Engineering Analysis/Error Estimation

Consider a scenario where the matrix representation of a system A differs from the actual implementation of the system by a factor of &Delta;A. In other words, our system uses the matrix:


 * $$A + \Delta A$$

From the study of Control Systems, we know that the values of the eigenvectors can affect the stability of the system. For that reason, we would like to know how a small error in A will affect the eigenvalues.

First off, we assume that &Delta;A is a small shift. The definition of "small" in this sense is arbitrary, and will remained open. Keep in mind that the techniques discussed here are more accurate the smaller &Delta;A is.

If &Delta;A is the error in the matrix A, then &Delta;&lambda; is the error in the eigenvalues and &Delta;v is the error in the eigenvectors. The characteristic equation becomes:


 * $$(A + \Delta A)(v + \Delta v) = (\lambda + \Delta \lambda)(v + \Delta v)$$

We have an equation now with two unknowns: &Delta;&lambda; and &Delta;v. In other words, we don't know how a small change in A will affect the eigenvalues and eigenvectors. If we multiply out both sides, we get:


 * $$Av + \Delta A v + A \Delta v + O(\Delta^2) = \lambda v + \Delta \lambda v + v \Delta \lambda + O(\Delta^2)$$

This situation seems hopeless, until we multiply both sides by the corresponding left-eigenvector w from the left:


 * $$w^TAv + w^T\Delta A v + w^Tv \Delta A = w^T\lambda v + w^T\Delta \lambda v + w^T v \Delta \lambda + O(\Delta^2)$$

Terms where two &Delta;s (which are known to be small, by definition) are multiplied together, we can say are negligible, and ignore them. Also, we know from our right-eigenvalue equation that:


 * $$w^TA = \lambda w^T$$

Another fact is that the right-eigenvectors and left eigenvectors are orthogonal to each other, so the following result holds:


 * $$w^T v = 0$$

Substituting these results, where necessary, into our long equation above, we get the following simplification:


 * $$w^T \Delta A v = \Delta \lambda w^T\Delta v$$

And solving for the change in the eigenvalue gives us:


 * $$\Delta \lambda = \frac{w^T \Delta A v}{w^T \Delta v}$$

This approximate result is only good for small values of &Delta;A, and the result is less precise as the error increases.