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453 lines
30 KiB
Markdown
<!-- source-page: 861 -->
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$\mathbf{M} = \mathbf{I}$ would also be eigenvectors. For example, we could use
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$$
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\boldsymbol {\phi} _ {1} = \left[ \begin{array}{c} \frac {1}{\sqrt {2}} \\ \frac {1}{\sqrt {2}} \\ 0 \end{array} \right] \quad \text { and } \quad \boldsymbol {\phi} _ {2} = \left[ \begin{array}{c} \frac {1}{\sqrt {2}} \\ - \frac {1}{\sqrt {2}} \\ 0 \end{array} \right]
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$$
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That these are indeed eigenvectors corresponding to $\lambda_{1} = \lambda_{2} = 2$ can again be checked as in Example 10.1. It should be noted that any eigenvectors $\phi_{1}$ and $\phi_{2}$ provide a basis for the unique two-dimensional eigenspace that corresponds to $\lambda_{1}$ and $\lambda_{2}$ .
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The solution to (10.4) for all p required eigenvalues and corresponding eigenvectors was established in (10.5). Using the relations in (10.10) and (10.11), we may now write
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$$
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\boldsymbol {\Phi} ^ {T} \mathbf {K} \boldsymbol {\Phi} = \boldsymbol {\Lambda} \tag {10.12}
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$$
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and $\mathbf{\Phi}^T\mathbf{M}\mathbf{\Phi} = \mathbf{I}$ (10.13)
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where the p columns of $\Phi$ are the eigenvectors. It is very important to note that (10.12) and (10.13) are conditions that the eigenvectors must satisfy, but that if the M-orthonormality and K-orthogonality are satisfied, the p vectors need not necessarily be eigenvectors unless p = n. In other words, assume that X stores p vectors, p < n, and that $X^{T}KX = D$ and $X^{T}MX = I$ ; then the vectors in X and the diagonal elements in D may or may not be eigenvectors and eigenvalues of (10.4). However, if p = n, then $X = \Phi$ and $D = \Lambda$ because only the eigenvectors span the complete n-dimensional space and diagonalize the matrices K and M. To underline this observation we present the following example.
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EXAMPLE 10.3: Consider the eigenproblem $\mathbf{K}\phi = \lambda \mathbf{M}\phi$ , where
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$$
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\mathbf {K} = \left[ \begin{array}{r r r} 2 & - 1 & 0 \\ - 1 & 4 & - 1 \\ 0 & - 1 & 2 \end{array} \right]; \quad \mathbf {M} = \left[ \begin{array}{r r r} \frac {1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac {1}{2} \end{array} \right]
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$$
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and the two vectors
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$$
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\mathbf {v} _ {1} = \left[ \begin{array}{c} 1 \\ \frac {1}{\sqrt {2}} \\ 0 \end{array} \right]; \quad \mathbf {v} _ {2} = \left[ \begin{array}{c} 1 \\ - \frac {1}{\sqrt {2}} \\ 0 \end{array} \right]
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$$
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Show that the vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ satisfy the orthogonality relations in (10.12) and (10.13) [i.e., the relations in (10.11) and (10.10)] but that they are not eigenvectors.
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For the check we let $\mathbf{v}_1$ and $\mathbf{v}_2$ be the columns in $\Phi$ , and we evaluate (10.12) and (10.13). Thus, we obtain
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$$
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\left[ \begin{array}{c c c} 1 & \frac {1}{\sqrt {2}} & 0 \\ 1 & - \frac {1}{\sqrt {2}} & 0 \end{array} \right] \left[ \begin{array}{c c c} 2 & - 1 & 0 \\ - 1 & 4 & - 1 \\ 0 & - 1 & 2 \end{array} \right] \left[ \begin{array}{c c} 1 & 1 \\ \frac {1}{\sqrt {2}} & - \frac {1}{\sqrt {2}} \\ 0 & 0 \end{array} \right] = \left[ \begin{array}{c c} (4 - \sqrt {2}) & 0 \\ 0 & (4 + \sqrt {2}) \end{array} \right] \tag {a}
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$$
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and
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$$
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\left[ \begin{array}{c c c} 1 & \frac {1}{\sqrt {2}} & 0 \\ 1 & - \frac {1}{\sqrt {2}} & 0 \end{array} \right] \left[ \begin{array}{c c c} \frac {1}{2} & & \\ & 1 & \\ & & \frac {1}{2} \end{array} \right] \left[ \begin{array}{c c} 1 & 1 \\ \frac {1}{\sqrt {2}} & - \frac {1}{\sqrt {2}} \\ 0 & 0 \end{array} \right] = \left[ \begin{array}{c c} 1 & 0 \\ 0 & 1 \end{array} \right]
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$$
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Hence, the orthogonality relations are satisfied. To show that $\mathbf{v}_1$ and $\mathbf{v}_2$ are not eigenvectors, we employ (10.8). For example,
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$$
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\mathbf {K} \mathbf {v} _ {1} = \left[ \begin{array}{l} 2 - \frac {1}{\sqrt {2}} \\ - 1 + \frac {4}{\sqrt {2}} \\ - \frac {1}{\sqrt {2}} \end{array} \right]; \quad \mathbf {M} \mathbf {v} _ {1} = \left[ \begin{array}{l} \frac {1}{2} \\ \frac {1}{\sqrt {2}} \\ 0 \end{array} \right]
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$$
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However, the vector $Kv_{1}$ cannot be equal to the vector $\alpha Mv_{1}$ , where $\alpha$ is a scalar; i.e., $Kv_{1}$ is not parallel to $Mv_{1}$ , and therefore $v_{1}$ is not an eigenvector. Similarly, $v_{2}$ is not an eigenvector and the values $(4 - \sqrt{2})$ and $(4 + \sqrt{2})$ calculated in (a) are not eigenvalues. The actual eigenvalues and corresponding eigenvectors are given in Example 10.4.
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In the preceding presentation we considered the properties of the eigenvectors of the problem $K\phi = \lambda M\phi$ , and we should now briefly comment on the properties of the eigenvectors calculated in the solution of the other eigenvalue problems of interest. The comment is simple: the orthogonality relations discussed here hold equally for the eigenvectors of the problems encountered in buckling analysis and heat transfer analysis. That is, we also have in buckling analysis, using the notation in (10.6),
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$$
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\left. \begin{array}{l l} \boldsymbol {\Phi} _ {i} ^ {T} {} ^ {t - \Delta t} \mathbf {K} \boldsymbol {\Phi} _ {j} = \delta_ {i j}; & \boldsymbol {\Phi} _ {i} ^ {T} {} ^ {t} \mathbf {K} \boldsymbol {\Phi} _ {j} = \lambda_ {i} \delta_ {i j} \\ \boldsymbol {\Phi} ^ {T} {} ^ {t - \Delta t} \mathbf {K} \boldsymbol {\Phi} = \mathbf {I}; & \boldsymbol {\Phi} ^ {T} {} ^ {t} \mathbf {K} \boldsymbol {\Phi} = \boldsymbol {\Lambda} \end{array} \right\} \tag {10.14}
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$$
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and in heat transfer analysis, using the notation in (10.7), we have
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$$
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\left. \begin{array}{l l} \boldsymbol {\Phi} _ {i} ^ {T} \mathbf {C} \boldsymbol {\Phi} _ {j} = \delta_ {i j}; & \boldsymbol {\Phi} _ {i} ^ {T} \mathbf {K} \boldsymbol {\Phi} _ {j} = \lambda_ {i} \delta_ {i j} \\ \boldsymbol {\Phi} ^ {T} \mathbf {C} \boldsymbol {\Phi} = \mathbf {I}; & \boldsymbol {\Phi} ^ {T} \mathbf {K} \boldsymbol {\Phi} = \boldsymbol {\Lambda} \end{array} \right\} \tag {10.15}
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$$
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As for the eigenproblem $K\phi = \lambda M\phi$ , the proof of the relations in (10.14) and (10.15) depends on the fact that the generalized eigenproblems can be transformed to a standard form. We discuss this matter further in Section 10.2.5.
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# 10.2.2 The Characteristic Polynomials
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# of the Eigenproblem $K\phi = \lambda M\phi$ and of Its Associated Constraint Problems
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An important property of the eigenvalues of the problem $K\Phi = \lambda M\Phi$ is that they are the roots of the characteristic polynomial,
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$$
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p (\lambda) = \det (\mathbf {K} - \lambda \mathbf {M}) \tag {10.16}
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$$
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We can show that this property derives from the basic relation in (10.8). Rewriting (10.8) in the form
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$$
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(\mathbf {K} - \lambda_ {i} \mathbf {M}) \boldsymbol {\phi} _ {i} = \mathbf {0} \tag {10.17}
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$$
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we observe that (10.8) can be satisfied only for nontrivial $\phi_{i}$ (i.e., $\phi_{i}$ not being equal to a null vector) provided that the matrix $\mathbf{K} - \lambda_{i}\mathbf{M}$ is singular. This means that if we factorize $\mathbf{K} - \lambda_{i}\mathbf{M}$ into a unit lower triangular matrix $\mathbf{L}$ and an upper triangular matrix $\mathbf{S}$ using Gauss elimination, we have $s_{nn} = 0$ . However, since
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$$
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p \left(\lambda_ {i}\right) = \det \mathbf {L S} = \prod_ {i = 1} ^ {n} s _ {i i} \tag {10.18}
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$$
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it follows that $p(\lambda_{i}) = 0$ . Furthermore, if $\lambda_{i}$ has multiplicity m, we also have $s_{n-1,n-1} = \cdots = s_{n-m+1,n-m+1} = 0$ . We should note that in the factorization of $K - \lambda_{i}M$ , interchanges may be needed, in which case the factorization of $K - \lambda_{i}M$ with its rows and possibly its columns interchanged is obtained (each row and each column interchange then introduces a sign change in the determinant which must be taken into account; see Section 2.2). If no interchanges are carried out, or row and corresponding column interchanges are performed, which in practice is nearly always possible (but see Example 10.4 for a case where it is not possible), the coefficient matrix remains symmetric. In this case we can write for (10.18)
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$$
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p \left(\lambda_ {i}\right) = \det \mathbf {L D L} ^ {T} = \prod_ {i = 1} ^ {n} d _ {i i} \tag {10.19}
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$$
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where $LDL^{T}$ is the factorization of K - $\lambda_{i}M$ or of the matrix derived from it by interchanging rows and corresponding columns, i.e., using a different ordering for the system degrees of freedom (see Section 8.2.5). The condition $s_{nn} = 0$ is now $d_{nn} = 0$ , and when $\lambda_{i}$ has multiplicity m, the last m elements in D are zero.
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In Section 8.2.5 we discussed the Sturm sequence property of the characteristic polynomials of the constraint problems associated with the problem $K\phi = \lambda\phi$ . The same properties that we observed in that discussion are applicable also to the characteristic polynomials of the constraint problems associated with the problem $K\phi = \lambda M\phi$ . The proof follows from the fact that the generalized eigenproblem $K\phi = \lambda M\phi$ can be transformed to a standard eigenproblem for which the Sturm sequence property of the characteristic polynomials holds. Referring the proof to Section 10.2.5, Example 10.11, let us summarize the important result.
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The eigenproblem of the $r$ th associated constraint problem corresponding to $\mathbf{K}\phi = \lambda \mathbf{M}\phi$ is given by
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$$
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\mathbf {K} ^ {(r)} \boldsymbol {\phi} ^ {(r)} = \lambda^ {(r)} \mathbf {M} ^ {(r)} \boldsymbol {\phi} ^ {(r)} \tag {10.20}
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$$
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where all matrices are of order $n - r$ and $\mathbf{K}^{(r)}$ and $\mathbf{M}^{(r)}$ are obtained by deleting from $\mathbf{K}$ and $\mathbf{M}$ the last $r$ rows and columns. The characteristic polynomial of the $r$ th associated constraint problem is
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$$
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p ^ {(r)} (\lambda^ {(r)}) = \det (\mathbf {K} ^ {(r)} - \lambda^ {(r)} \mathbf {M} ^ {(r)}) \tag {10.21}
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$$
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and as for the special case $\mathbf{M} = \mathbf{I}$ , the eigenvalues of the $(r + 1)$ st constraint problem separate those of the $r$ th constraint problem; i.e., as stated in (8.38), we again have
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$$
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\lambda_ {1} ^ {(r)} \leq \lambda_ {1} ^ {(r + 1)} \leq \lambda_ {2} ^ {(r)} \leq \lambda_ {2} ^ {(r + 1)} \leq \dots \leq \lambda_ {n - r - 1} ^ {(r)} \leq \lambda_ {n - r - 1} ^ {(r + 1)} \leq \lambda_ {n - r} ^ {(r)} \tag {10.22}
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$$
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Consider the following example.
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<!-- source-page: 864 -->
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EXAMPLE 10.4: Consider the eigenproblem $\mathbf{K}\phi = \lambda \mathbf{M}\phi$ , where
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$$
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\mathbf {K} = \left[ \begin{array}{r r r} 2 & - 1 & 0 \\ - 1 & 4 & - 1 \\ 0 & - 1 & 2 \end{array} \right]; \quad \mathbf {M} = \left[ \begin{array}{c c c} \frac {1}{2} & & \\ & 1 & \\ & & \frac {1}{2} \end{array} \right]
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$$
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(a) Calculate the eigenvalues using the characteristic polynomial as defined in (10.16).
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(b) Solve for the eigenvectors $\Phi_i$ , $i = 1, 2, 3$ , by using the relation in (10.17) and the M-orthonormality condition of the eigenvectors.
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(c) Calculate the eigenvalues of the associated constraint problems and show that the eigenvalue separation property given in (10.22) holds.
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Using (10.16), we obtain the characteristic polynomial
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$$
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p (\lambda) = (2 - \frac {1}{2} \lambda) (4 - \lambda) (2 - \frac {1}{2} \lambda) - (- 1) (- 1) (2 - \frac {1}{2} \lambda) - (- 1) (- 1) (2 - \frac {1}{2} \lambda)
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$$
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Hence, $p(\lambda) = -\frac{1}{4}\lambda^3 + 3\lambda^2 - 11\lambda + 12$
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and we have $\lambda_{1} = 2;$ $\lambda_{2} = 4;$ $\lambda_{3} = 6$
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To obtain the corresponding eigenvectors, we use the relation in (10.17). For $\lambda_{1}$ we have
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$$
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\left[ \begin{array}{r r r} 1 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 1 \end{array} \right] \boldsymbol {\phi} _ {1} = \mathbf {0} \tag {a}
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$$
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The coefficient matrix $\mathbf{K} - \lambda_1\mathbf{M}$ in (a) can be factorized into $\mathbf{LDL}^T$ without interchanges. Using the procedure described in Section 8.2.2, we obtain
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$$
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\left[ \begin{array}{r r r} 1 & & \\ - 1 & 1 & \\ 0 & - 1 & 1 \end{array} \right] \left[ \begin{array}{r r r} 1 & & \\ & 1 & \\ & & 0 \end{array} \right] \left[ \begin{array}{r r r} 1 & - 1 & 0 \\ & 1 & - 1 \\ & & 1 \end{array} \right] \boldsymbol {\phi} _ {1} = \boldsymbol {0} \tag {b}
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$$
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We note that $d_{33} = 0.0$ . To evaluate $\phi_1$ , we obtain from (b),
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$$
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\left[ \begin{array}{c c c} 1 & - 1 & 0 \\ & 1 & - 1 \\ & & 0 \end{array} \right] \boldsymbol {\phi} _ {1} = \mathbf {0}
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$$
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Using also $\phi_1^T\mathbf{M}\phi_1 = 1$ , we have
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$$
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\phi_ {1} ^ {T} = \left[ \frac {1}{\sqrt {2}} \quad \frac {1}{\sqrt {2}} \quad \frac {1}{\sqrt {2}} \right]
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$$
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To obtain $\Phi_2$ and $\Phi_3$ , we proceed in an analogous way. Evaluating $\mathbf{K} - \lambda_2\mathbf{M}$ , we obtain from (10.17),
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$$
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\left[ \begin{array}{c c c} 0 & - 1 & 0 \\ - 1 & 0 & - 1 \\ 0 & - 1 & 0 \end{array} \right] \boldsymbol {\phi} _ {2} = \mathbf {0}
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$$
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In this case we cannot factorize the coefficient matrix preserving symmetry; i.e., we need to interchange only the first and second rows (and not the corresponding columns). This row
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interchange results into the relation
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$$
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\left[ \begin{array}{c c c} - 1 & 0 & - 1 \\ 0 & - 1 & 0 \\ 0 & - 1 & 0 \end{array} \right] \boldsymbol {\phi} _ {2} = \mathbf {0}
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$$
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Factorizing the coefficient matrix into a unit lower triangular matrix $\mathbf{L}$ and an upper triangular matrix $\mathbf{S}$ , we obtain
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$$
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\left[ \begin{array}{l l l} 1 & & \\ 0 & 1 & \\ 0 & 1 & 1 \end{array} \right] \left[ \begin{array}{r r r} - 1 & 0 & - 1 \\ & - 1 & 0 \\ & & 0 \end{array} \right] \boldsymbol {\phi} _ {2} = \mathbf {0}
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$$
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and hence, $s_{33} = 0$ . To solve for $\phi_2$ , we use
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$$
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\left[ \begin{array}{c c c} - 1 & 0 & - 1 \\ & - 1 & 0 \\ & & 0 \end{array} \right] \boldsymbol {\phi} _ {2} = \boldsymbol {0}
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$$
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and $\phi_2^T\mathbf{M}\phi_2 = 1$ . Thus, we obtain
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$$
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\phi_ {2} ^ {T} = \left[ \begin{array}{l l l} - 1 & 0 & 1 \end{array} \right]
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$$
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To calculate $\phi_{3}$ , we evaluate K - $\lambda_{3}M$ and have
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$$
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\left[ \begin{array}{c c c} - 1 & - 1 & 0 \\ - 1 & - 2 & - 1 \\ 0 & - 1 & - 1 \end{array} \right] \boldsymbol {\phi} _ {3} = \mathbf {0}
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$$
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The coefficient matrix can be factorized into $LDL^{T}$ without interchanges; i.e., we have
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$$
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\left[ \begin{array}{l l l} 1 & & \\ 1 & 1 & \\ 0 & 1 & 1 \end{array} \right] \left[ \begin{array}{l l l} - 1 & & \\ & - 1 & \\ & & 0 \end{array} \right] \left[ \begin{array}{l l l} 1 & 1 & 0 \\ & 1 & 1 \\ & & 1 \end{array} \right] \boldsymbol {\Phi} _ {3} = \boldsymbol {0}
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$$
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We note that $d_{33} = 0$ . To calculate $\phi_3$ we use
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$$
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\left[ \begin{array}{c c c} - 1 & - 1 & 0 \\ & - 1 & - 1 \\ & & 0 \end{array} \right] \boldsymbol {\phi} _ {3} = \mathbf {0}
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$$
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and $\phi_3^T\mathbf{M}\phi_3 = 1$ . Hence,
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$$
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\boldsymbol {\phi} _ {3} ^ {T} = \left[ \frac {1}{\sqrt {2}} - \frac {1}{\sqrt {2}} \quad \frac {1}{\sqrt {2}} \right]
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$$
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The eigenvalues of the first associated constraint problem are obtained from the solution of
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$$
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\left[ \begin{array}{c c} 2 & - 1 \\ - 1 & 4 \end{array} \right] \boldsymbol {\phi} ^ {(1)} = \lambda^ {(1)} \left[ \begin{array}{c c} \frac {1}{2} & 0 \\ 0 & 1 \end{array} \right] \boldsymbol {\phi} ^ {(1)}
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$$
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Hence, $p^{(1)}(\lambda^{(1)}) = \frac{1}{2}\lambda^{(1)^2} - 4\lambda^{(1)} + 7$
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and $\lambda_1^{(1)} = 4 - \sqrt{2};\qquad \lambda_2^{(1)} = 4 + \sqrt{2}$
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Also $\lambda^{(2)} = 4$ , and hence the eigenvalue separation property given in (10.22) is in this case:
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<!-- source-page: 866 -->
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1. For the eigenvalues of the first and second associated constraint problems,
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$$
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4 - \sqrt {2} < 4 < 4 + \sqrt {2}
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$$
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2. For the eigenvalues of $\mathbf{K}\phi = \lambda \mathbf{M}\phi$ and the first associated constraint problem,
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$$
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2 < (4 - \sqrt {2}) < 4 < (4 + \sqrt {2}) < 6
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$$
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An important fact that follows from the property of the separation of eigenvalues as expressed in (10.22) is the following. Assume that we can factorize the matrix $\mathbf{K} - \mu \mathbf{M}$ into $\mathbf{LDL}^T$ ; i.e., none of the associated constraint problems has a zero eigenvalue. For simplicity of discussion let us first assume that all eigenvalues are distinct; i.e., there are no multiple eigenvalues. The important fact is that in the decomposition of $\mathbf{K} - \mu \mathbf{M}$ , the number of negative elements in $\mathbf{D}$ is equal to the number of eigenvalues smaller than $\mu$ . Conversely, if $\lambda_i < \mu < \lambda_{i+1}$ , there are exactly $i$ negative diagonal elements in $\mathbf{D}$ . The proof is obtained using the separation property in (10.22) and is relatively easily outlined by the following considerations. Referring to Fig. 10.1, assume that in the sketch of the characteristic polynomials we connect by straight lines all eigenvalues $\lambda_1^{(r)}, r = 0, 1, \ldots$ , with $\lambda_1^{(0)} = \lambda_1$ , and call the resulting curve $C_1$ . Similarly, we establish curves $C_2, C_3, \ldots$ , as indicated in Fig. 10.1. Consider now that $\lambda_i < \mu < \lambda_{i+1}$ and draw a vertical line corresponding to $\mu$ in the figure of the characteristic polynomials; i.e., this line establishes where $\mu$ lies in relation to the eigenvalues of the associated constraint problems. We note that the line corresponding to $\mu$ must cross the curves $C_1, \ldots, C_i$ and because of the eigenvalue
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<details>
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<summary>line</summary>
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| λ | p(λ) | p(λ¹) | p(λ²) | p(λ³) | p(λ⁴) | p(λ⁵) |
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|-------|------|-------|-------|-------|-------|-------|
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| λ₁ | λ₁ | λ₁⁽¹⁾ | λ₁⁽²⁾ | λ₁⁽³⁾ | λ₁⁽⁴⁾ | λ₁⁽⁵⁾ |
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| λ₂ | λ₂ | λ₂⁽¹⁾ | λ₂⁽²⁾ | λ₂⁽³⁾ | λ₂⁽⁴⁾ | λ₂⁽⁵⁾ |
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| λ₃ | λ₃ | λ₃⁽¹⁾ | λ₃⁽²⁾ | λ₃⁽³⁾ | λ₃⁽⁴⁾ | λ₃⁽⁵⁾ |
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| λ₄ | λ₄ | λ₄⁽¹⁾ | λ₄⁽²⁾ | λ₄⁽³⁾ | λ₄⁽⁴⁾ | λ₄⁽⁵⁾ |
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| λ₅ | λ₅ | λ₅⁽¹⁾ | λ₅⁽²⁾ | λ₅⁽³⁾ | λ₅⁽⁴⁾ | λ₅⁽⁵⁾ |
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</details>
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Figure 10.1 Construction of curves $C_{i}$ for the characteristic polynomials of the problem $K\phi = \lambda M\phi$ and of the associated constraint problems
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<!-- source-page: 867 -->
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separation property cannot cross the curves $C_{i+1}, \ldots, C_n$ . However, since
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$$
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p ^ {(r)} (\mu) = \prod_ {i = 1} ^ {n - r} d _ {i i} \tag {10.23}
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$$
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and since each crossing of $\mu$ with an envelope $C_k$ corresponds to a negative element appearing in $\mathbf{D}$ , we have exactly $i$ negative elements in $\mathbf{D}$ .
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These considerations also hold in the case of multiple eigenvalues; i.e., in Fig. 10.1 we would merely find that some eigenvalues are equal, but the argument given above would not change.
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The property that the number of negative elements in D is equal to the number of eigenvalues smaller than $\mu$ can be used directly in the solution of eigenvalues (see Section 11.4.3). Namely, by assuming a shift $\mu$ and checking whether $\mu$ is smaller or larger than the required eigenvalue, we can successively reduce the interval in which the eigenvalue must lie. We demonstrate the solution procedure in the following example.
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EXAMPLE 10.5: Use the fact that the number of negative elements in D, where $LDL^{T} = K - \mu M$ , is equal to the number of eigenvalues smaller than $\mu$ in order to calculate $\lambda_{2}$ of $K\phi = \lambda M\phi$ , where
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$$
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\mathbf {K} = \left[ \begin{array}{r r r} 2 & - 1 & 0 \\ - 1 & 4 & - 1 \\ 0 & - 1 & 2 \end{array} \right]; \quad \mathbf {M} = \left[ \begin{array}{c c c} \frac {1}{2} & & \\ & 1 & \\ & & \frac {1}{2} \end{array} \right]
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$$
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The three eigenvalues of the problem have already been calculated in Example 10.4. We now proceed in the following systematic steps.
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1. Let us assume $\mu = 1$ and evaluate $LDL^{T}$ of K - $\mu M$ ,
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$$
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\mathbf {K} - \mu \mathbf {M} = \left[ \begin{array}{r r r} \frac {3}{2} & - 1 & 0 \\ - 1 & 3 & - 1 \\ 0 & - 1 & \frac {3}{2} \end{array} \right]
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$$
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Hence, $\mathbf{L}\mathbf{D}\mathbf{L}^T = \begin{bmatrix} 1 & & \\ -\frac{2}{3} & 1 & \\ 0 & -\frac{3}{7} & 1 \end{bmatrix} \begin{bmatrix} \frac{3}{2} & & \\ & \frac{7}{3} & \\ & & \frac{15}{14} \end{bmatrix} \begin{bmatrix} 1 & -\frac{2}{3} & 0 \\ & 1 & -\frac{3}{7} \\ & & 1 \end{bmatrix}$
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Since all elements in D are larger than zero, we have $\lambda_{1} > 1$ .
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2. We now try $\mu = 8$ , where
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$$
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\mathbf {K} - \mu \mathbf {M} = \left[ \begin{array}{c c c} - 2 & - 1 & 0 \\ - 1 & - 4 & - 1 \\ 0 & - 1 & - 2 \end{array} \right]
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$$
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and $\mathbf{LDL}^{T}=\begin{bmatrix}1&\\ \frac{1}{2}&1\\ 0&\frac{2}{7}&1\end{bmatrix}\begin{bmatrix}-2&\\ &-\frac{7}{2}&\\ &&-\frac{12}{7}\end{bmatrix}\begin{bmatrix}1&\frac{1}{2}&0\\ &1&\frac{2}{7}\\ &&1\end{bmatrix}$
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Since all three diagonal elements are smaller than zero, it follows that $\lambda_3 < 8$ .
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3. The next estimate $\mu$ should logically lie between 1 and 8; we choose $\mu = 5$ , for which
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$$
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\mathbf {K} - \mu \mathbf {M} = \left[ \begin{array}{c c c} - \frac {1}{2} & - 1 & 0 \\ - 1 & - 1 & - 1 \\ 0 & - 1 & - \frac {1}{2} \end{array} \right]
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$$
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<!-- source-page: 868 -->
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and
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$$
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\mathbf {L D L} ^ {T} = \left[ \begin{array}{c c c} 1 & & \\ 2 & 1 & \\ 0 & - 1 & 1 \end{array} \right] \left[ \begin{array}{c c c} - \frac {1}{2} & & \\ & 1 & \\ & & - \frac {3}{2} \end{array} \right] \left[ \begin{array}{c c c} 1 & 2 & 0 \\ & 1 & - 1 \\ & & 1 \end{array} \right]
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$$
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Since two negative elements are in $\mathbf{D}$ , we have $\lambda_2 < 5$ .
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4. The next estimate must lie between 1 and 5. Let us use $\mu = 3$ , in which case
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$$
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\mathbf {K} - \mu \mathbf {M} = \left[ \begin{array}{c c c} \frac {1}{2} & - 1 & 0 \\ - 1 & 1 & - 1 \\ 0 & - 1 & \frac {1}{2} \end{array} \right]
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$$
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and $\mathbf{L}\mathbf{D}\mathbf{L}^T = \begin{bmatrix} 1 & & \\ -2 & 1 & \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} \frac{1}{2} & & \\ & -1 & \\ & & \frac{3}{2} \end{bmatrix} \begin{bmatrix} 1 & -2 & 0 \\ & 1 & 1 \\ & & 1 \end{bmatrix}$
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Hence $\lambda_{2} > 3$ , because there is only one negative element in $\mathbf{D}$ .
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The pattern of the solution procedure has now been established. So far we know that $3 < \lambda_{2} < 5$ . In order to obtain a closer estimate on $\lambda_{2}$ we would continue choosing a shift $\mu$ in the interval 3 to 5 and investigate whether the new shift is smaller or larger than $\lambda_{2}$ . By always choosing an appropriate new shift, the required eigenvalue can be determined very accurately (see Section 11.4.3). It should be noted that we did not need to use interchanges in the factorizations of K - $\mu$ M carried out above.
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# 10.2.3 Shifting
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An important procedure that is used extensively in the solution of eigenvalues and eigenvectors is shifting. The purpose of shifting is to accelerate the calculations of the required eigensystem. In the solution $K\phi = \lambda M\phi$ , we perform a shift $\rho$ on K by calculating
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$$
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\hat {\mathbf {K}} = \mathbf {K} - \rho \mathbf {M} \tag {10.24}
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$$
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and we then consider the eigenproblem
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$$
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\hat {\mathbf {K}} \boldsymbol {\psi} = \mu \mathbf {M} \boldsymbol {\psi} \tag {10.25}
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$$
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To identify how the eigenvalues and eigenvectors of $\mathbf{K}\phi = \lambda \mathbf{M}\phi$ are related to those of the problem $\hat{\mathbf{K}}\psi = \mu \mathbf{M}\psi$ we rewrite (10.25) in the form
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$$
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\mathbf {K} \boldsymbol {\psi} = \gamma \mathbf {M} \boldsymbol {\psi} \tag {10.26}
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$$
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where $\gamma = \rho + \mu$ . However, (10.26) is, in fact, the eigenproblem $K\phi = \lambda M\phi$ , and since the solution of this problem is unique, we have
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$$
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\lambda_ {i} = \rho + \mu_ {i}; \quad \phi_ {i} = \psi_ {i} \tag {10.27}
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$$
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In other words, the eigenvectors of $\hat{K}\psi = \mu M\psi$ are the same as the eigenvectors of $K\phi = \lambda M\phi$ , but the eigenvalues have been decreased by $\rho$ . A frequent application of shifting occurs in the calculation of rigid body modes when an algorithm is to be used that is not designed explicitly to calculate zero eigenvalues. We illustrate such an application in the following example.
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<!-- source-page: 869 -->
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EXAMPLE 10.6: Consider the eigenproblem
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$$
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\left[ \begin{array}{c c} 3 & - 3 \\ - 3 & 3 \end{array} \right] \boldsymbol {\phi} = \lambda \left[ \begin{array}{l l} 2 & 1 \\ 1 & 2 \end{array} \right] \boldsymbol {\phi} \tag {a}
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$$
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Calculate the eigenvalues and eigenvectors. Then impose a shift $\rho = -2$ and solve again for the eigenvalues and corresponding eigenvectors.
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To calculate the eigenvalues we use the characteristic polynomial
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$$
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p (\lambda) = \det (\mathbf {K} - \lambda \mathbf {M}) = 3 \lambda^ {2} - 1 8 \lambda
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$$
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and thus obtain $\lambda_{1}=0$ , $\lambda_{2}=6$ . To calculate $\phi_{1}$ and $\phi_{2}$ we use the relation in (10.17) and the mass orthonormality condition $\phi_{i}^{T}\mathbf{M}\phi_{i}=1$ . We have
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$$
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\left[ \begin{array}{c c} 3 & - 3 \\ - 3 & 3 \end{array} \right] \boldsymbol {\phi} _ {1} = \mathbf {0}; \quad \text { hence, } \boldsymbol {\phi} _ {1} = \left[ \begin{array}{c} \frac {1}{\sqrt {6}} \\ \frac {1}{\sqrt {6}} \end{array} \right] \tag {b}
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$$
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and $\left[ \begin{array}{ll} -9 & -9\\ -9 & -9 \end{array} \right]\pmb{\phi}_{2} = \pmb{0};\qquad \text{hence,}\pmb{\phi}_{2} = \left[ \begin{array}{c}\frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}} \end{array} \right]$ (c)
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Imposing a shift of $\rho = -2$ , we obtain the problem
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$$
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\left[ \begin{array}{c c} 7 & - 1 \\ - 1 & 7 \end{array} \right] \boldsymbol {\phi} = \lambda \left[ \begin{array}{l l} 2 & 1 \\ 1 & 2 \end{array} \right] \boldsymbol {\phi} \tag {d}
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$$
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Proceeding as before, we have
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$$
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p (\lambda) = \lambda^ {2} - 1 0 \lambda + 1 6
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$$
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and obtain as the roots $\lambda_1 = 2, \lambda_2 = 8$ . Hence the eigenvalues have increased by 2; i.e., they have decreased by $\rho$ .
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The eigenvectors would be calculated using (10.17). However, we note that this relation again yields the equations in (b) and (c), and therefore the eigenvectors of the problem in (d) are those of the problem in (a).
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An important observation resulting from the above discussion is that, in principle, we need only solution algorithms to calculate the eigenvalues and corresponding eigenvectors of the problem $K\phi = \lambda M\phi$ when all eigenvalues are larger than zero. This follows, because if rigid body modes are present, we may always operate on a shifted stiffness matrix that renders all eigenvalues positive.
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Extensive applications of shifting are given in Chapter 11, where the various eigensystem solution algorithms are discussed.
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# 10.2.4 Effect of Zero Mass
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When using a lumped mass matrix, M is diagonal with positive and possibly some zero diagonal elements. If all elements $m_{ii}$ are larger than zero, the eigenvalues $\lambda_{i}$ can usually not be obtained without the use of an eigenvalue solution algorithm as described in Chapter 11.
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However, if M has some zero diagonal elements, say r diagonal elements in M are zero, we can immediately say that the problem $K\phi = \lambda M\phi$ has the eigenvalues $\lambda_{n} = \lambda_{n-1} = \cdots = \lambda_{n-r+1} = \infty$ and can also construct the corresponding eigenvectors by inspection.
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To obtain the above result, let us recall the fundamental objective in an eigensolution. It is important to remember that all we require is a vector $\phi$ and scalar $\lambda$ that satisfy the equation
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$$
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\mathbf {K} \phi = \lambda \mathbf {M} \phi \tag {10.4}
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$$
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where $\phi$ is nontrivial; i.e., $\phi$ is a vector with at least one element in it nonzero. In other words, if we have a vector $\phi$ and scalar $\lambda$ that satisfy (10.4), then $\lambda$ and $\phi$ are an eigenvalue $\lambda_{i}$ and eigenvector $\phi_{i}$ , respectively, where it should be noted that it does not matter how $\phi$ and $\lambda$ have been obtained. If, for example, we can guess $\phi$ and $\lambda$ , we should certainly take advantage of it. This may be the case when rigid body modes are present in the structural element assemblage. Thus, if we know that the element assemblage can undergo a rigid body mode, we have $\lambda_{1} = 0$ and need to seek $\phi_{1}$ to satisfy the equation $\mathbf{K}\phi_{1} = \mathbf{0}$ . In general, the solution of $\phi_{1}$ must be obtained using an equation solver, but in a simple finite element assemblage we may be able to identify $\phi_{1}$ by inspection.
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In the case of $r$ zero diagonal elements in a diagonal mass matrix $\mathbf{M}$ , we can always immediately establish $r$ eigenvalues and corresponding eigenvectors. Rewriting the eigenproblem in (10.4) in the form
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$$
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\mathbf {M} \phi = \mu \mathbf {K} \phi \tag {10.28}
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$$
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where $\mu = \lambda^{-1}$ , we find that if $m_{kk} = 0$ , we have an eigenpair $(\mu_i, \phi_i) = (0, \mathbf{e}_k)$ ; i.e.,
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$$
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\boldsymbol {\Phi} _ {i} ^ {T} = \left[ \begin{array}{l l l l l l l l} 0 & 0 & \dots & 0 & 1 & 0 & \dots & 0 \end{array} \right]; \quad \mu_ {i} = 0 \tag {10.29}
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$$
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kth element
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That $\phi_{i}$ and $\mu_{i}$ in (10.29) are indeed an eigenvector and eigenvalue of (10.28) is verified by simply substituting into (10.28) and noting that $(\mu_{i}, \phi_{i})$ is a nontrivial solution. Since $\mu = \lambda^{-1}$ , we therefore found that an eigenpair of $\mathbf{K}\phi = \lambda\mathbf{M}\phi$ is given by $(\lambda_{i}, \phi_{i}) = (\infty, \mathbf{e}_{k})$ . Considering the case of $r$ zero diagonal elements in $\mathbf{M}$ , it follows that there are $r$ infinite eigenvalues, and the corresponding eigenvectors can be taken to be unit vectors with each unit vector having the 1 in a location corresponding to a zero mass element in $\mathbf{M}$ . Since $\lambda_{n}$ is then an eigenvalue of multiplicity $r$ , the corresponding eigenvectors are not unique (see Section 10.2.1). In addition, we note that the length of an eigenvector cannot be fixed using the condition of $\mathbf{M}$ -orthonormality. We demonstrate how we establish the eigenvalues and eigenvectors by means of a brief example.
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EXAMPLE 10.7: Consider the eigenproblem
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$$
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\left[ \begin{array}{c c c c} 2 & - 1 & & \\ - 1 & 2 & - 1 & \\ & - 1 & 2 & - 1 \\ & & - 1 & 1 \end{array} \right] \boldsymbol {\Phi} = \lambda \left[ \begin{array}{c c c c} 0 & & & \\ & 2 & & \\ & & 0 & \\ & & & 1 \end{array} \right] \boldsymbol {\Phi}
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$$
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Establish $\lambda_3, \lambda_4$ and $\phi_3, \phi_4$ .
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