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

Jordan blocks + kernels

  1. Oct 1, 2011 #1
    For some of my homework exercises, the lecturer has specified the values for a kernel of a square matrix minus the scalar product of the eigenvector and the identity matrix.

    Mathematically, I am given

    kernel(A - λI) = some integer value where A is a nxn square matrix

    lambda is the eigenvalue obtained from the characteristic polynomial

    I is the identity matrix

    My question is, what is the logical and mathematical connection between kernel(A - λI) and finding the equivalent jordan block ?
  2. jcsd
  3. Oct 7, 2011 #2
    The kernel of an operator (or matrix) is a vectorspace. The operator transforms the elements of the kernel to the null vector. For the Jordan form, the following dimensions should be known

    dim kernel (A-λI)k,

    where k=1...r and r is the algebraic multiplicity of λ.

    For k=1,

    dim kernel (A-λI)

    gives the number of blocks of λ in the J form.

    If and only if,

    dim kernel (A-λI)=r

    for all λ, then the matrix can be diagonalized, since each block is 1-dimensional.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook