Symmetric Matrix Eigenvalue Problems - Numerical Methods in Engineering with MATLAB - page 369

EXAMPLE 9.9
Use the Sturm sequence property to show that the smallest eigenvalueof A is in the
interval(0

. 25 , 0 . 5), where

⎡
⎣

⎤
⎦

2 − 1 00

− 12 − 1 0

A =

0 − 12 − 1

00 − 12

5 and c i − 1 = Solution Taking λ = 0 . 5, wehave d i − λ = 1 . 1 and the Sturm sequence

in Eqs. (9.49) becomes

P 0 (0 . 5) = 1

P 1 (0 . 5) = 1 . 5

P 2 (0 . 5) = 1 . 5(1 . 5) − 1 = 1 . 25

P 3 (0 . 5) = 1 . 5(1 . 25) − 1 . 5 = 0 . 375

P 4 (0 . 5) = 1 . 5(0 . 375) − 1 . 25 =− 0 . 6875

Since the sequence containsonesign change, thereexists oneeigenvaluesmaller
than 0.5.
Repeating the process with

75, c i − 1 = λ = 0 . 25 ( d i − λ = 1 . 1), we get

P 0 (0 . 25) = 1

P 1 (0 . 25) = 1 . 75

P 2 (0 . 25) = 1 . 75(1 . 75) − 1 = 2 . 0625

P 3 (0 . 25) = 1 . 75(2 . 0625) − 1 . 75 = 1 . 8594

P 4 (0 . 25) = 1 . 75(1 . 8594) − 2 . 0625 = 1 . 1915

There are nosign changes in the sequence, so that all the eigenvalues are greater than
0.25. Wethusconcludethat0 . 25 <λ 1 < 0 . 5.

Gerschgorin's Theorem

Gerschgorin's theoremis useful in determining the global bounds on the eigenvalues
of an n n matrix A . The term “global”means the boundsthatenclose all the eigen-

values.We givehere a simplifiedversion of the theorem forasymmetric matrix.

×

If λ is an eigenvalueof A , then

a i − r i ≤ λ ≤ a i + r i , i = 1 , 2 ,..., n