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Fundamentals of Stress and Vibration                1. Mathematics for Structural mechanics
                 [A Practical guide for aspiring Designers / Analysts]
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                Eigen Vector Equation =  AX = λX  =   A − λI X = 0 , where  X  is the Eigen vector.

                                                 X        X ′
                Here, there are two Eigen vectors:    1    and     1
                                                 X 2      X ′ 2
                Let us now find the first Eigen vector. Expanding the Eigen vector equation, we have:

                    50 − λ     10     X      0
                =         1              1   =        - - - - (1.71)
                      10     25 − λ 1  X 2   0

                where  X   and  X   are components of the first Eigen vector.
                         1
                                 2
                Here, the values of  X   and  X   can not be determined, as, the equations are homogenous, that is,
                                    1
                                            2
                RHS of the equations are zero. Only the relationship between  X   and  X   can be got by solving
                                                                           1
                                                                                    2
                either equations (1.72) or (1.73) that is got by simplifying equation (1.71).
                  50 − λ  X +  10 ∗ X   = 0    - - - - (1.72)
                        1
                                     2
                           1
                 (10 ∗ X ) + (25 − λ )X = 0    - - - - (1.73)
                                   1
                       1
                                      2
                Let us simplify equation (1.72) to compute a relationship between  X   and  X  .
                                                                                1
                                                                                         2
                         10 ∗ X                10 ∗ X
                               2
                                                     2
                 X = −             =  X = −               =  X = 2.849 X
                  1
                                       1
                                                              1
                                                                         2
                         50 − λ              50 − 53.51
                               1
                                                                                        2.849
                Substituting (X = 1), we get the components of the first Eigen vector to be:      1
                              2
                The second Eigen vector is got by replacing (λ ) with (λ ) and the first Eigen vector with the second
                                                          1
                                                                   2
                in equations (1.71).
                    50 − λ     10     X ′    0
                =         2              1   =        - - - - (1.74)
                      10     25 − λ 2 X ′ 2  0
                Simplifying equation (1.74), we get:
                                     ′
                           ′
                  50 − λ  X +  10 ∗ X   = 0    - - - - (1.75)
                        2
                           1
                                     2
                                      ′
                       ′
                 (10 ∗ X ) + (25 − λ )X = 0    - - - - (1.76)
                       1
                                   2
                                      2


                              QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,   Page 57
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