# Eigenvalue perturbation

Eigenvalue perturbation

Eigenvalue perturbation is a perturbation approach to finding eigenvalues and eigenvectors of systems perturbed from one with known eigenvectors and eigenvalues. It also allows one to determine the sensitivity of the eigenvalues and eigenvectors with respect to changes in the system.

Example

Suppose we have solutions to the generalized eigenvalue problem,:$\left[K_0\right] mathbf\left\{x\right\}_\left\{0i\right\} = lambda_\left\{0i\right\} \left[M_0\right] mathbf\left\{x\right\}_\left\{0i\right\}. qquad \left(1\right)$That is, we know $lambda_\left\{0i\right\}$ and $mathbf\left\{x\right\}_\left\{0i\right\}$ for $i=1,dots,N$.Now suppose we want to change the matrices by a small amount. That is, we want to let:$\left[K\right] = \left[K_0\right] + \left[delta K\right]$and:$\left[M\right] = \left[M_0\right] + \left[delta M\right]$where all of the $delta$ terms are much smaller than the corresponding term. We expect answers to be of the form:$lambda_i = lambda_\left\{0i\right\}+deltalambda_\left\{0i\right\}$and:$mathbf\left\{x\right\}_i = mathbf\left\{x\right\}_\left\{0i\right\}+deltamathbf\left\{x\right\}_\left\{0i\right\}.$

teps

We assume that the matrices are symmetric and positive definite and assume we have scaled the eigenvectors such that:$mathbf\left\{x\right\}_\left\{0j\right\}^T \left[M_0\right] mathbf\left\{x\right\}_\left\{0i\right\} = delta_i^j qquad\left(2\right)$where $delta_i^j$ is the Kronecker delta.

Now we want to solve the equation:$\left[K\right] mathbf\left\{x\right\}_i = lambda_i \left[M\right] mathbf\left\{x\right\}_i$.Substituting, we get:$\left( \left[K_0\right] + \left[delta K\right] \right)\left(mathbf\left\{x\right\}_\left\{0i\right\} + delta mathbf\left\{x\right\}_i\right) = \left(lambda_\left\{0i\right\}+deltalambda_\left\{i\right\}\right)\left( \left[M_0\right] + \left[delta M\right] \right)\left(mathbf\left\{x\right\}_\left\{0i\right\}+deltamathbf\left\{x\right\}_\left\{0i\right\}\right)$.which expands to:$\left[K_0\right] mathbf\left\{x\right\}_\left\{0i\right\}+ \left[delta K\right] mathbf\left\{x\right\}_\left\{0i\right\} + \left[K_0\right] delta mathbf\left\{x\right\}_i + \left[delta K\right] delta mathbf\left\{x\right\}_i$:::$= lambda_\left\{0i\right\} \left[M_0\right] mathbf\left\{x\right\}_\left\{0i\right\}+ lambda_\left\{0i\right\} \left[M_0\right] deltamathbf\left\{x\right\}_i + lambda_\left\{0i\right\} \left[delta M\right] mathbf\left\{x\right\}_\left\{0i\right\} + deltalambda_i \left[M_0\right] mathbf\left\{x\right\}_\left\{0i\right\}$::::$\left\{\right\} + lambda_\left\{0i\right\} \left[delta M\right] deltamathbf\left\{x\right\}_i + deltalambda_i \left[delta M\right] mathbf\left\{x\right\}_\left\{0i\right\} + deltalambda_i \left[M_0\right] deltamathbf\left\{x\right\}_i + deltalambda_i \left[delta M\right] deltamathbf\left\{x\right\}_i$.Canceling from (1) leaves:$\left[delta K\right] mathbf\left\{x\right\}_\left\{0i\right\} + \left[K_0\right] delta mathbf\left\{x\right\}_i + \left[delta K\right] delta mathbf\left\{x\right\}_i$:::$= lambda_\left\{0i\right\} \left[M_0\right] deltamathbf\left\{x\right\}_i + lambda_\left\{0i\right\} \left[delta M\right] mathbf\left\{x\right\}_\left\{0i\right\} + deltalambda_i \left[M_0\right] mathbf\left\{x\right\}_\left\{0i\right\}$::::$\left\{\right\} + lambda_\left\{0i\right\} \left[delta M\right] deltamathbf\left\{x\right\}_i + deltalambda_i \left[delta M\right] mathbf\left\{x\right\}_\left\{0i\right\} + deltalambda_i \left[M_0\right] deltamathbf\left\{x\right\}_i + deltalambda_i \left[delta M\right] deltamathbf\left\{x\right\}_i$.Removing the higher-order terms, this simplifies to:$\left[K_0\right] deltamathbf\left\{x\right\}_i+ \left[delta K\right] mathbf\left\{x\right\}_\left\{0i\right\} = lambda_\left\{0i\right\} \left[M_0\right] delta mathbf\left\{x\right\}_i + lambda_\left\{0i\right\} \left[delta M\right] mathrm\left\{x\right\}_\left\{0i\right\} + delta lambda_i \left[M_0\right] mathbf\left\{x\right\}_\left\{0i\right\}. qquad\left(3\right)$

We note that, when the matrix is symmetric, the unperturbed eigenvectors are orthogonal and so we use them as a basis for the perturbed eigenvectors. That is, we want to construct

:$delta mathbf\left\{x\right\}_i = sum_\left\{j=1\right\}^N epsilon_\left\{ij\right\} mathbf\left\{x\right\}_\left\{0j\right\} qquad\left(4\right)$

where the $epsilon_\left\{ij\right\}$ are small constants that are to be determined. Substituting (4) into (3) and rearranging gives:$\left[K_0\right] sum_\left\{j=1\right\}^N epsilon_\left\{ij\right\} mathbf\left\{x\right\}_\left\{0j\right\} + \left[delta K\right] mathbf\left\{x\right\}_\left\{0i\right\} = lambda_\left\{0i\right\} \left[M_0\right] sum_\left\{j=1\right\}^N epsilon_\left\{ij\right\} mathbf\left\{x\right\}_\left\{0j\right\} + lambda_\left\{0i\right\} \left[delta M\right] mathbf\left\{x\right\}_\left\{0i\right\} + deltalambda_i \left[M_0\right] mathbf\left\{x\right\}_\left\{0i\right\} qquad \left(5\right)$.

Or::$sum_\left\{j=1\right\}^N epsilon_\left\{ij\right\} \left[K_0\right] mathbf\left\{x\right\}_\left\{0j\right\} + \left[delta K\right] mathbf\left\{x\right\}_\left\{0i\right\} = lambda_\left\{0i\right\} \left[M_0\right] sum_\left\{j=1\right\}^N epsilon_\left\{ij\right\} mathbf\left\{x\right\}_\left\{0j\right\} + lambda_\left\{0i\right\} \left[delta M\right] mathbf\left\{x\right\}_\left\{0i\right\} + deltalambda_i \left[M_0\right] mathbf\left\{x\right\}_\left\{0i\right\}$.

By equation (1)::$sum_\left\{j=1\right\}^N epsilon_\left\{ij\right\} lambda_\left\{0j\right\} \left[M_0\right] mathbf\left\{x\right\}_\left\{0j\right\} + \left[delta K\right] mathbf\left\{x\right\}_\left\{0i\right\} = lambda_\left\{0i\right\} \left[M_0\right] sum_\left\{j=1\right\}^N epsilon_\left\{ij\right\} mathbf\left\{x\right\}_\left\{0j\right\} + lambda_\left\{0i\right\} \left[delta M\right] mathbf\left\{x\right\}_\left\{0i\right\} + deltalambda_i \left[M_0\right] mathbf\left\{x\right\}_\left\{0i\right\}$.

Because the eigenvectors are orthogonal, we can remove the summations by left multiplying by $mathbf\left\{x\right\}_\left\{0i\right\}^T$:

:$mathbf\left\{x\right\}_\left\{0i\right\}^T epsilon_\left\{ii\right\} lambda_\left\{0i\right\} \left[M_0\right] mathbf\left\{x\right\}_\left\{0i\right\} + mathbf\left\{x\right\}_\left\{0i\right\}^T \left[delta K\right] mathbf\left\{x\right\}_\left\{0i\right\} = lambda_\left\{0i\right\} mathbf\left\{x\right\}_\left\{0i\right\}^T \left[M_0\right] epsilon_\left\{ii\right\} mathbf\left\{x\right\}_\left\{0i\right\} + lambda_\left\{0i\right\}mathbf\left\{x\right\}_\left\{0i\right\}^T \left[delta M\right] mathbf\left\{x\right\}_\left\{0i\right\} + deltalambda_imathbf\left\{x\right\}_\left\{0i\right\}^T \left[M_0\right] mathbf\left\{x\right\}_\left\{0i\right\}$.

By use of equation (1) again:

:$mathbf\left\{x\right\}_\left\{0i\right\}^T \left[K_0\right] epsilon_\left\{ii\right\} mathbf\left\{x\right\}_\left\{0i\right\} + mathbf\left\{x\right\}_\left\{0i\right\}^T \left[delta K\right] mathbf\left\{x\right\}_\left\{0i\right\} = lambda_\left\{0i\right\} mathbf\left\{x\right\}_\left\{0i\right\}^T \left[M_0\right] epsilon_\left\{ii\right\} mathbf\left\{x\right\}_\left\{0i\right\} + lambda_\left\{0i\right\}mathbf\left\{x\right\}_\left\{0i\right\}^T \left[delta M\right] mathbf\left\{x\right\}_\left\{0i\right\} + deltalambda_imathbf\left\{x\right\}_\left\{0i\right\}^T \left[M_0\right] mathbf\left\{x\right\}_\left\{0i\right\} ~~\left(6\right)$.

The two terms containing $epsilon_\left\{ii\right\}$ are equal because left-multiplying (1) by $mathbf\left\{x\right\}_\left\{0i\right\} ^T$ gives

:$mathbf\left\{x\right\}_\left\{0i\right\}^T \left[K_0\right] mathbf\left\{x\right\}_\left\{0i\right\} = lambda_\left\{0i\right\}mathbf\left\{x\right\}_\left\{0i\right\}^T \left[M_0\right] mathbf\left\{x\right\}_\left\{0i\right\}$.

Canceling those terms in (6) leaves

:$mathbf\left\{x\right\}_\left\{0i\right\}^T \left[delta K\right] mathbf\left\{x\right\}_\left\{0i\right\} = lambda_\left\{0i\right\} mathbf\left\{x\right\}_\left\{0i\right\}^T \left[delta M\right] mathbf\left\{x\right\}_\left\{0i\right\} + deltalambda_i mathbf\left\{x\right\}_\left\{0i\right\}^T \left[M_0\right] mathbf\left\{x\right\}_\left\{0i\right\}$.

Rearranging gives

:$deltalambda_i = frac\left\{mathbf\left\{x\right\}^T_\left\{0i\right\}\left( \left[delta K\right] - lambda_\left\{0i\right\} \left[delta M\right] \right)mathbf\left\{x\right\}_\left\{0i\left\{mathbf\left\{x\right\}_\left\{0i\right\}^T \left[M_0\right] mathbf\left\{x\right\}_\left\{0i$

But by (2), this denominator is equal to 1. Thus

:$deltalambda_i = mathbf\left\{x\right\}^T_\left\{0i\right\}\left( \left[delta K\right] - lambda_\left\{0i\right\} \left[delta M\right] \right)mathbf\left\{x\right\}_\left\{0i\right\}$

Then, by left multiplying equation (6) by $mathbf\left\{x\right\}_\left\{0k\right\}$ (for $i eq k$):

:$epsilon_\left\{ik\right\} = frac\left\{mathbf\left\{x\right\}^T_\left\{0i\right\}\left( \left[delta K\right] - lambda_\left\{0k\right\} \left[delta M\right] \right)mathbf\left\{x\right\}_\left\{0k\left\{lambda_\left\{0k\right\}-lambda_\left\{0i, qquad i eq k.$

Or by changing the name of the indices:

:$epsilon_\left\{ij\right\} = frac\left\{mathbf\left\{x\right\}^T_\left\{0i\right\}\left( \left[delta K\right] - lambda_\left\{0j\right\} \left[delta M\right] \right)mathbf\left\{x\right\}_\left\{0j\left\{lambda_\left\{0j\right\}-lambda_\left\{0i, qquad i eq j.$

To find $epsilon_\left\{ii\right\}$, use

:$mathbf\left\{x\right\}^T_i \left[M\right] mathbf\left\{x\right\}_i = 1 Rightarrow epsilon_\left\{ii\right\}=-frac\left\{1\right\}\left\{2\right\}mathbf\left\{x\right\}^T_\left\{0i\right\} \left[delta M\right] mathbf\left\{x\right\}_\left\{0i\right\}.$

Summary

:$lambda_i = lambda_\left\{0i\right\} + mathbf\left\{x\right\}^T_\left\{0i\right\} \left( \left[delta K\right] - lambda_\left\{0i\right\} \left[delta M\right] \right) mathbf\left\{x\right\}_\left\{0i\right\}$and:$mathbf\left\{x\right\}_i = mathbf\left\{x\right\}_\left\{0i\right\}\left(1 - frac\left\{1\right\}\left\{2\right\} mathbf\left\{x\right\}^T_\left\{0i\right\} \left[delta M\right] mathbf\left\{x\right\}_\left\{0i\right\}\right) + sum_\left\{j=1atop j eq i\right\}^N frac\left\{mathbf\left\{x\right\}^T_\left\{0j\right\}\left( \left[delta K\right] - lambda_\left\{0i\right\} \left[delta M\right] \right)mathbf\left\{x\right\}_\left\{0i\left\{lambda_\left\{0i\right\}-lambda_\left\{0jmathbf\left\{x\right\}_\left\{0j\right\}$

Results

This means it is possible to efficiently do a sensitivity analysis on $lambda_i$ as a function of changes in the entries of the matrices. (Recall that the matrices are symmetric and so changing $K_\left\{\left(kell\right)\right\}$ will also change $K_\left\{\left(ell k\right)\right\}$, hence the $\left(2-delta_k^ell\right)$ term.):$frac\left\{partial lambda_i\right\}\left\{partial K_\left\{\left(kell\right) = frac\left\{partial\right\}\left\{partial K_\left\{\left(kell\right)left\left(lambda_\left\{0i\right\} + mathbf\left\{x\right\}^T_\left\{0i\right\} \left( \left[delta K\right] - lambda_\left\{0i\right\} \left[delta M\right] \right) mathbf\left\{x\right\}_\left\{0i\right\} ight\right) = x_\left\{0i\left(k\right)\right\} x_\left\{0i\left(ell\right)\right\} \left(2-delta_k^ell\right)$and:$frac\left\{partial lambda_i\right\}\left\{partial M_\left\{\left(kell\right) = frac\left\{partial\right\}\left\{partial M_\left\{\left(kell\right)left\left(lambda_\left\{0i\right\} + mathbf\left\{x\right\}^T_\left\{0i\right\} \left( \left[delta K\right] - lambda_\left\{0i\right\} \left[delta M\right] \right) mathbf\left\{x\right\}_\left\{0i\right\} ight\right) =lambda_i x_\left\{0i\left(k\right)\right\} x_\left\{0i\left(ell\right)\right\}\left(2-delta_k^ell\right)$.Similarly:$frac\left\{partialmathbf\left\{x\right\}_i\right\}\left\{partial K_\left\{\left(kell\right) = sum_\left\{j=1atop j eq i\right\}^N frac\left\{x_\left\{0j\left(k\right)\right\} x_\left\{0i\left(ell\right)\right\}\left(2-delta_k^ell\right)\right\}\left\{lambda_\left\{0i\right\}-lambda_\left\{0jmathbf\left\{x\right\}_\left\{0j\right\}$and:$frac\left\{partial mathbf\left\{x\right\}_i\right\}\left\{partial M_\left\{\left(kell\right) = -mathbf\left\{x\right\}_\left\{0i\right\}frac\left\{x_\left\{0i\left(k\right)\right\}x_\left\{0i\left(ell\right)\left\{2\right\}\left(2-delta_k^ell\right) - sum_\left\{j=1atop j eq i\right\}^N frac\left\{lambda_\left\{0i\right\}x_\left\{0j\left(k\right)\right\} x_\left\{0i\left(ell\right)\left\{lambda_\left\{0i\right\}-lambda_\left\{0jmathbf\left\{x\right\}_\left\{0j\right\}\left(2-delta_k^ell\right)$.

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