The equations of motion of the system can be written as:

(- \omega^{2} [M] + [K])\left\{\overline{z} \right\} = 0                          (A)

or since λ = ω²

(-\lambda [M] + [K])\left\{\overline{z}\right\} = 0                          (B)

where λ is an eigenvalue and \left\{\overline{z}\right\} is an eigenvector. In numerical form, for this example:

Since the mass matrix is diagonal in this case, decomposing it into [M] = [U]^{T} [U] is very easy since:

[U] = [M]^{\frac{1}{2}} = \begin{bmatrix} \sqrt{m_{11}} & 0 & 0 \\ 0 & \sqrt{m_{22}} & 0 \\ 0 & 0 & \sqrt{m_{33}} \end{bmatrix} = \begin{bmatrix} \sqrt{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \sqrt{2} \end{bmatrix}                          (D)

The inverse of [U],  [U]^{-1} is found by simply inverting the individual terms:

[U]^{-1} = \begin{bmatrix} \frac{1}{\sqrt{m_{11}}} & 0 & 0 \\ 0 & \frac{1}{\sqrt{m_{22}}} & 0 \\ 0 & 0 & \frac{1}{\sqrt{m_{33}}} \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{\sqrt{2}} \end{bmatrix}                          (E) \\ \mathrm{and} \\ [U]^{-T} = [U]^{-1} = \begin{bmatrix} \frac{1}{\sqrt{2}} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{\sqrt{2}} \end{bmatrix}                          (F)

From Eq. (7.19), the eigenvalue problem to be solved is

([\underline{B}] – \lambda [I]) \left\{\overline{y} \right\} = 0                          (G)

where from Eq. (7.20):

[\underline{B}] = [U]^{-T} [K] [U]^{-1}                          (H)

and λ = ω² .

A more convenient form of Eq. (G), using the modal matrix, [\Psi ], is

Evaluating [\underline{B}] from Eq. (H)

The end result of the Jacobi diagonalization that we are aiming for is, from Eq. (7.27),

[\Psi ]^{T} [\underline{B}] [\Psi ] = [\Lambda ].                          (N)

So if we can find a series of transformations that eventually change [\underline{B}]   \mathrm{into}   [\Lambda], the product of these transformations must be the modal matrix, [\Psi ]. The first transformation, aimed at reducing the b_{12}   \mathrm{and}   b_{21} terms to zero, is

giving \theta _{1} = 0.5416 .

Substituting for \theta _{1},

This has reduced the b^{(1)}_{12}   \mathrm{and}   b^{(1)}_{21} terms to zero, as expected. We next reduce the b^{(1)}_{23}   \mathrm{and}   b^{(1)}_{32} terms to zero by the transformation:

giving

Applying Eq. (S): [\underline{B}]^{(2)} = [T_{2}]^{T} [\underline{B}]^{(1)} [T_{2}] , gives

This has reduced the b^{(1)}_{23}   \mathrm{and}   b^{(1)}_{32} terms to zero, but the previously zero b^{(1)}_{12}   \mathrm{and}   b^{(1)}_{21} terms have changed to 0.06291.

Three more transformations were applied, in the same way, with results as follows:

Third transformation, \theta _{3} = -0.6145 \\ [T_{3}] = \begin{bmatrix} \cos\theta_{3} & 0 & -\sin\theta_{3} \\ 0 & 1 & 0 \\ \sin\theta_{3} & 0 & \cos\theta_{3} \end{bmatrix} ;          [\underline{B} ]^{(3)} = 10^{3} \begin{bmatrix} 0.9025 & 0.05140 & 0 \\ 0.05140 & 3.9569 & 0.03626 \\ 0 & 0.03626 & 0.1404 \end{bmatrix}

Fourth transformation, \theta _{4} = – 0.0168; \\ [T_{4}] = \begin{bmatrix} \cos\theta_{4} & -\sin\theta_{4} & 0 \\ \sin\theta_{4} & \cos\theta_{4} & 0 \\ \sin\theta_{3} & 0 & 1 \end{bmatrix} ;          [\underline{B} ]^{(4)} = 10^{3} \begin{bmatrix} 0.9017 & 0 & -0.0006 \\ 0 & 3.9577 & 0.03626 \\ 0 & 0.03626 & 0.1404 \end{bmatrix}

Fifth transformation, \theta _{5} = 0.0095; \\ [T_{5}] = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\theta_{5} & -\sin\theta_{5} \\ 0 & \sin\theta_{5} & \cos\theta_{5} \end{bmatrix} ;          [\underline{B} ]^{(5)} = 10^{3} \begin{bmatrix} 0.9017 & 0 & -0.0006 \\ 0 & 3.9581 & 0 \\ -0.0006 & 0 & 0.1401 \end{bmatrix}

This could be continued for one or two more transformations, to further reduce the off-diagonal terms, but it can already be seen that [\underline{B} ]^{(5)} has practically converged to a diagonal matrix of the eigenvalues. From Example 7.1, representing the same system, these were, in ascending order:

λ_{1} = 140.1,         λ_{2} = 901.7,         λ_{3} = 3958

It will be seen that they appear in [\underline{B} ]^{(5)} in the different order: λ_{2}, λ_{3}, λ_{1}.

The matrix of eigenvectors, [Ψ], is now given by Eq. (7.29).

[\Psi] = [T_{1}][T_{2}][T_{3}]…                     (7.29)

These are the eigenvectors of matrix [\underline{B}], not those of the system, since it will be remembered that the latter were transformed by [U]^{-1} in order to make [\underline{B}] symmetric. Since there were five transformations, [T_{1}]   \mathrm{through}   [T_{5}], in this case, the eigenvectors [Ψ] are given by:

or numerically:

The final matrix product above, in the order in which it emerged from the diagonalization, is the modal matrix:

To find the eigenvectors, \left\{\overline{z} \right\}, of the physical system, we must use Eq. (7.14):

All the vectors can be transformed together by writing Eq. (7.14) as:

where [Φ] is the modal matrix of eigenvectors \left\{\overline{z} \right\}   \mathrm{and}   [Ψ] the modal matrix of eigenvectors \left\{\overline{y}\right\}. From Eq. (Y):

These are, finally, the eigenvectors of the system, \left\{\overline{z} \right\}_{i}, in the same order as the eigenvalues, i.e. \left\{\overline{z} \right\}_{2}, \left\{\overline{z} \right\}_{3}, \left\{\overline{z} \right\}_{1}. These would normally be rearranged into the order of ascending frequency.
As a check, we would expect the product [\Phi ]^{T} [M][\Phi] to equal the unit matrix, [I] , and this is the case, to within four significant figures, showing that the eigenvectors emerge in orthonormal form, without needing to be scaled.

Another check is that the product [\Phi ]^{T} [K][\Phi] should be a diagonal matrix of the eigenvalues, which is also the case to within four significant figures.

It should be noted that the presentation of the numerical results above, to about four significant figures, does not reflect the accuracy actually used in the computation.