Symmetric Matrix Eigenvalue Problems - Numerical Methods in Engineering with MATLAB - page 349

30 2

| z | = + 10 2 + 20 2 = 37 . 417

⎡
⎣

⎤
⎦

⎡
⎣

⎤
⎦

− 30 . 0 − 0 . 80177

z 1

v = =

417 =

10 . 0 0 . 267 26

| z | 37 .

20 . 0 0 . 534 52

Second iteration:

⎡
⎣

⎤
⎦

⎡
⎣

⎤
⎦ =

⎡
⎣

⎤
⎦

− 30 10 20 − 0 . 80177 37 . 416

z = Sv =

10 0 − 50 0 . 267 26 − 24 . 053

20 − 50 − 10 0 . 534 52 − 34 . 744

37

| z | = . 416 2 + 24 . 053 2 + 34 . 744 2 = 56 . 442

⎡
⎣

⎤
⎦

⎡
⎣

⎤
⎦

37 . 416 0 . 662 91

z 1

v = =

442 =

− 24 . 053 − 0 . 426 15

| z | 56 .

− 34 . 744 − 0 . 615 57

Third iteration:

⎡
⎣

⎤
⎦

⎡
⎣

⎤
⎦ =

⎡
⎣

⎤
⎦

− 30 10 20 0 . 66291 − 36 . 460

z = Sv =

10 0 − 50 − 0 . 42615 20 . 362

20 − 50 − 10 − 0 . 61557 40 . 721

36

| z | = . 460 2 + 20 . 362 2 + 40 . 721 2 = 58 . 328

⎡
⎣

⎤
⎦

⎡
⎣

⎤
⎦

− 36 . 460 − 0 . 62509

z
|

1

v = =

328 =

20 . 362 0 . 34909

z | 58 .

40 . 721 0 . 69814

Atthis point the approximation of the eigenvalue we seek is 328 MPa (the

negativesign is determinedbythe sign reversalof z betweeniterations). This is
actually close to the second-largest eigenvalue

λ =− 58 .

39 MPa! By continuing the
iterative process we would eventually end up with the largest eigenvalue

λ 2 =− 58 .

λ 3 = 70 . 94

MPa. Butsince are rather close, the convergence is too slow from this

pointon for manuallabor. Here is aprogram that does the calculationsforus:

| λ 2 | and | λ 3 |

% Example 9.4 (Power method)
S = [-30 10 20; 10 40 -50; 20 -50 -10];
v=[1;0;0];
fori=1:100
vOld=v;z=S*v;zMag=sqrt(dot(z,z));
v = z/zMag; vSign = sign(dot(vOld,v));