Eigenvalue Perturbation - Steps

Steps

We assume that the matrices are symmetric and positive definite and assume we have scaled the eigenvectors such that

where is the Kronecker delta.

Now we want to solve the equation

Substituting, we get

which expands to


\begin{align}
\left\mathbf{x}_{0i} & + \mathbf{x}_{0i} + \delta \mathbf{x}_i + \delta \mathbf{x}_i \\
& = \lambda_{0i}\mathbf{x}_{0i}+ \lambda_{0i}\delta\mathbf{x}_i + \lambda_{0i}\mathbf{x}_{0i} + \delta\lambda_i\mathbf{x}_{0i} \\
& {} + \lambda_{0i}\delta\mathbf{x}_i + \delta\lambda_i\mathbf{x}_{0i} + \delta\lambda_i\delta\mathbf{x}_i + \delta\lambda_i\delta\mathbf{x}_i.
\end{align}

Canceling from (1) leaves


\begin{align}
\left\mathbf{x}_{0i} & + \delta \mathbf{x}_i + \delta \mathbf{x}_i \\
& = \lambda_{0i}\delta\mathbf{x}_i + \lambda_{0i}\mathbf{x}_{0i} + \delta\lambda_i\mathbf{x}_{0i} \\
& {} + \lambda_{0i}\delta\mathbf{x}_i + \delta\lambda_i\mathbf{x}_{0i} + \delta\lambda_i\delta\mathbf{x}_i + \delta\lambda_i\delta\mathbf{x}_i.
\end{align}

Removing the higher-order terms, this simplifies to

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

where the are small constants that are to be determined. Substituting (4) into (3) and rearranging gives

Or:

By equation (1):

Because the eigenvectors are orthogonal, we can remove the summations by left multiplying by :

By use of equation (1) again:

The two terms containing are equal because left-multiplying (1) by gives

Canceling those terms in (6) leaves

Rearranging gives

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

Then, by left-multiplying equation (5) by (for ):

Or by changing the name of the indices:

To find, use

Read more about this topic:  Eigenvalue Perturbation

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