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 an invertible matrix P such that A=PJP^-1

  1. Apr 10, 2013 #1
    1. The problem statement, all variables and given/known data
    [tex]
    A=\left[\begin{array}{cccc}
    5 & 4 & 2 & 1\\
    0 & 1 & -1 & -1\\
    -1 & -1 & 3 & 0\\
    1 & 1 & -1 & 2
    \end{array}\right]
    [/tex]

    1. Find the Jordan form of A
    2. Find a matrix P such that A = PJP-1

    2. Relevant equations



    3. The attempt at a solution

    I've already found, and verified the Jordan form it is,

    [tex]
    J=\left[\begin{array}{cccc}
    4 & 1 & 0 & 0\\
    0 & 4 & 0 & 0\\
    0 & 0 & 2 & 0\\
    0 & 0 & 0 & 1
    \end{array}\right]
    [/tex]

    from the characteristic polynomial [itex] q(x) = (x-4)^{2}(x-2)(x-1) [/itex].

    To find P, I found a basis vector for each of null spaces of each eigenvalue.
    [tex] \ker{(A-4I)} [/tex]
    Basis: [itex] (1,0,-1,1) [/itex]

    [tex] \ker{(A-4I)^{2}} [/tex]
    Basis: [itex] (0,0,-1,1), (1,0,0,0) [/itex]

    [tex] \ker{(A-2I)} [/tex]
    Basis: [itex] (1,-1,0,1) [/itex]

    [tex] \ker{A-I)} [/tex]
    Basis: [itex] (-1,1,0,0) [/itex]

    (these have all been verified with Mathematica)

    Now I'm supposed to construct the columns of P by picking a set of 4 linearly independent vectors, which they all are. Here is my struggle. In choosing between the two basis vectors for [itex] \ker{(A-4I)^{2}} [/itex], choosing [itex] (1,0,0,0) [/itex] works fine i.e.

    [tex]
    P=\left[\begin{array}{cccc}
    1 & 1 & 1 & -1\\
    0 & 0 & -1 & 1\\
    -1 & 0 & 0 & 0\\
    1 & 0 & 1 & 0
    \end{array}\right]
    [/tex]
    but if I choose the other vector,

    [tex]
    P=\left[\begin{array}{cccc}
    1 & 0 & 1 & -1\\
    0 & 0 & -1 & 1\\
    -1 & -1 & 0 & 0\\
    1 & 1 & 1 & 0
    \end{array}\right]
    [/tex]
    and P doesn't satisfy [itex] A=PJP^{-1} [/itex]. Oddly, if I take [itex] (0,0,1,-1) [/itex] it does work. Does anyone see if my logic is going awry somewhere? (0,0,1,-1) is just the negative of (0,0,-1,1) so shouldn't they in fact both work, satisfying A=PJP-1?
    Sorry for the long post.
     
  2. jcsd
  3. Apr 11, 2013 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If ##(A-4I)^2 v = 0## but ##(A - 4I) v \neq 0##, then let ##u = (A - 4I)v.## These give ##Av = u + 4v## and ##Au = 4u##. So, if you include u and v in a basis, they will give you the upper left Jordan block. The matrix P constructed from u, v and the other eigenvectors will do what you want.
     
  4. Apr 12, 2013 #3
    Great, thanks.
     
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 an invertible matrix P such that A=PJP^-1
Loading...