Page 131 - Compact Numerical Methods For Computers

120 Compact numerical methods for computers
the eigenvalues of A are found as the diagonal elements S , i = 1, 2, . . . , n, and
ii
the matrices U and V both contain complete sets of eigenvectors. These sets (the
columns of each matrix) are not necessarily identical since, if any two eigenvalues
are equal (also denoted as being degenerate or of multiplicity greater than one),
any linear combination of their corresponding eigenvectors is also an eigenvector
of the same eigenvalue. If the original two eigenvectors are orthogonal, then
orthogonal linear combinations are easily generated by means of orthogonal
matrices such as the plane rotations of §3.3. This is an important point to keep in
mind; recently a computer service bureau wasted much time and money trying to
find the ‘bug’ in a program taken from a Univac 1108 to an IBM 370 because the
eigenvectors corresponding to degenerate eigenvalues were computed differently
by the two computers.
Property (iii) above will now be demonstrated. First note that the eigenvalues of
a symmetric positive definite matrix are positive. For, from (7.9) and equation
(10.1), we have
T T T
0 < y Ay = y XEX y

(10.8)
T
thus implying that all the e must be positive or else a vector w = X y could be
j
devised such that w j = 1, w i = 0 for i j corresponding to the non-positive
eigenvalue, thus violating the condition (7.9) for definiteness. Hereafter, E and S
will be ordered so that
S i > S i +l > 0 (10.9)
e i > e i + l · (10.10)
This enables S and E to be used interchangably once their equivalence has been
demonstrated.
Now consider pre-multiplying equation (10.1) by A. Thus we obtain
2 2
A X = AAX = AXE = XEE = XE (10.11)
while from symmetry (10.7) and the decomposition (2.53)
2 T 2
A V = A AV = VS . (10.12)
2
Since (10.11) and (10.12) are both eigenvalue equations for A , S 2 and E 2 are
identical to within ordering, and since all e are positive, the orderings (10.9) and
i
(10.10) imply
S = E. (10.13)
Now it is necessary to show that
AV = VS. (10.14)
T
From (10.1), letting Q = X V, we obtain
T
AV = XEX V = XEQ = XSQ. (10.15)
However, from (10.11) and (10.12), we get
2 2
QS = S Q. (10.16)