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Math Help - eigen value geometric multiplicity problem

  1. #1
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    eigen value geometric multiplicity problem

    find the jordan form of

    A=\left(\begin{array}{cccc}
    4 & 1 & 1 & 1\\
    -1 & 2 & -1 & -1\\
    6 & 1 & -1 & 1\\
    -6 & -1 & 4 & 2\end{array}\right)

    ?

    my latex around the matrices is not working and it meesed up the whole thing
    so i removed it

    |tI-A|=\left|\begin{array}{cccc}
    t-4 & -1 & -1 & -1\\
    1 & t-2 & 1 & 1\\
    -6 & -1 & t+1 & -1\\
    6 & 1 & -4 & t-2\end{array}\right|[/TEX]


    =(t+3)\left|\begin{array}{cccc}
    1 & 1 & 1 & 1\\
    1 & t-2 & 1 & 1\\
    -6 & -1 & t+1 & -1\\
    6 & 1 & -4 & t-2\end{array}\right|=(t-3)^{2}(t-2)

    i did R_{1}+R_{2}+R_{3}+R_{4}->R_{1} and took out (t+3) and got the polinomial above
    so i need to have one block jordan of eigen value 2
    so when i calculate
    dim ker(A-2I)

    A-2I=\left(\begin{array}{cccc}
    2 & 1 & 1 & 1\\
    -1 & 0 & -1 & -1\\
    6 & 1 & -3 & 1\\
    -6 & -1 & 4 & 0\end{array}\right)

    2R_{2}+R_{1}->R_{1}

    =\left(\begin{array}{cccc}
    0 & 1 & -1 & -1\\
    -1 & 0 & -1 & -1\\
    6 & 1 & -3 & 1\\
    -6 & -1 & 4 & 0\end{array}\right)

    6R_{2}+R_{3}->R_{3}

    =\left(\begin{array}{cccc}
    0 & 1 & -1 & -1\\
    -1 & 0 & -1 & -1\\
    0 & 1 & -9 & -5\\
    -6 & -1 & 4 & 0\end{array}\right)

    -6R_{2}+R_{4}>R_{4}

    =\left(\begin{array}{cccc}
    0 & 1 & -1 & -1\\
    -1 & 0 & -1 & -1\\
    0 & 1 & -9 & -5\\
    0 & -1 & 10 & 6\end{array}\right)

    R_{3}+R_{4}>R_{4}

    =\left(\begin{array}{cccc}
    0 & 1 & -1 & -1\\
    -1 & 0 & -1 & -1\\
    0 & 1 & -9 & -5\\
    0 & 0 & 1 & 1\end{array}\right)
    R_{4}+R_{2}>R_{2}
    R_{4}+R_{1}>R_{1}

    =\left(\begin{array}{cccc}
    0 & 1 & 0 & 0\\
    -1 & 0 & 0 & 0\\
    0 & 1 & -9 & -5\\
    0 & 0 & 1 & 1\end{array}\right)

    -R_{1}+R_{3}>R_{3}

    =\left(\begin{array}{cccc}
    0 & 1 & 0 & 0\\
    -1 & 0 & 0 & 0\\
    0 & 0 & -9 & -5\\
    0 & 0 & 1 & 1\end{array}\right)


    9R_{4}+R_{3}>R_{3}


    \left(\begin{array}{cccc}
    0 & 1 & 0 & 0\\
    -1 & 0 & 0 & 0\\
    0 & 0 & 0 & 4\\
    0 & 0 & 1 & 1\end{array}\right)=0

    so from here if my vector is (x,y,z,t) i get
    4t=0 ->t=0
    z+t=0 -> z=0
    -x=0 -> x=0
    y=0
    so i get
    that the geometric multiplicity of t=2 is zero
    and it should be 1
    where is my mistake
    ?
    Last edited by transgalactic; November 7th 2011 at 06:35 AM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: eigen value geometric multiplicity problem

    The eigenvalues of A are \lambda_1=3 (triple) and \lambda_2=-2 (simple), so \dim \ker (A+2I)=1 and \dim \ker (A-3I)=4-\textrm{rank}(A-3I)=\ldots=4-2=2 . Hence the Jordan form of A is:

    J=\begin{bmatrix}{3}&{1}&{0}&\;\; 0\\0&{3}&{0} & \;\;0\\{0}&{0}&{3}&\;\;0\\0&0&0&-2\end{bmatrix}
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Re: eigen value geometric multiplicity problem

    Quote Originally Posted by transgalactic View Post
    my latex around the matrices is not working and it meesed up the whole thing so i removed it
    Look at the horrible code you have written. Only with a miracle as the multiplication of the loaves and fishes you could get a correct expression.
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