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Hence<br />
The first eigen vector is<br />
Similarly for ω 2 = 0.438<br />
ϕ 21 = 0.475<br />
−2 = −0.237<br />
⎧ ⎫ ⎧ ⎫<br />
⎨ϕ 11 ⎬ ⎨<br />
ϕ 1 =<br />
⎩<br />
ϕ<br />
⎭ = 1 ⎬<br />
⎩<br />
21 −0.237<br />
⎭<br />
⎛⎡<br />
⎤ ⎡ ⎤⎞<br />
⎧ ⎫ ⎧ ⎫<br />
⎝⎣ 3 −2 ⎦ − 0.438 2 ⎣ 1 0 ⎨ϕ ⎦⎠<br />
12 ⎬ ⎨<br />
−2 2<br />
0 3<br />
⎩<br />
ϕ<br />
⎭ = 0⎬<br />
⎩<br />
22 0<br />
⎭<br />
⎡<br />
⎤ ⎧ ⎫ ⎧ ⎫<br />
⎣ 2.808 −2 ⎨ 1 ⎬ ⎨<br />
⎦<br />
−2 1.425<br />
⎩<br />
ϕ<br />
⎭ = 0⎬<br />
⎩<br />
22 0<br />
⎭<br />
This gives one equation to solve for ϕ 22 (the first row equation is only used)<br />
2.808 − 2ϕ 22 = 0<br />
Hence<br />
The second eigen vector is<br />
ϕ 22 = −2.808<br />
−2<br />
= 1.404<br />
⎧ ⎫ ⎧ ⎫<br />
⎨ϕ 12 ⎬ ⎨<br />
ϕ 2 =<br />
⎩<br />
ϕ<br />
⎭ = 1 ⎬<br />
⎩<br />
22 1.404<br />
⎭<br />
Now step 4 is applied, which is mass normalization of the shape vectors (or the eigenvectors)<br />
µ 1 = ϕ T 1 [M] ϕ 1<br />
Hence<br />
Similarly, µ 2 is found<br />
Hence<br />
⎧ ⎫<br />
⎨ 1 ⎬<br />
µ 1 =<br />
⎩<br />
−0.237<br />
⎭<br />
= 1. 169<br />
⎧ ⎫<br />
⎨ 1 ⎬<br />
µ 2 =<br />
⎩<br />
1.404<br />
⎭<br />
= 6.914<br />
⎤ ⎧ ⎫<br />
⎣ 1 0 ⎨ 1 ⎬<br />
⎦<br />
0 3<br />
⎩<br />
−0.237<br />
⎭<br />
T ⎡<br />
µ 2 = ϕ T 2 [M] ϕ 2<br />
⎤ ⎧ ⎫<br />
⎣ 1 0 ⎨ 1 ⎬<br />
⎦<br />
0 3<br />
⎩<br />
1.404<br />
⎭<br />
T ⎡<br />
11