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Jordan Normal Form

  1. Feb 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the Jordan form and use it to find the matrix exponential.

    2. Relevant equations



    3. The attempt at a solution

    Let A =\begin{bmatrix}
    -3 &-1 &-1 &-1 \\
    -1 & -3 & 1 & 1 \\
    1 & -1 & -5 & -1 \\
    1 & -1 & -1 & -5
    \end{bmatrix}

    det(A-λI)=λ4+16λ3+96λ2+256λ+256

    (λ+4)4
    → λ=-4 (Multiplicity 4)


    A+4I=\begin{bmatrix}
    1 &-1 &-1 &-1 \\
    -1 & 1 & 1 & 1 \\
    1 & -1 & -1 & -1 \\
    1 & -1 & -1 & -1
    \end{bmatrix}

    =\begin{bmatrix}
    1 &-1 &-1 &-1 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0
    \end{bmatrix}

    \begin{bmatrix}
    v1 \\
    v2 \\
    v3 \\
    v4
    \end{bmatrix} =
    v1\begin{bmatrix}
    1 \\
    1 \\
    0 \\
    0
    \end{bmatrix}+v2\begin{bmatrix}
    1 \\
    0 \\
    1 \\
    0
    \end{bmatrix} + v3\begin{bmatrix}
    1 \\
    0 \\
    0 \\
    1
    \end{bmatrix}



    Therefore, the Jordan Normal Form must look like:


    J=\begin{bmatrix}
    -4 &0 &0 &0 \\
    0 & -4 & 0 & 0 \\
    0 & 0 & -4 & 1 \\
    0 & 0 & 0 & -4
    \end{bmatrix}


    Consider now that:

    JE1=-4E1
    JE2=-4E2
    JE3=-4E3
    JE4=E3-4E4


    But this is equivalent to:

    AE1=-4E1
    AE2=-4E2
    AE3=-4E3
    AE4=E3-4E4


    AE4=E3-4E4→ (A+4I)E4=E3


    Now,

    ker[(A+4I)2]=0

    (A+4I)2E4=E3
    0*E4=0

    Therefore, E4 can be anything that is Linearly Independent of E1,E2, and E3.

    Let E4=\begin{bmatrix}
    1 \\
    1 \\
    1 \\
    1
    \end{bmatrix}


    (A+4I)E4=E3
    \begin{bmatrix}
    1 &-1 &-1 &-1 \\
    -1 & 1 & 1 & 1 \\
    1 & -1 & -1 & -1 \\
    1 & -1 & -1 & -1
    \end{bmatrix}*\begin{bmatrix}
    1 \\
    1 \\
    1 \\
    1
    \end{bmatrix}

    =\begin{bmatrix}
    -2 \\
    2 \\
    -2 \\
    -2
    \end{bmatrix}

    I'm lost after this point
     
  2. jcsd
  3. Mar 1, 2012 #2

    vela

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    $$\begin{bmatrix}
    1 &-1 &-1 &-1 \\
    -1 & 1 & 1 & 1 \\
    1 & -1 & -1 & -1 \\
    1 & -1 & -1 & -1
    \end{bmatrix} \ne
    \begin{bmatrix}
    1 &-1 &-1 &-1 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0
    \end{bmatrix}$$ I know what you mean, but don't use an equal sign there. The matrices aren't equal.

    This isn't correct. ##(A+4I)^2 = 0##, so the kernel is all of ##\mathbb{R}^4##.

    This isn't correct either. It's true that ##(A+4I)^2\vec{x}=0## is satisfied by any ##\vec{x}##, but it's not true that the vector will also satisfy ##(A+4I)\vec{x} = \vec{E}_3##.

    You found the kernel of A+4I consists of vectors of the form
    $$v1\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix} +
    v2\begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} +
    v3\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}$$ What you want to do is find a particular linear combination ##\vec{v}## so that ##(A+4I)\vec{x} = \vec{v}## will have a solution. The vector ##\vec{v}## corresponds to what you called E3, and the solution ##\vec{x}##, to E4.
     
  4. Mar 2, 2012 #3
    Ok, I went back and tried this:

    Let
    E4=
    \begin{bmatrix}
    0\\
    1\\
    1\\
    0
    \end{bmatrix}

    Then,

    (A+4I)E4=E3
    \begin{bmatrix}
    1 & -1 & -1 & -1\\
    -1 & 1 & 1 & 1\\
    1 & -1 & -1 & -1\\
    1 & -1 & -1 & -1
    \end{bmatrix}*\begin{bmatrix}
    0 \\
    1 \\
    1 \\
    0
    \end{bmatrix}
    =\begin{bmatrix}
    -2\\
    2 \\
    -2 \\
    -2
    \end{bmatrix}


    Form a matrix T such that T-1AT=J


    So, T=\begin{bmatrix}
    1 & 1 & -2 & 1\\
    1 & 0 & 2 & 1\\
    0 & 1 & -2 & 1\\
    0 & 0 & -2 & 1
    \end{bmatrix}

    and T-1=\begin{bmatrix}
    1 & 0 & -1 & 0\\
    0 & 0 & 1 & -1\\
    -1/4 & 1/4 & 1/4 & -1/4\\
    -1/2 & 1/2 & 1/2 & 1/2
    \end{bmatrix}


    T-1AT=\begin{bmatrix}
    -4 & 0 & 0 & 0\\
    0 & -4 & & 0\\
    0 & 0 & -4 & 1\\
    0 & 0 & 0 & -4
    \end{bmatrix}
    =J


    If this is correct, computing the matrix exponential is easy, if its correct...
     
  5. Mar 2, 2012 #4

    vela

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    Did you mean
    $$T = \begin{bmatrix}
    1 & 1 & -2 & 1\\
    1 & 0 & 2 & 1\\
    0 & 1 & -2 & 1\\
    0 & 0 & -2 & 1
    \end{bmatrix}$$ or
    $$T= \begin{bmatrix}
    1 & 1 & -2 & 0\\
    1 & 0 & 2 & 1\\
    0 & 1 & -2 & 1\\
    0 & 0 & -2 & 0
    \end{bmatrix}?$$
     
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