The Chemistry Maths Book, Second Edition

540 Chapter 19The matrix eigenvalue problem

It is always possible to find two linear combinations of the vectors that are

orthogonal. Letx

1

andx

2

be nonorthogonal vectors, withx

1

T

x

2

1 ≠ 10. Letx′

2

be the

linear combination

x′

2

1 = 1 x

2

1 − 1 cx

1

in which the parameter cis chosen such thatx′

2

be orthogonal tox

1

; that is,x

1

T

x′

2

1 = 10.

Then

x

1

T

x′

2

1 = 1 x

1

T

x

2

1 − 1 cx

1

T

x

1

1 = 1 0if

and new vector is orthogonal tox

1

. This is an example of the

widely-used Schmidt orthogonalizationmethod.

EXAMPLE 19.8Orthogonalization of vectors

The eigenvectorsx

2

andx

3

of Example 19.4(ii), belonging to the degenerate eigenvalue

λ 1 = 11 , are not orthogonal. We have, ignoring the arbitrary multiplersc

2

andc

3

,

Then

is orthogonal tox

2

. Including an arbitrary multiplier, the new vector is

and, withx

2

(andx

1

), can now be normalized. The orthonormal eigenvectors of the

real symmetric matrix

are

0 Exercise 19

xx

12

1

3

1

1

1

1

6

1

1

2

=

























=

−

























,,′′ =−

























x

3

1

2

1

1

0

211

121

112

























′= ′ −

























x

33

1

1

0

c

′=−

















=

−























xx

xx

xx

x

33

23

22

1

1

2

3

T

T



−

−

























=

−

























9

6

1

1

2

12

12

0

xx xx

22 23

11 2

1

1

2

6112

TT

=−

−

























(), ()==−

11

2

3

9

−

























=

′=−

















xx

xx

xx

x

22

12

11

1

T

T

c=

xx

xx

12

11

T

T