1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the Jordan decomposition

  1. Oct 6, 2013 #1
    1. The problem statement, all variables and given/known data
    I have a linear transformation T defined by
    $$
    T(v_{1})=v_{1}+iv_{2}\\
    T(v_{2})=-iv_{1}+v_{2}\\
    $$
    and I want to find a triangular matrix B of T and an invertible matrix S such that SB=AS where A is the matrix of T with respect to the basis ##\{v_{1},v_{2}\}##.


    3. The attempt at a solution
    The matrix A is
    $$
    \begin{pmatrix}
    1 & -i \\
    i & 1 \\
    \end{pmatrix}.
    $$
    From the proof of the triangular form theorem, I know that I need to find a new basis ##\{w_{1},w_{2}\}## such that ##Tw_{1}=0## and ##(T-2)w_{2}=0## since the minimal polynomial is ##m(T)=T(T-2)=0## but I must be messing something up because there is no way I can create a triangular matrix B from ##w_{1}## and ##w_{2}##. If I find a linear combination of ##v_{1},v_{2}## equal to ##w_{1}## such that ##Tw_{1}=0## then the first column of my matrix B would always be 0.
     
  2. jcsd
  3. Oct 6, 2013 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    It's okay to have zeros on the diagonal.
     
  4. Oct 6, 2013 #3
    Hmm. I am choosing the new basis ##w_{1}=v_{1}-iv_{2}## and ##w_{2}=v_{1}+iv_{2}## which gives me
    $$
    B=\begin{pmatrix}
    0 & 0 \\
    0 & 2 \\
    \end{pmatrix}
    $$
    and
    $$
    S=\begin{pmatrix}
    1 & 1 \\
    -i & i \\
    \end{pmatrix}
    $$
    which gives me the SB=AS that I need but then when I do the Jordan decomposition is it okay to have
    $$
    D=\begin{pmatrix}
    0 & 0 \\
    0 & 2 \\
    \end{pmatrix}
    $$
    and
    $$
    N=\begin{pmatrix}
    0 & 0 \\
    0 & 0 \\
    \end{pmatrix}
    $$
    so that T=D+N?
     
  5. Oct 6, 2013 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I'm afraid I'm not familiar with what you're trying to do. D, I assume, is supposed to be a diagonal matrix. What is N defined to be?
     
  6. Oct 6, 2013 #5
    I am trying to find the Jordan Decomposition of T. That is what it is called in my book. From a quick search it doesn't look like it goes by the name of Jordan Decomposition on wikipedia.

    In my book D is defined to be a diagonalizable linear transformation while N is defined to be a nilpotent linear transformation such that T=D+N.
     
  7. Oct 6, 2013 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Your N is definitely nilpotent, so that decomposition looks fine to me. Since the transformation you were given doesn't have degenerate eigenvalues, the associated eigenvectors will form a basis, and you should expect the similarity transformation to yield a diagonal matrix and to therefore have N=0.

    If you had degenerate eigenvalues, it's then possible that the eigenvectors won't form a basis, and you'd have to resort to generalized eigenvectors. In this case, the similarity transformation would yield an almost-diagonal matrix, the so-called Jordan normal form, and N would have non-zero entries just above some of the diagonal elements.
     
  8. Oct 6, 2013 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    It seems to go by the name Jordan-Chevalley decomposition.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Finding the Jordan decomposition
  1. Jordan Decomposition (Replies: 1)

Loading...