Hence, $\tilde{\mathbf{K}} = [2];\quad \tilde{\mathbf{M}} = [12]$ $$ \rho_ {1} = \frac {1}{6}; \quad \mathbf {x} _ {1} = \left[ \frac {1}{2 \sqrt {3}} \right] $$ and $\overline{\Phi}\overline{1} = \left[\frac{1}{2\sqrt{3}}\quad \frac{1}{\sqrt{3}}\quad \frac{1}{\sqrt{3}}\quad \frac{1}{\sqrt{3}}\right]$ Hence we have, as expected, $\rho_{1} > \lambda_{1}$ . The Ritz analysis procedure presented above is a very general tool, and, as pointed out earlier, various analysis methods known under different names can actually be shown to be Ritz analyses. In Section 10.3.3 we present the component mode synthesis as a Ritz analysis. In the following we briefly want to show that the technique of static condensation as described in Section 10.3.1 is, in fact, also a Ritz analysis. In the static condensation analysis we assumed that all mass can be lumped at q degrees of freedom. Therefore, as an approximation to the eigenproblem $K\phi = \lambda M\phi$ , we obtained the following problem: $$ \left[ \begin{array}{l l} \mathbf {K} _ {a a} & \mathbf {K} _ {a c} \\ \mathbf {K} _ {c a} & \mathbf {K} _ {c c} \end{array} \right] \left[ \begin{array}{l} \boldsymbol {\phi} _ {a} \\ \boldsymbol {\phi} _ {c} \end{array} \right] = \lambda \left[ \begin{array}{l l} \mathbf {M} _ {a} & \mathbf {0} \\ \mathbf {0} & \mathbf {0} \end{array} \right] \left[ \begin{array}{l} \boldsymbol {\phi} _ {a} \\ \boldsymbol {\phi} _ {c} \end{array} \right] \tag {10.42} $$ with $q$ finite and $n - q$ infinite eigenvalues, which correspond to the massless degrees of freedom (see Section 10.2.4). To calculate the finite eigenvalues, we used static condensation on the massless degrees of freedom and arrived at the eigenproblem $$ \mathbf {K} _ {a} \boldsymbol {\phi} _ {a} = \lambda \mathbf {M} _ {a} \boldsymbol {\phi} _ {a} \tag {10.45} $$ where $K_{a}$ is defined in (10.46). However, this solution is actually a Ritz analysis of the lumped mass model considered in (10.42). The Ritz basis vectors are the displacement patterns associated with the $\phi_{a}$ degrees of freedom when the $\phi_{c}$ degrees of freedom are released. Solving the equations $$ \left[ \begin{array}{l l} \mathbf {K} _ {a a} & \mathbf {K} _ {a c} \\ \mathbf {K} _ {c a} & \mathbf {K} _ {c c} \end{array} \right] \left[ \begin{array}{l} \mathbf {F} _ {a} \\ \mathbf {F} _ {c} \end{array} \right] = \left[ \begin{array}{l} \mathbf {I} \\ \mathbf {0} \end{array} \right] \tag {10.51} $$ in which $\mathbf{F}_a = \mathbf{K}_a^{-1}$ , we find that the Ritz basis vectors to be used in (10.80), (10.81), and (10.84) are $$ \Psi = \left[ \begin{array}{c} \mathbf {I} \\ \mathbf {F} _ {c} \mathbf {K} _ {a} \end{array} \right] \tag {10.85} $$ To verify that a Ritz analysis with the base vectors in (10.85) yields in fact (10.45), we evaluate (10.80) and (10.81). Substituting for $\Psi$ and K in (10.80), we obtain $$ \tilde {\mathbf {K}} = \left[ \begin{array}{l l} \mathbf {I} & \left(\mathbf {F} _ {c} \mathbf {K} _ {a}\right) ^ {T} \end{array} \right] \left[ \begin{array}{l l} \mathbf {K} _ {a a} & \mathbf {K} _ {a c} \\ \mathbf {K} _ {c a} & \mathbf {K} _ {c c} \end{array} \right] \left[ \begin{array}{l} \mathbf {I} \\ \mathbf {F} _ {c} \mathbf {K} _ {a} \end{array} \right] \tag {10.86} $$ which, using (10.51), reduces to $$ \tilde {\mathbf {K}} = \mathbf {K} _ {a} \tag {10.87} $$ Similarly, substituting for $\Psi$ and M in (10.81), we have $$ \tilde {\mathbf {M}} = \left[ \begin{array}{l l} \mathbf {I} & \left(\mathbf {F} _ {c} \mathbf {K} _ {a}\right) ^ {T} \end{array} \right] \left[ \begin{array}{l l} \mathbf {M} _ {a} & \mathbf {0} \\ \mathbf {0} & \mathbf {0} \end{array} \right] \left[ \begin{array}{l} \mathbf {I} \\ \mathbf {F} _ {c} \mathbf {K} _ {a} \end{array} \right] \tag {10.88} $$ or $$ \tilde {\mathbf {M}} = \mathbf {M} _ {a} \tag {10.89} $$ Hence, in the static condensation we actually perform a Ritz analysis of the lumped mass model. It should be noted that in the analysis we calculate the q finite eigenvalues exactly (i.e., $\rho_{i} = \lambda_{i}$ for $i = 1, \ldots, q$ ) because the Ritz basis vectors span the q-dimensional subspace corresponding to the finite eigenvalues. In practice, the evaluation of the vectors $\Psi$ in (10.85) is not necessary (and would be costly), and instead the Ritz analysis is better carried out using $$ \boldsymbol {\Psi} = \left[ \begin{array}{l} \mathbf {F} _ {a} \\ \mathbf {F} _ {c} \end{array} \right] \tag {10.90} $$ Since the vectors in (10.90) span the same subspace as the vectors in (10.85), the same eigenvalues and eigenvectors are calculated employing either set of base vectors. Specifically, using (10.90), we obtain in the Ritz analysis the reduced eigenproblem $$ \mathbf {F} _ {a} \mathbf {x} = \lambda \mathbf {F} _ {a} \mathbf {M} _ {a} \mathbf {F} _ {a} \mathbf {x} \tag {10.91} $$ To show that this eigenproblem is indeed equivalent to the problem in (10.45), we premultiply both sides in (10.91) by $K_{a}$ and use the transformation $x = K_{a}\tilde{x}$ , giving $K_{a}\tilde{x} = \lambda M_{a}\tilde{x}$ , i.e., the problem in (10.45). EXAMPLE 10.17: Use the Ritz analysis procedure to perform static condensation of the massless degrees of freedom in the problem $K\phi = \lambda M\phi$ considered in Example 10.12. We first need to evaluate the Ritz basis vectors given in (10.90). This was done in Example 10.13, where we found that $$ \mathbf {F} _ {a} = \left[ \begin{array}{l l} 2 & 2 \\ 2 & 4 \end{array} \right]; \quad \mathbf {F} _ {c} = \left[ \begin{array}{l l} 1 & 1 \\ 2 & 3 \end{array} \right] $$ The Ritz reduction given in (10.91) thus yields the eigenproblem $$ \left[ \begin{array}{l l} 2 & 2 \\ 2 & 4 \end{array} \right] \mathbf {x} = \lambda \left[ \begin{array}{l l} 1 2 & 1 6 \\ 1 6 & 2 4 \end{array} \right] \mathbf {x} $$ Finally, we should note that the use of the Ritz basis vectors in (10.85) [(or in (10.90)] is also known as the Guyan reduction (see R. J. Guyan [A]). In the Guyan scheme the Ritz vectors are used to operate on a lumped mass matrix with zero elements on the diagonal as in (10.88) or on general full lumped or consistent mass matrices. In this reduction the $\phi_{a}$ degrees of freedom are frequently referred to as dynamic degrees of freedom. # 10.3.3 Component Mode Synthesis As for the static condensation procedure, the component mode synthesis is, in fact, a Ritz analysis, and the method might have been presented in the previous section as a specific application. However, as was repeatedly pointed out, the most important aspect in a Ritz analysis is the selection of appropriate Ritz basis vectors because the results can be only as good as the Ritz basis vectors allow them to be. The specific scheme used in the component mode synthesis is of particular interest, which is the reason we want to devote a separate section to the discussion of the method. The component mode synthesis has been developed to a large extent as a natural consequence of the analysis procedure followed in practice when large and complex structures are analyzed. The general practical procedure is that different groups perform the analyses of different components of the structure under consideration. For example, in a plant analysis, one group may analyze a main pipe and another group a piping system attached to it. In a first preliminary analysis, both groups work separately and model the effects of the other components on the specific component that they consider in an approximate manner. For example, in the analysis of the two piping systems referred to above, the group analyzing the side branch may assume full fixity at the point of intersection with the main pipe, and the group analyzing the main pipe may introduce a concentrated spring and mass to allow for the side branch. The advantage of considering the components of the structure separately is primarily one of time scheduling; i.e., the separate groups can work on the analyses and designs of the components at the same time. It is primarily for this reason that the component mode synthesis is very appealing in the analysis and design of large structural systems. Assume that the preliminary analyses of the components have been carried out and that the complete structure shall now be analyzed. It is at this stage that the component mode synthesis is a natural procedure to use. Namely, with the mode shape characteristics of each component known, it appears natural to use this information in estimating the frequencies and mode shapes of the complete structure. The specific procedure may vary (see R. R. Craig, Jr. [A]), but, in essence, the mode shapes of the components are used in a Rayleigh-Ritz analysis to calculate approximate mode shapes and frequencies of the complete structure. Consider for illustration that each component structure was obtained by fixing all its boundary degrees of freedom and denote the stiffness matrices of the component structures by $K_{I}, K_{II}, \ldots, K_{M}$ (see Example 10.18). Assume that only component structures L - 1 and L connect, $L = 2, \ldots, M$ ; then we can write for the stiffness matrix of the complete structure, $$ \mathbf {K} = \left[ \begin{array}{c c c c c c} \mathbf {K} _ {1} & & & & & \\ & \ddots & & & & \\ & & \cdot \mathbf {K} _ {\mathrm{II}} & & & \\ & & & \ddots & & \\ & & & & \ddots & \\ & & & & & \mathbf {K} _ {\mathrm{M}} \end{array} \right] \tag {10.92} $$ Using an analogous notation for the mass matrices, we also have $$ \mathbf {M} = \left[ \begin{array}{c c c c c c c} \mathbf {M} _ {\mathrm{I}} & \cdot & & & & & \\ & \cdot & \cdot & & & & \\ & & \cdot \mathbf {M} _ {\mathrm{II}} & \cdot & & & \\ & & & \cdot & \cdot & & \\ & & & & \cdot & \cdot & \\ & & & & & & \mathbf {M} _ {\mathrm{M}} \end{array} \right] \tag {10.93} $$ Assume that the lowest eigenvalues and corresponding eigenvectors of each component structure have been calculated; i.e., we have for each component structure, $$ \left. \begin{array}{c} \mathbf {K} _ {\mathrm{I}} \boldsymbol {\Phi} _ {\mathrm{I}} = \mathbf {M} _ {\mathrm{I}} \boldsymbol {\Phi} _ {\mathrm{I}} \boldsymbol {\Lambda} _ {\mathrm{I}} \\ \mathbf {K} _ {\mathrm{II}} \boldsymbol {\Phi} _ {\mathrm{II}} = \mathbf {M} _ {\mathrm{II}} \boldsymbol {\Phi} _ {\mathrm{II}} \boldsymbol {\Lambda} _ {\mathrm{II}} \\ \vdots \quad \vdots \\ \mathbf {K} _ {\mathrm{M}} \boldsymbol {\Phi} _ {\mathrm{M}} = \mathbf {M} _ {\mathrm{M}} \boldsymbol {\Phi} _ {\mathrm{M}} \boldsymbol {\Lambda} _ {\mathrm{M}} \end{array} \right\} \tag {10.94} $$ where $\Phi_L$ and $\Lambda_L$ are the matrices of calculated eigenvectors and eigenvalues of the $L$ th component structure. In a component mode synthesis, approximate mode shapes and frequencies can be obtained by performing a Rayleigh-Ritz analysis with the following assumed loads on the right-hand side of (10.79), $$ \mathbf {R} = \left[ \begin{array}{c c c c} \boldsymbol {\Phi} _ {\mathrm{I}} & \mathbf {0} & \mathbf {0} & \dots \\ \mathbf {0} & \mathbf {I} _ {\mathrm{I}, \mathrm{II}} & \mathbf {0} \\ \boldsymbol {\Phi} _ {\mathrm{II}} & \mathbf {0} & \mathbf {0} & \dots \\ \mathbf {0} & \mathbf {0} & \mathbf {I} _ {\mathrm{II}, \mathrm{III}} \\ \vdots & \vdots & \vdots & \dots \\ \boldsymbol {\Phi} _ {\mathrm{M}} & \mathbf {0} & & \end{array} \right] \tag {10.95} $$ where $I_{L-1,L}$ is a unit matrix of order equal to the connection degrees of freedom between component structures L - 1 and L. The unit matrices correspond to loads that are applied to the connection degrees of freedom of the component structures. Since in the derivation of the mode shape matrices used in (10.95) the component structures were fixed at their boundaries, the unit loads have the effect of releasing these connection degrees of freedom. If, on the other hand, the connection degrees of freedom have been included in the analysis of the component structures, we may dispense with the unit matrices in R. An important consideration is the accuracy that can be expected in the above component mode synthesis. Since a Ritz analysis is performed, all accuracy considerations discussed in Section 10.3.2 are directly applicable; i.e., the analysis yields upper bounds to the exact eigenvalues of the problem $K\phi = \lambda M\phi$ . However, the actual accuracy achieved in the solution is not known, although it can be evaluated, for example, as described in Section 10.4. The fact that the solution accuracy is highly dependent on the vectors used in R (i.e., the Ritz basis vectors) is, as in all Ritz analyses, the main defect of the method. However, in practice, reasonable accuracy can often be obtained because the eigenvectors corresponding to the smallest eigenvalues of the component structures are used in R. We demonstrate the analysis procedure in the following example. EXAMPLE 10.18: Consider the eigenproblem $\mathbf{K}\phi = \lambda \mathbf{M}\phi$ , where $$ \mathbf {K} = \left[ \begin{array}{c c c c c} 2 & - 1 & & & \\ - 1 & 2 & - 1 & & \\ & - 1 & 2 & - 1 & \\ & & - 1 & \boxed {2} & - 1 \\ & & & - 1 & 1 \end{array} \right]; \quad \mathbf {M} = \left[ \begin{array}{c c c c c} 1 & & & & \\ & 1 & & \\ & & 1 & & \\ & & & \boxed {1} & \\ & & & & \frac {1}{2} \end{array} \right] $$ Use the substructure eigenproblems indicated by the dashed lines in K and M to establish the load matrix given in (10.95) for a component mode synthesis analysis. Then calculate eigenvalue and eigenvector approximations. Here we have for substructure I, $$ \mathbf {K} _ {\mathrm{I}} = \left[ \begin{array}{c c} 2 & - 1 \\ - 1 & 2 \end{array} \right]; \quad \mathbf {M} _ {\mathrm{I}} = \left[ \begin{array}{c c} 1 & 0 \\ 0 & 1 \end{array} \right] $$ with the eigensolution $$ \lambda_ {1} = 1, \quad \lambda_ {2} = 3; \quad \phi_ {1} = \left[ \begin{array}{l} \frac {\sqrt {2}}{2} \\ \frac {\sqrt {2}}{2} \end{array} \right], \quad \phi_ {2} = \left[ \begin{array}{l} - \frac {\sqrt {2}}{2} \\ \frac {\sqrt {2}}{2} \end{array} \right] $$ and for substructure II, $$ \mathbf {K} _ {\mathrm{II}} = \left[ \begin{array}{c c} 2 & - 1 \\ - 1 & 1 \end{array} \right]; \quad \mathbf {M} _ {\mathrm{II}} = \left[ \begin{array}{c c} 1 & 0 \\ 0 & \frac {1}{2} \end{array} \right] $$ with the eigensolution $$ \lambda_ {1} = 2 - \sqrt {2}, \quad \lambda_ {2} = 2 + \sqrt {2}; \quad \phi_ {1} = \left[ \begin{array}{c} \frac {\sqrt {2}}{2} \\ 1 \end{array} \right], \quad \phi_ {2} = \left[ \begin{array}{c} - \frac {\sqrt {2}}{2} \\ 1 \end{array} \right] $$ Thus we have for the matrix $\mathbf{R}$ in (10.95), $$ \mathbf {R} = \left[ \begin{array}{c c c} \frac {\sqrt {2}}{2} & - \frac {\sqrt {2}}{2} & 0 \\ \frac {\sqrt {2}}{2} & \frac {\sqrt {2}}{2} & 0 \\ 0 & 0 & 1 \\ \frac {\sqrt {2}}{2} & - \frac {\sqrt {2}}{2} & 0 \\ 1 & 1 & 0 \end{array} \right] $$ Now performing the Ritz analysis as given in (10.79) to (10.84), we obtain $$ \tilde {\mathbf {K}} = \left[ \begin{array}{c c c} 2 2. 4 0 & 5. 3 2 8 & 7. 2 4 3 \\ 5. 3 2 8 & 2. 2 5 7 & 1. 5 8 6 \\ 7. 2 4 3 & 1. 5 8 6 & 3 \end{array} \right] $$ $$ \tilde {\mathbf {M}} = \left[ \begin{array}{l l l} 2 2 2. 4 & 5 0. 6 9 & 7 7. 6 9 \\ 5 0. 6 9 & 1 1. 9 4 & 1 7. 5 9 \\ 7 7. 6 9 & 1 7. 5 9 & 2 7. 5 \end{array} \right] $$ and hence, $\pmb{\rho} = \begin{bmatrix} 0.098 & & \\ & 2.83 & \\ & & 1.82 \end{bmatrix}$ $$ \overline {{{\Phi}}} = \left[ \begin{array}{c c c} 0. 2 0 7 & - 0. 7 7 3 & 0. 0 0 6 9 0 \\ 0. 1 8 1 & 0. 0 9 8 4 & - 0. 0 6 5 5 \\ 0. 5 0 9 & 1. 4 7 & 0. 4 4 3 \\ 0. 5 9 4 & - 0. 3 8 5 & - 0. 1 6 6 \\ 0. 6 5 5 & 0. 5 7 4 & - 0. 9 7 8 \end{array} \right] $$ The exact eigenvalues are $$ \lambda_ {1} = 0. 0 9 7 8 9 \quad \lambda_ {2} = 0. 8 2 4; \quad \lambda_ {3} = 2. 0 0; \quad \lambda_ {4} = 3. 1 8; \quad \lambda_ {5} = 3. 9 0 $$ and hence we note that we obtained in $\rho_{1}$ a good approximation to $\lambda_{1}$ , but $\rho_{2}$ and $\rho_{3}$ do not represent approximations to eigenvalues. # 10.3.4 Exercises # 10.7. Consider the eigenproblem $$ \left[ \begin{array}{r r r} 6 & - 1 & 0 \\ - 1 & 4 & - 1 \\ 0 & - 1 & 2 \end{array} \right] \boldsymbol {\phi} = \lambda \left[ \begin{array}{r r r} 0 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & 1 \end{array} \right] \boldsymbol {\phi} $$ Perform the static condensation as usually performed [see (10.46)] and then by the Rayleigh-Ritz analysis procedure [see (10.51)]. 10.8. Consider the eigenproblem in Exercise 10.1. Perform a Rayleigh-Ritz analysis with the two vectors $$ \boldsymbol {\psi} _ {1} = \left[ \begin{array}{l} 1 \\ 1 \\ 1 \end{array} \right]; \quad \boldsymbol {\psi} _ {2} = \left[ \begin{array}{c} 1 \\ - 1 \\ 1 \end{array} \right] $$ to calculate an approximation to the smallest eigenvalue and corresponding eigenvector. 10.9. It is being claimed that if, in the solution of the generalized eigenproblem $K\phi = \lambda M\phi$ , the Ritz vectors are $$ \psi_ {1} = \phi_ {1} + 2 \phi_ {2} $$ $$ \psi_ {2} = 3 \phi_ {1} - \phi_ {2} $$ where $\phi_{1}$ and $\phi_{2}$ are the eigenvectors corresponding to $\lambda_{1}$ and $\lambda_{2}$ , then the Rayleigh-Ritz analysis will give the exact eigenvalues $\lambda_{1}$ and $\lambda_{2}$ and the corresponding eigenvectors $\phi_{1}$ and $\phi_{2}$ . Show explicitly that this result is indeed obtained. 10.10. Consider the following spring system. (a) Evaluate the exact smallest frequency of the system. (b) Evaluate an approximation of the smallest frequency by using the component mode synthesis technique in Section 10.3.3. Use only the eigenvector of the smallest frequency of each component in the system. 

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Component I k m* K m m m K K K K = 10 k = 1 m = 2 m* = 1 Component II

# 10.4 SOLUTION ERRORS An important part of an eigenvalue and vector solution is to estimate the accuracy with which the required eigensystem has been calculated. Since an eigensystem solution is necessarily iterative, the solution should be terminated once convergence within the prescribed tolerances giving the actual accuracy has been obtained. When one of the approximate solution techniques outlined in Section 10.3 is used, an estimate of the actual solution accuracy obtained is of course also important. # 10.4.1 Error Bounds In order to identify the accuracy that has been obtained in an eigensolution, we recall that the equation to be solved is $$ \mathbf {K} \boldsymbol {\phi} = \lambda \mathbf {M} \boldsymbol {\phi} \tag {10.96} $$ Let us first assume that using any one solution procedure we obtained an approximation $\overline{\lambda}$ and $\overline{\phi}$ to an eigenpair. Then without regard to how the values have been obtained, we can evaluate a residual vector that gives important information about the accuracy with which $\overline{\lambda}$ and $\overline{\phi}$ approximate the eigenpair. The results are given in (10.101) to (10.104). We then present also error bound calculations useful in solutions based on inverse iterations and a simple error measure. # Standard Eigenproblem Consider first that $\mathbf{M} = \mathbf{I}$ . In that case we can write $$ \mathbf {r} = \mathbf {K} \overline {{{\phi}}} - \overline {{{\lambda}}} \overline {{{\phi}}} \tag {10.97} $$ and using the relations in (10.12) and (10.13), we have $$ \mathbf {r} = \Phi (\Lambda - \bar {\lambda} \mathbf {I}) \Phi^ {T} \overline {{\phi}} \tag {10.98} $$ or because $\bar{\lambda}$ is not equal but only close to an eigenvalue, we have $$ \overline {{{\boldsymbol {\Phi}}}} = \boldsymbol {\Phi} (\boldsymbol {\Lambda} - \overline {{{\lambda}}} \mathbf {I}) ^ {- 1} \boldsymbol {\Phi} ^ {T} \mathbf {r} \tag {10.99} $$ Hence, because $\| \overline{\phi}\| _2 = 1$ , taking norms we obtain $$ 1 \leq \left\| (\mathbf {\Lambda} - \overline {{{\lambda}}} \mathbf {I}) ^ {- 1} \right\| _ {2} \left\| \mathbf {r} \right\| _ {2} \tag {10.100} $$ But since $\| (\mathbf{\Lambda} - \overline{\lambda}\mathbf{I})^{-1}\| _2 = \max_i\frac{1}{|\lambda_i - \overline{\lambda}|}$ we have $\min_{i}|\lambda_{i} - \overline{\lambda} |\leq \| \mathbf{r}\|_{2}$ (10.101) Therefore, a conclusive statement can be made about the accuracy with which $\lambda$ approximates an eigenvalue $\lambda_{i}$ by evaluating $\| \mathbf{r}\| _2$ as expressed in (10.101). This is quite different from the information that could be obtained from the evaluation of the residual vector $\mathbf{r}$ in the solution of the static equilibrium equations. Although the relation in (10.101) establishes that $\bar{\lambda}$ is close to an eigenvalue provided that $\|r\|_{2}$ is small, it should be recognized that the relation does not tell which eigenvalue is approximated. In fact, to identify which specific eigenvalue has been approximated, it is necessary to use the Sturm sequence property (see Section 10.2.2 and the following example). EXAMPLE 10.19: Consider the eigenproblem $\mathbf{K}\phi = \lambda \phi$ , where $$ \mathbf {K} = \left[ \begin{array}{r r r} 3 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 3 \end{array} \right] $$ The eigensolution is $\lambda_1 = 1, \lambda_2 = 3, \lambda_3 = 4$ , and $$ \boldsymbol {\phi} _ {1} = \frac {1}{\sqrt {6}} \left[ \begin{array}{l} 1 \\ 2 \\ 1 \end{array} \right]; \quad \boldsymbol {\phi} _ {2} = \frac {1}{\sqrt {2}} \left[ \begin{array}{c} 1 \\ 0 \\ - 1 \end{array} \right]; \quad \boldsymbol {\phi} _ {3} = \frac {1}{\sqrt {3}} \left[ \begin{array}{c} 1 \\ - 1 \\ 1 \end{array} \right] $$ Assume that we calculated $$ \overline {{{\lambda}}} = 3. 1 \quad \text { and } \quad \overline {{{\Phi}}} = \left[ \begin{array}{c} 0. 7 \\ 0. 1 4 1 4 \\ - 0. 7 \end{array} \right] $$ as approximations to $\lambda_{2}$ and $\phi_{2}$ . Apply the error bound relation in (10.101). We have in this case $$ \mathbf {r} = \left[ \begin{array}{r r r} 3 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 3 \end{array} \right] \left[ \begin{array}{l} 0. 7 \\ 0. 1 4 1 4 \\ - 0. 7 \end{array} \right] - 3. 1 \left[ \begin{array}{l l l} 1 & & \\ & 1 & \\ & & 1 \end{array} \right] \left[ \begin{array}{l} 0. 7 \\ 0. 1 4 1 4 \\ - 0. 7 \end{array} \right] $$ Hence $\mathbf{r} = \begin{bmatrix} -0.2114\\ -0.1555\\ -0.2114 \end{bmatrix};\qquad \| \mathbf{r}\| _2 = 0.3370$ The relation in (10.101) now gives $$ \left| \lambda_ {2} - \bar {\lambda} \right| \leq 0. 3 3 7 0 $$ which is indeed true because $\overline{\lambda} - \lambda_2 = 0.1$ . Assume now that we have calculated only $\overline{\lambda}$ and $\overline{\Phi}$ and do not know which eigenvalue and eigenvector they approximate. In this case we can use the relation in (10.101) to establish bounds on the unknown exact eigenvalue in order to apply Sturm sequence checks (see Section 10.2.2). For the example considered here we have $$ 2. 7 6 3 0 \leq \lambda_ {i} \leq 3. 4 3 7 0 $$ Let us use as a lower bound 2.7 and as an upper bound 3.5. The $LDL^{T}$ triangular factorization of K - $\mu I$ gives, at $\mu = 2.7$ , $$ \begin{array}{l} \left[ \begin{array}{c c c} 0. 3 & - 1 & 0 \\ - 1 & - 0. 7 & - 1 \\ 0 & - 1 & 0. 3 \end{array} \right] \\ = \left[ \begin{array}{c c c} 1 & & \\ - 3. 3 3 3 & 1 & \\ 0 & 0. 2 4 7 9 & 1 \end{array} \right] \left[ \begin{array}{c c c} 0. 3 & & \\ & - 4. 0 3 3 & \\ & & 0. 3 2 4 8 \end{array} \right] \left[ \begin{array}{c c c} 1 & - 3. 3 3 3 & 0 \\ & 1 & 0. 2 4 7 9 \\ & & 1 \end{array} \right] \tag {a} \\ \end{array} $$ and at $\mu = 3.5$ , $$ \left[ \begin{array}{c c c} - 0. 5 & - 1 & 0 \\ - 1 & - 1. 5 & - 1 \\ 0 & - 1 & - 0. 5 \end{array} \right] = \left[ \begin{array}{c c c} 1 & & \\ 2 & 1 & \\ 0 & - 2 & 1 \end{array} \right] \left[ \begin{array}{c c c} - 0. 5 & & \\ & 0. 5 & \\ & & - 2. 5 \end{array} \right] \left[ \begin{array}{c c c} 1 & 2 & 0 \\ & 1 & - 2 \\ & & 1 \end{array} \right] \tag {b} $$ But there is one negative element in $\mathbf{D}$ in (a) and there are two negative elements in $\mathbf{D}$ in (b); hence, we can conclude that $2.7 < \lambda_2 < 3.5$ . Furthermore, it follows that $\overline{\lambda}$ and $\overline{\Phi}$ are approximations to $\lambda_2$ and $\Phi_2$ . Considering now the accuracy with which $\overline{\Phi}$ approximates an eigenvector, an analysis equivalent to this one not only requires the evaluation of $\|r\|_{2}$ but the spacing between the individual eigenvalues is also needed. In actual analysis this spacing is known only approximately because the eigenvalues have been evaluated only to a specific accuracy. Assume that $\overline{\lambda}$ and $\overline{\Phi}$ have been calculated, where $\| \overline{\Phi}\| _2 = 1$ , and that $\overline{\lambda}$ approximates the eigenvalues $\lambda_{i}, i = p, \ldots, q$ . For the error analysis we also assume that the eigenvalues $\lambda_{i}$ for all $i$ but $i \neq p, \ldots, q$ are known (although we would need to use the calculated eigenvalues here). The final result of the accuracy analysis is that if $|\lambda_{i} - \overline{\lambda}| \leq \| \mathbf{r}\| _2$ for $i = p, \ldots, q$ and $|\lambda_{i} - \overline{\lambda}| \geq s$ for all $i, i \neq p, \ldots, q$ , then there is a vector $\tilde{\Phi} = \alpha_p \Phi_p + \cdots + \alpha_q \Phi_q$ , for which $\| \overline{\Phi} - \tilde{\Phi}\| _2 \leq \| \mathbf{r}\| _2 / s$ (see Example 10.20). Therefore, if $\overline{\lambda}$ is an approximation to a single eigenvalue $\lambda_{i}$ , the corresponding vector $\overline{\Phi}$ is an approximation to $\Phi_{i}$ , where $$ \| \overline{\boldsymbol{\Phi}} - \alpha_{i}\boldsymbol{\Phi}_{i}\|_{2}\leq \frac{\|\boldsymbol{\mathbf{r}}\|_{2}}{s};\qquad s = \min_{\substack{\text{all} j\\ j\neq i}}|\lambda_{j} - \overline{\lambda} | \tag{10.102} $$ However, if $\overline{\lambda}$ is close to a number of eigenvalues $\lambda_{p},\ldots ,\lambda_{q}$ , then the analysis only shows that the corresponding vector $\overline{\Phi}$ is close to a vector that lies in the subspace corresponding to $\Phi_p,\ldots ,\Phi_q$ . In practical analysis (i.e., mode superposition in dynamic response calculations), this is most likely all that is required because the close eigenvalues may almost be dealt with as equal eigenvalues, in which case the calculated eigenvectors would also not be unique but lie in the subspace corresponding to the equal eigenvalues. In the following we first give the proof for the accuracy with which $\overline{\Phi}$ approximates an eigenvector and then demonstrate the results by means of examples. EXAMPLE 10.20: Assume that we have calculated $\overline{\lambda}$ , $\overline{\Phi}$ , with $\|\overline{\Phi}\|_2 = 1$ , as eigenvalue and eigenvector approximations and that $\mathbf{K}\overline{\Phi} - \overline{\lambda}\overline{\Phi} = \mathbf{r}$ . Consider the case in which $|\lambda_i - \overline{\lambda}| \leq \|\mathbf{r}\|_2$ for $i = 1, \ldots, q$ and $|\lambda_i - \overline{\lambda}| \geq s$ for $i = q + 1, \ldots, n$ . Show that $\|\overline{\Phi} - \tilde{\Phi}\|_2 \leq \|\mathbf{r}\|_2/s$ , where $\tilde{\Phi}$ is a vector in the subspace that corresponds to $\Phi_1, \ldots, \Phi_q$ . The calculated eigenvector approximation $\overline{\phi}$ can be written as $$ \overline {{{\phi}}} = \sum_ {i = 1} ^ {n} \alpha_ {i} \phi_ {i} $$ Using $\tilde{\phi} = \sum_{i=1}^{q} \alpha_i \phi_i$ , we have $$ \left\| \overline {{{\boldsymbol {\Phi}}}} - \bar {\boldsymbol {\Phi}} \right\| _ {2} = \left\| \sum_ {i = q + 1} ^ {n} \alpha_ {i} \boldsymbol {\Phi} _ {i} \right\| _ {2} $$ or, because $\Phi_i^T\Phi_j = \delta_{ij},$ $\| \overline{\Phi} -\tilde{\Phi}\| _2 = \left(\sum_{i = q + 1}^{n}\alpha_i^2\right)^{1 / 2}$ (a) But $\| \mathbf{r}\| _2 = \| \mathbf{K}\overline{\phi} -\overline{\lambda}\overline{\phi}\| _2$ $$ = \left\| \sum_ {i = 1} ^ {n} \alpha_ {i} \left(\lambda_ {i} - \bar {\lambda}\right) \phi_ {i} \right\| _ {2} $$ or $\| \mathbf{r}\| _2 = \left(\sum_{i = 1}^{n}\alpha_i^2 (\lambda_i - \overline{\lambda})^2\right)^{1 / 2}$ which gives $$ \| \mathbf {r} \| _ {2} \geq s \left(\sum_ {i = q + 1} ^ {n} \alpha_ {i} ^ {2}\right) ^ {1 / 2} \tag {b} $$ Hence, combining (a) and (b), we obtain $$ \| \overline {{{\boldsymbol {\phi}}}} - \tilde {\boldsymbol {\phi}} \| _ {2} \leq \frac {\| \mathbf {r} \| _ {2}}{s} $$ EXAMPLE 10.21: Consider the eigenproblem in Example 10.19. Assume that $\lambda_{1}$ and $\lambda_{3}$ are known (i.e., $\lambda_{1} = 1$ , $\lambda_{3} = 4$ ) and that $\overline{\lambda}$ and $\overline{\phi}$ given in Example 10.19 have been evaluated. (In actual analysis we would have only approximations to $\lambda_{1}$ and $\lambda_{3}$ , and all error bound calculations would be approximate.) Estimate the accuracy with which $\overline{\phi}$ approximates $\phi_{2}$ . For the estimate we use the relation in (10.102). In this case we have $$ \| \overline {{{\boldsymbol {\Phi}}}} - \alpha_ {2} \boldsymbol {\Phi} _ {2} \| _ {2} \leq \frac {0 . 3 3 7 0}{s} $$ with $s = \min_{i = 1,3}|\lambda_i - \overline{\lambda} |$ Hence, since $\overline{\lambda} = 3.1$ , we have $s = 0.9$ and $$ \| \overline {{{\boldsymbol {\phi}}}} - \alpha_ {2} \boldsymbol {\phi} _ {2} \| _ {2} \leq 0. 3 7 4 4 $$ Evaluating $\| \overline{\phi} - \phi_2\| _2$ exactly, we have $$ \left\| \overline {{{\boldsymbol {\Phi}}}} - \boldsymbol {\Phi} _ {2} \right\| _ {2} = \left[ \left(0. 7 - \frac {1}{\sqrt {2}}\right) ^ {2} + (0. 1 4 1 4 - 0) ^ {2} + \left(- 0. 7 + \frac {1}{\sqrt {2}}\right) ^ {2} \right] ^ {1 / 2} = 0. 1 4 1 8 $$ EXAMPLE 10.22: Consider the eigenproblem $\mathbf{K}\phi = \lambda \phi$ , where $$ \mathbf {K} = \left[ \begin{array}{c c} 1 0 0 & - 1 \\ - 1 & 1 0 0 \end{array} \right] $$ The eigenvalues and eigenvectors of the problem are $$ \lambda_ {1} = 9 9, \phi_ {1} = \frac {1}{\sqrt {2}} \left[ \begin{array}{l} 1 \\ 1 \end{array} \right]; \quad \lambda_ {2} = 1 0 1, \phi_ {2} = \frac {1}{\sqrt {2}} \left[ \begin{array}{l} 1 \\ - 1 \end{array} \right] $$ Assume that we have calculated eigenvalue and eigenvector approximations $\overline{\lambda} = 100$ , $\overline{\Phi} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ . Evaluate $\mathbf{r}$ and thus establish the relations given in (10.101) and (10.102). First, we calculate r as given in (10.97), $$ \mathbf {r} = \left[ \begin{array}{l l} 1 0 0 & - 1 \\ - 1 & 1 0 0 \end{array} \right] \left[ \begin{array}{l} 1 \\ 0 \end{array} \right] - 1 0 0 \left[ \begin{array}{l} 1 \\ 0 \end{array} \right] = \left[ \begin{array}{l} 0 \\ - 1 \end{array} \right] $$ Hence, $\| \mathbf{r}\| _2 = 1$ and (10.101) yields $$ \min _ {i} \left| \lambda_ {i} - \overline {{{\lambda}}} \right| \leq 1 \tag {a} $$ Therefore, we can conclude that an eigenvalue has been approximated with about 1 percent or less error. Since we know $\lambda_{1}$ and $\lambda_{2}$ , we can compare $\overline{\lambda}$ with $\lambda_{1}$ or $\lambda_{2}$ and find that (a) does indeed hold. Considering now the eigenvector approximation $\overline{\Phi}$ , we note that $\overline{\Phi}$ does not approximate either $\Phi_1$ or $\Phi_2$ . This is also reflected by evaluating the relation (10.102). Assuming that $\overline{\Phi}$ is an approximation to $\Phi_1$ , which gives $s = 1$ , we have $$ \left\| \overline {{{\boldsymbol {\phi}}}} - \alpha_ {1} \boldsymbol {\phi} _ {1} \right\| _ {2} \leq 1 $$