# How do you actually find the Jordan canonical form of a matrix?

Computing the matrix exponential $\exp^{At}$ is easy if A is already in diagonal form (i.e., all elements not on the diagonal are zero). The nxn matrix A can be transformed to a similar diagonal nxn matrix D if and only if A has n linearly independent eigenvectors. This is of great consequence because it simplifies the work required to solve a system of linear ODEs or when using the convolution integral.

If the nxn matrix A does not have n linearly independent eigenvectors, it cannot be represented by a similar diagonal matrix. A similar matrix has the same eigenvalues and eigenvectors as A. We then try to find what are called "generalized eigenvectors" until we have a total of n eigenvectors. Then we can transform matrix A to its Jordan form J = \inv(P)*A*P, where P is an invertible nxn matrix with the n eigenvectors (including the generalized eigenvectors) as columns.

If A does not have n linearly independent eigenvectors, then J will not be diagonal, but will be in a block-diagonal form that separates the different "normal modes" of the system into Jordan blocks. To find the matrix exponential of J, one finds the matrix exponential of each Jordan block separately.

For definitions and simple examples you can go to http://www.maths.surrey.ac.uk/explore/emmaspages/option3.html

or consult the book by Perko titled "Differential Equations and Dynamical Systems" available in the CDS library. --Luis Soto 18:51, 15 October 2008 (PDT)