# Question 7.3: Use Jacobi diagonalization to find the eigenvalues and eigen......

Use Jacobi diagonalization to find the eigenvalues and eigenvectors of the 3-DOF system used in Examples 7.1 and 7.2.

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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:

$\left(-λ \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix} + 10^{3} \begin{bmatrix} 3 & -2 & 0 \\ -2 & 3 & -1 \\ 0 & -1 & 1 \end{bmatrix}\right) \begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \\ \overline{z}_{3} \end{Bmatrix} = 0$                          (C)

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

$[\underline{B}] [\Psi ] = [\Psi ] [\Lambda ] (I)\\ \mathrm{where} \\ [\Psi ] = \left[\left\{\overline{y}\right\}_{1} \left\{\overline{y} \right\}_{2} \left\{\overline{y} \right\}_{3} \right] (J)\\ \mathrm{and} \\ [\Lambda] = \begin{bmatrix} \lambda_{1} & 0 & 0 \\ 0 & \lambda_{2} & 0 \\ 0 & 0 & \lambda_{3} \end{bmatrix} = \begin{bmatrix} \omega_{1}^{2} & 0 & 0 \\ 0 & \omega_{2}^{2} & 0 \\ 0 & 0 & \omega_{3}^{2} \end{bmatrix} (K)$

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

$[\underline{B}] = 10^{3} \begin{bmatrix} \frac{1}{\sqrt{2}} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} 3 & -2 & 0 \\ -2 & 3 & -1 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{\sqrt{2}} \end{bmatrix} = \begin{bmatrix} \frac{3}{2} & \frac{-2}{\sqrt{2}} & 0 \\ \frac{-2}{\sqrt{2}} & 3 & \frac{-1}{\sqrt{2}} \\ 0 & \frac{-1}{\sqrt{2}} & \frac{1}{2} \end{bmatrix} (L)\\ [\underline{B}] = 10^{3} \begin{bmatrix} 1.5 & -1.4142 & 0 \\ -1.4142 & 3 & -0.7071 \\ 0 & 0 & 0.5 \end{bmatrix} (M)$

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

$[B]^{(1)} = [T_{1}]^{T} [\underline{B}] [T_{1}] (O)\\ \mathrm{where} \\ [T_{1}] = \begin{bmatrix} \cos\theta_{1} & -\sin\theta_{1} & 0 \\ \sin\theta_{1} & \cos\theta_{1} & 0 \\ 0 & 0 & 1 \end{bmatrix} (P) \\ \mathrm{and} \\ \tan 2\theta_{1} = \frac{2b_{12}}{b_{11} – b_{22}} = \frac{2(-1.4142)}{1.5 – 3} = 1.8856$

giving $\theta _{1} = 0.5416 .$

Substituting for $\theta _{1},$

$[T_{1}] = \begin{bmatrix} 0.8569 & -0.5155 & 0 \\ 0.5155 & 0.8569 & 0 \\ 0 & 0 & 1 \end{bmatrix} (Q) \\ \mathrm{Then,} \\ [\underline{B}]^{(1)} = [T_{1}]^{T} [\underline{B}] [T_{1}] = 10^{3} \begin{bmatrix} 0.8569 & -0.5155 & 0 \\ 0.5155 & 0.8569_{1} & 0 \\ 0 & 0 & 1 \end{bmatrix}^{T} \begin{bmatrix} 1.5 & -1.4142 & 0 \\ -1.4142 & 3 & -0.7071 \\ 0 & 0 & 0.5 \end{bmatrix} \begin{bmatrix} 0.8569 & -0.5155 & 0 \\ 0.5155 & 0.8569 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ [\underline{B}]^{(1)} = 10^{3} \begin{bmatrix} 0.6492 & 0 & -0.3645 \\ 0 & 3.8507 & -0.6059 \\ -0.3645 & -0.6059 & 0.5000 \end{bmatrix} (R)$

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:

$[\underline{B}]^{(2)} = [T_{2}]^{T} [\underline{B}]^{(1)} [T_{2}] (S)\\ \mathrm{where} \\ [T_{2}] = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\theta_{2} & -\sin\theta_{2} \\ 0 & \sin\theta_{2} & \cos\theta_{2} \end{bmatrix} (T)\\ \mathrm{and} \\ \tan 2\theta_{2} = \frac{2b_{23}^{(1)}}{b_{22}^{(1)} – b_{33}^{(1)}} = \frac{2(-0.6059)}{3.8507 – 0.5000} = 0.36166 (U)$

giving

$\theta _{2} = -0.17351 \\ [T_{2}] = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0.9849 & 0.1726 \\ 0 & -0.1726 & 0.9849 \end{bmatrix} (V)$

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

$[\underline{B}]^{(2)} = 10^{3} \begin{bmatrix} 0.6492 & 0.06291 & -0.3590 \\ 0.06291 & 3.9569 & 0 \\ -0.3590 & 0 & 0.3938 \end{bmatrix} (W)$

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:

$[\Psi] = [T_{1}][T_{2}][T_{3}][T_{4}][T_{5}] (X)$

or numerically:

$[\Psi] = \begin{bmatrix} 0.8569 & -0.5155 & 0 \\ 0.5155 & 0.8569 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0.9849 & 0.1726 \\ 0 & -0.1726 & 0.9849 \end{bmatrix} \begin{bmatrix} 0.8170 & 0 & 0.5766 \\ 0 & 1 & 0 \\ -0.5766 & 0 & 0.8170 \end{bmatrix} \\ \times \begin{bmatrix} 0.9998 & 0.01682 & 0 \\ -0.01682 & 0.9998 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0.9999 & -0.0095 \\ 0 & 0.0095 & 0.9999 \end{bmatrix} \\ = \begin{bmatrix} 0.7598 & -0.4910 & 0.4260 \\ 0.3216 & 0.8534 & 0.4099 \\ -0.5649 & -0.1745 & 0.8064 \end{bmatrix}$

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

$[\Psi] = \left[\left\{\overline{y}\right\}_{2} \left\{\overline{y}\right\}_{3} \left\{\overline{y}\right\}_{1} \right]$

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

$\left\{\overline{z} \right\} = [U]^{-1} \left\{\overline{y}\right\}$            (7.14)

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

$[\Phi] = [U]^{-1} [\Psi] (Y)$

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):

$[\Phi] = [U]^{-1} [\Psi] = \begin{bmatrix} \frac{1}{\sqrt{2}} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} 0.7598 & -0.4910 & 0.4260 \\ 0.3216 & 0.8534 & 0.4099 \\ -0.5649 & -0.1745 & 0.8064 \end{bmatrix} \\ = \begin{bmatrix} 0.5373 & -0.3472 & 0.3012 \\ 0.3216 & 0.8535 & 0.4099 \\ -0.3994 & -0.1234 & 0.5703 \end{bmatrix}$                      (Z)

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.

$[\Phi ]^{T} [M][\Phi] = \begin{bmatrix} 1.000 & 0 & 0 \\ 0 & 1.000 & 0 \\ 0 & 0 & 1.000 \end{bmatrix}$

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.

$[\Phi ]^{T} [K][\Phi] = \begin{bmatrix} 901.7 & 0 & 0 \\ 0 & 3958 & 0 \\ 0 & 0 & 140.1 \end{bmatrix}$

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.

Question: 7.2

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