Results 1 to 2 of 2

Thread: Reducing a matrix to its Jordan normal form

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    Reducing a matrix to its Jordan normal form

    I must reduce the following matrix to its Jordan normal form: $\displaystyle M= \begin {bmatrix} 2&1&-1&0&1&0 \\ 0&2&0&1&1&0 \\ 0&0&2&0&1&0 \\ 0&0&0&2&1&0 \\ 0&0&0&0&2&1 \\ 0&0&0&0&0&-1 \end{bmatrix}$.
    My attempt: I first calculate its characteristic polynomial which gave me $\displaystyle p(\lambda )= (2-\lambda)^5(-1-\lambda)$.
    So there are 2 different eigenvalues, namely $\displaystyle \lamdba _1=2$ and $\displaystyle \lambda _2 =-1$.
    Now, I think I have to find the solution of $\displaystyle (M-\lambda _1I)^5 v_1=0$.
    I did lots of arithmetics to reach that $\displaystyle (M-\lambda _1I)^5 =\begin {bmatrix} 0&0&0&0&0&-27 \\ 0&0&0&0&0&-27 \\ 0&0&0&0&0&-27 \\ 0&0&0&0&0&-27 \\ 0&0&0&0&0&81 \\ 0&0&0&0&0&-243 \end{bmatrix}$ so that it gives me $\displaystyle v_1=\vec 0$... Mathematica doesn't agree on $\displaystyle (M-\lambda _1I)^5$, it seems I've made some errors in some values in the last column, but the matrix has the same form and therefore $\displaystyle v_1$ is still worth $\displaystyle 0$ which doesn't give me anything useful I believe.
    I don't know how to proceed.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,718
    Thanks
    3003
    You don't want to solve $\displaystyle (M- \lambda_1I)^5v=0$. You want to start by solving $\displaystyle (M- \lambda_1I)v= 0$ (such solutions will, of course, also satisfy $\displaystyle (M- \lambda_1I)^5v= 0$). From the Jordan Normal form, it looks like you should be able to find two independent solutions, call them $\displaystyle v_1$ and $\displaystyle v_2$. Those correspond to the two "$\displaystyle \lambda_1$"s on the diagonal with "0" above them. Then look for "generalized eigenvectors" that satisfy $\displaystyle (M- \lamba_1I)v= v_1$ and $\displaystyle (M- \lambda_1I)v= v_2$ (those will also satisfy $\displaystyle (M- \lambda_1I)^5v= 0$).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Jordan Normal Form Question
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: Jan 17th 2011, 03:29 PM
  2. Reducing 2x4 matrix to row-echelon form
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Feb 16th 2010, 02:16 AM
  3. Jordan Normal Form quick help!!!
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Jul 1st 2009, 10:06 AM
  4. Matrix - Jordan Normal/Canonical Form
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Mar 2nd 2009, 06:33 AM
  5. Jordan normal form
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Jun 8th 2008, 06:28 PM

Search Tags


/mathhelpforum @mathhelpforum