Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Jordan Normal Form & Wronskian Derivative

  1. Dec 4, 2008 #1
    I haven' been able to find good explanations of either of these:

    Part 1:

    Jordan Normal Form: Is this it?

    An n*n matrix A is not diagonizable (ie. A=PDP^-1) because it has linearly dependent eigenvectors (no. of eigenvectors is less than n). However, it can be expressed in a similar form A=PJP^-1 , where J is the Jordan Normal Form ie. matrix of eigenvalues on main diagonal and 1's on super diagonal next to duplicate eigenvalues.

    If that is correct, what use is this form of A?

    Part 2:

    How does one compute the derivative of a Wronskian, and what use is this? (I know it must be differentiable since it is a function of differentiable functions)
  2. jcsd
  3. Dec 4, 2008 #2


    User Avatar
    Homework Helper

    It's easier to multiply a matrix in its Jordan form. So since every complex nxn matrix is similar to a Jordan form, that means we can multiply it easily if we know it's Jordan form. Unfortunately it's not an easy task to determine the Jordan form of a matrix, though we can limit the range of possibilities of its Jordan form if we know it's minimal polynomial.

    I'm puzzled as to why you want to take the derivative of a Wronskian.
  4. Dec 4, 2008 #3
    Thanks for the reply.

    However, regarding what you said, a lot of it is beyond my level at this stage. Im just trying to get the basic idea of JNM and what it does.

    The derivative of a Wronksian: I know its on my course, and have seen a method for it, but its use is beyond me!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook