458 Appendix
Sometimes complications arise, as for the matrix
A =
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
1 0 0
0 0 0
0 0 1
(30)
with eigenvalues 1, 1, and 0. Using the method given above to
find the eigenvector for λ 3 = 0 by setting x 1
(3) = 1 yields no solution. Setting x 2
(3) = 1, however, yields a correct solution for the
eigenvector, (0, 1, 0). Because this has no ê 1 component, we
could not have set x 1
(3) = 1 and found the other components.
This example illustrates a complication that arises for a degenerate, or repeated, eigenvalue: e.g., λ 1 = λ 2 = 1. In this case,
the eigenvalue corresponds not to an eigenvector but to an
entire plane, and any vector contained within it is an eigenvector.
Two eigenvectors spanning this plane can be found by finding
the eigenvector of the nondegenerate eigenvalue, and then
choosing two independent vectors orthogonal to it. Because the
eigenvector for the nondegenerate eigenvalue is (0, 1, 0), two
possible orthogonal eigenvectors for the degenerate eigenvalue
are (1, 0, 0) and (0, 0, 1).
A.5.3 Symmetric matrix eigenvalues, eigenvectors,
diagonalization, and decomposition
The eigenvalues and eigenvectors of a symmetric matrix have
interesting properties. An n × n matrix H has a characteristic
polynomial of degree n, each of whose n roots is an eigenvalue.
Consider two eigenvalues and their associated eigenvectors
Hx (i) = λ i x (i) , Hx (j) = λ j x (j) .
(31)
Multiplication of the first equation by x ( j)T (the transpose of
x ( j ) ) and the second equation by x (i)T yields
x ( j)T Hx (i) = λ i x ( j)T x (i) , x (i)T Hx ( j) = λ j x (i)T x ( j) .
(32)
Transposing both sides of the second part of Eqn 32 and subtracting it from the first gives
x ( j)T Hx (i) − x ( j)T H T x (i) = (λ i − λ j )x ( j)T x ( i) .
(33)
Because H is symmetric, it equals its transpose, H = H T , so the
left-hand side is zero
0 = (λ i − λ j )x ( j)T x (i) .
(34)
Thus, if i ≠ j and the two eigenvalues are different, their associated eigenvectors must be orthogonal so that their scalar
product x ( j)T x (i) is zero. Thus, for a symmetric matrix, eigenvectors associated with distinct eigenvalues are orthogonal.
This result lets us diagonalize a symmetric matrix. To illustrate this for a 3 × 3 case, consider a matrix U whose columns
are the eigenvectors of the symmetric matrix H
U
x
x
x
x
x
x
x
x
x
.
( )
( )
( )
( )
( )
( )
( )
( )
( )
=
⎛
⎝
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
1
1
1
2
1
3
2
1
2
2
2
3
3
1
3
2
3
3
(35)
If the eigenvalues of H are distinct, the eigenvectors of H, and
hence the columns of the eigenvector matrix, are orthogonal,
so U is an orthogonal matrix satisfying U −1 = U T .
The entire set of eigenvalue–eigenvector pairs, each of which
satisfy Hx
(i)
= λ i x
(i)
, can be written as the matrix equation
HU = UΛ,
(36)
where Λ is the diagonal matrix with eigenvalues on the diagonal
Λ
.
=
⎛
⎝
⎜
⎜ ⎜
⎞
⎠
⎟
⎟ ⎟
λ
λ
λ
1
2
3
0 0
0
0
0 0
(37)
Premultiplying both sides of Eqn 36 by the inverse of the
eigenvector matrix yields
U −1 HU = U T HU = Λ,
(38)
which shows how the eigenvector matrix can be used to
diagonalize a symmetric matrix. This result can also be stated
as
H = UΛU T ,
(39)
which illustrates how a symmetric matrix can be decomposed
into a diagonal eigenvalue matrix and the orthogonal eigenvector matrix. Similar results apply for complex Hermitian
matrices.
We will see that if a matrix contains the components of
vectors expressed in a coordinate system, the physical problem
under discussion can be simplified by diagonalizing the matrix.
This corresponds to rewriting the problem in its “natural” coordinate system, whose basis set is the eigenvectors, an idea
used in discussing stresses in the earth (Section 2.3.4) and the
seismic moment tensor (Section 4.4.5).
A.6 Vector calculus
A.6.1 Scalar and vector fields
Many phenomena in seismology depend on how physical
quantities vary in space. Some, like density or temperature,
are scalar fields, scalar valued functions of the position vector x
denoted by expressions like φ(x) or φ(x 1 , x 2 , x 3 ). Similarly,
a vector that varies in space is described by a vector field. For
example, seismic waves are described by the variation in the
displacement vector
Sometimes complications arise, as for the matrix
A =
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
1 0 0
0 0 0
0 0 1
(30)
with eigenvalues 1, 1, and 0. Using the method given above to
find the eigenvector for λ 3 = 0 by setting x 1
(3) = 1 yields no solution. Setting x 2
(3) = 1, however, yields a correct solution for the
eigenvector, (0, 1, 0). Because this has no ê 1 component, we
could not have set x 1
(3) = 1 and found the other components.
This example illustrates a complication that arises for a degenerate, or repeated, eigenvalue: e.g., λ 1 = λ 2 = 1. In this case,
the eigenvalue corresponds not to an eigenvector but to an
entire plane, and any vector contained within it is an eigenvector.
Two eigenvectors spanning this plane can be found by finding
the eigenvector of the nondegenerate eigenvalue, and then
choosing two independent vectors orthogonal to it. Because the
eigenvector for the nondegenerate eigenvalue is (0, 1, 0), two
possible orthogonal eigenvectors for the degenerate eigenvalue
are (1, 0, 0) and (0, 0, 1).
A.5.3 Symmetric matrix eigenvalues, eigenvectors,
diagonalization, and decomposition
The eigenvalues and eigenvectors of a symmetric matrix have
interesting properties. An n × n matrix H has a characteristic
polynomial of degree n, each of whose n roots is an eigenvalue.
Consider two eigenvalues and their associated eigenvectors
Hx (i) = λ i x (i) , Hx (j) = λ j x (j) .
(31)
Multiplication of the first equation by x ( j)T (the transpose of
x ( j ) ) and the second equation by x (i)T yields
x ( j)T Hx (i) = λ i x ( j)T x (i) , x (i)T Hx ( j) = λ j x (i)T x ( j) .
(32)
Transposing both sides of the second part of Eqn 32 and subtracting it from the first gives
x ( j)T Hx (i) − x ( j)T H T x (i) = (λ i − λ j )x ( j)T x ( i) .
(33)
Because H is symmetric, it equals its transpose, H = H T , so the
left-hand side is zero
0 = (λ i − λ j )x ( j)T x (i) .
(34)
Thus, if i ≠ j and the two eigenvalues are different, their associated eigenvectors must be orthogonal so that their scalar
product x ( j)T x (i) is zero. Thus, for a symmetric matrix, eigenvectors associated with distinct eigenvalues are orthogonal.
This result lets us diagonalize a symmetric matrix. To illustrate this for a 3 × 3 case, consider a matrix U whose columns
are the eigenvectors of the symmetric matrix H
U
x
x
x
x
x
x
x
x
x
.
( )
( )
( )
( )
( )
( )
( )
( )
( )
=
⎛
⎝
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
1
1
1
2
1
3
2
1
2
2
2
3
3
1
3
2
3
3
(35)
If the eigenvalues of H are distinct, the eigenvectors of H, and
hence the columns of the eigenvector matrix, are orthogonal,
so U is an orthogonal matrix satisfying U −1 = U T .
The entire set of eigenvalue–eigenvector pairs, each of which
satisfy Hx
(i)
= λ i x
(i)
, can be written as the matrix equation
HU = UΛ,
(36)
where Λ is the diagonal matrix with eigenvalues on the diagonal
Λ
.
=
⎛
⎝
⎜
⎜ ⎜
⎞
⎠
⎟
⎟ ⎟
λ
λ
λ
1
2
3
0 0
0
0
0 0
(37)
Premultiplying both sides of Eqn 36 by the inverse of the
eigenvector matrix yields
U −1 HU = U T HU = Λ,
(38)
which shows how the eigenvector matrix can be used to
diagonalize a symmetric matrix. This result can also be stated
as
H = UΛU T ,
(39)
which illustrates how a symmetric matrix can be decomposed
into a diagonal eigenvalue matrix and the orthogonal eigenvector matrix. Similar results apply for complex Hermitian
matrices.
We will see that if a matrix contains the components of
vectors expressed in a coordinate system, the physical problem
under discussion can be simplified by diagonalizing the matrix.
This corresponds to rewriting the problem in its “natural” coordinate system, whose basis set is the eigenvectors, an idea
used in discussing stresses in the earth (Section 2.3.4) and the
seismic moment tensor (Section 4.4.5).
A.6 Vector calculus
A.6.1 Scalar and vector fields
Many phenomena in seismology depend on how physical
quantities vary in space. Some, like density or temperature,
are scalar fields, scalar valued functions of the position vector x
denoted by expressions like φ(x) or φ(x 1 , x 2 , x 3 ). Similarly,
a vector that varies in space is described by a vector field. For
example, seismic waves are described by the variation in the
displacement vector