆
λ
2
> 0. There are several ways
to achieve this. Here we assume the following
To illustrate the calculation of
Λ
, we consider
the simplest multiple case of
m
= 2. We assume
that the two eigenvalues are repeated if
T
∆
λ
1
=
f
∆
x
=
1
(1.55)
λ
−
λ
11
2
1
≤
δ
(1.49)
λ
1
λ
−
λ
∆
λ
=
f
T
∆
x
= −
(1.56)
1
2
1
2
22
λ
1
with
δ
being a small positive tolerance. In this
case we have
f
T
∆ =
x
(1.57)
0
12
Φ
=
(
)
ϕ ϕ
1
,
(1.50)
2
Equation (1.57) implies that the matrix in Equa-
tion (1.54) is diagonal, hence the eigenvalues are
equivalent to the diagonal terms as stated in Eqs.
(1.55) and (1.56). In Equation (1.55) we consid-
ered an increase of 1 for the lowest eigenvalue.
If the two eigenvalues are different, the increase
assigned to the second eigenvalue in Equation
(1.56) will be slightly lower than 1. This is to
reduce the difference between the two repeated
eigenvalues.
Substituting Equation (1.53) in Eqs. (1.55) to
(1.57), we obtain a set of three equations which
can be solved to yield the three unknown coef-
ficients
Λ
1
∗
,
Λ
2
∗
, and
Λ
2
∗
. These equations can
be summarized as
We also introduce the following positive
semidefinite symmetric matrix
∗
∗
Λ
Λ
Λ
∗
=
11
12
(1.51)
∗
∗
Λ
Λ
12
22
based on which we define
Λ
∗
Λ
=
(1.52)
∗
trace(
Λ
)
to ensure that trace(
Λ
) = 1.
Substituting Equation (1.51) in Equation (1.48)
and using Eqs. (1.45) and (1.44), we may rewrite
Equation (1.48) in the following form
1
f v
T
+
Γ
f
T
f
f
T
f
f
T
f
Λ
Λ
Λ
∗
11
11 11
11 22
11 12
11
λ
−
λ
f
T
f
f
T
f
∗
=
1
−
+
Γ
f v
T
2
1
22 22
22 12
22
λ
22
1
symm.
f
T
f
2
∗
T
Γ
f v
12 12
12
∆
x
=
Λ
∗
f
+
Λ
∗
f
+
Λ
∗
f
−
Γ
v
(1.53)
2
12
11 11
12 12
22 22
(1.58)
The change in the multiple eigenvalues ∆
λ
1
and ∆
λ
2
due to ∆
x
can be calculated by solving the
By solving Equation (1.58) one finds
Λ
*
which is used in Equation (1.52) to find
Λ
and
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