Geography Reference
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Question: “What is the role of strain in the context of
the pair of matrices
{
E
l
,
G
l
}
and
{
E
r
,
G
r
}
, respectively?”
Answer: “
G
l
,
G
r
}
are symmetric, positive-definite matrices. Thus, according
to a standard lemma of matrix algebra, both matrices can
be simultaneously diagonalized, one matrix being the unit
matrix. With the reference to the general eigenvalue we
experienced for the Cauchy-Green deformation tensor, we
arrive at Lemma
1.7
.”
{
E
l
,
E
r
}
are symmetric matrices arid
{
Lemma 1.7 (Left and right general eigenvalue problem of the Euler-Lagrange deformation ten-
sor).
For the pair of symmetric matrices
{
E
l
,
G
l
}
or
{
E
r
,
G
r
}
,where
{
G
l
,
G
r
}
are positive-definite
matrices, a simultaneous diagonalization, namely
F
l
E
l
F
l
=diag(
K
1
,K
2
)
,
F
l
G
l
F
l
=I
2
versus F
r
E
r
F
r
=diag(
κ
1
,κ
2
)
,
F
r
G
r
F
r
=I
2
,
(1.113)
is immediately obtained from the left and right general eigenvalue-eigenvector problems
E
l
F
l
−
G
l
F
l
diag(
K
1
,K
2
)=0
E
r
F
r
−
G
r
F
r
diag(
κ
1
,κ
2
)=0
⇔
⇔
(E
l
−
K
i
G
l
)
f
li
=0(
∀
i
∈{
1
,
2
}
)
(
r
−
κ
i
G
r
)
f
r
i
=0(
∀
i
∈{
1
,
2
}
)
⇔
and
⇔
(1.114)
|
E
l
− K
i
G
l
|
=0(F
l
G
l
F
l
=I
2
)
,
|
E
r
− κ
i
G
r
|
=0(F
r
G
r
F
r
=I
2
)
,
K
1
,
2
=
K
+
,−
=
2
{
tr[E
l
G
−
l
]
κ
1
,
2
=
κ
+
,−
=
2
{
tr[E
r
G
−
r
]
±
±
(tr[E
l
G
−
l
])
2
±
(tr[E
r
G
−
r
])
2
4det[E
l
G
−
l
]
±
−
}
,
−
4det[E
r
G
−
r
]
}
.
End of Lemma.
In order to visualize the eigenspace of both the left and the right Euler-Lagrange deformation
tensor E
l
and E
r
relative to the left and right metric tensors G
l
and G
r
, we are forced to compute
in addition the left and right eigenvectors (namely the left and right eigencolumns, also called
eigendirectories) of the pairs
{
E
l
,
G
l
}
md
{
E
r
,
G
r
}
, respectively. Lemma
1.8
summarizes the
results.
Lemma 1.8 (Left and right eigenvectors of the left and the right Euler-Lagrange deformation
tensor).
For the pair of symmetric matrices
{
E
l
,
G
l
}
or
{
E
r
,
G
r
}
. an explicit form of the left eigencolumns
and the right eigencolumns is
1st left eigencolumns
,K
1
:
F
11
F
21
=
1
G
11
(
e
22
−
K
1
G
12
)
×
K
1
G
22
)
2
−
2
G
12
(
e
12
−
K
1
G
12
)(
e
22
−
K
1
G
22
)+
G
22
(
e
12
−
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