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Upper triangular matrix

  1. Jan 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex] A = \begin{pmatrix}4 & 1 & 0\\0 & 3 & 1\\-1 & -1 & 2\end{pmatrix}[/itex], which has characteristic polynomial [itex](3-x)^{3}[/itex]. Find a matrix X such that:

    [itex]X^{-1}AX = \begin{pmatrix}3 & 1 & 0\\0 & 3 & 1\\0 & 0 & 3\end{pmatrix}[/itex]



    3. The attempt at a solution
    If [itex]X^{-1}AX = \begin{pmatrix}3 & 1 & 0\\0 & 3 & 1\\0 & 0 & 3\end{pmatrix} =: B[/itex]

    AX = XB

    Suppose that the columns of X are [itex]v_{1}, v_{2}, v_{3}[/itex]
    The first column of B has the coordinates of [itex]Av_{1}[/itex]

    [itex]Av_{1}=3v_{1}+0v_{2}+0v_{3}=3v_{3}~~~~(A-3I)v_{1}=0[/itex]

    [itex] v_{1} = \begin{pmatrix}a\\d\\g\end{pmatrix}[/itex]

    [itex] (A-3I) = \begin{pmatrix}1 & 1 & 0\\0 & 0 & 1\\-1 & -1 & -1\end{pmatrix} → \begin{pmatrix}1 & 1 & 0\\0 & 0 & 1\\0 & 0 & -1\end{pmatrix} → \begin{pmatrix}1 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{pmatrix}[/itex]

    let d=1, a=-1, g=0

    [itex] v_{1} = \begin{pmatrix}-1\\1\\0\end{pmatrix}[/itex]

    [itex]Av_{2} = 1v_{1}+3v_{2}+0v_{3} = v_{1} + 3v_{2}[/itex]

    [itex](A-3I)v_{2} = v_{1}[/itex]

    [itex] v_{2} = \begin{pmatrix}b\\e\\h\end{pmatrix}[/itex]

    [itex] \begin{pmatrix}1 & 1 & 0\\0 & 0 & 1\\-1 & -1 & -1\end{pmatrix} \begin{pmatrix}b\\e\\h\end{pmatrix} = \begin{pmatrix}-1\\1\\0\end{pmatrix}[/itex]

    [itex] \begin{pmatrix}1 & 1 & 0 & | ~~-1\\0 & 0 & 1 & |~~ 1\\-1 & -1 & -1 & |~~ 0\end{pmatrix} → \begin{pmatrix}1 & 1 & 0 & | ~~-1\\0 & 0 & 1 & | ~~1\\0 & 0 & -1 & | ~~-1\end{pmatrix} → \begin{pmatrix}1 & 1 & 0 & | ~~-1\\0 & 0 & 1 & |~~ 1\\0 & 0 & 0 & |~~ 0\end{pmatrix} [/itex]

    [itex] v_{2} = \begin{pmatrix}0\\-1\\0\end{pmatrix}[/itex]



    [itex]Av_{3} = 0v_{1}+1v_{2}+3v_{3} = v_{2} + 3v_{3}[/itex]

    [itex](A-3I)v_{3} = v_{2}[/itex]

    [itex] v_{3} = \begin{pmatrix}c\\f\\i\end{pmatrix}[/itex]

    [itex] \begin{pmatrix}1 & 1 & 0\\0 & 0 & 1\\-1 & -1 & -1\end{pmatrix} \begin{pmatrix}c\\f\\i\end{pmatrix} = \begin{pmatrix}0\\-1\\0\end{pmatrix}[/itex]

    [itex] \begin{pmatrix}1 & 1 & 0 & | ~~0\\0 & 0 & 1 & | ~~-1\\-1 & -1 & -1 & |~~ 0\end{pmatrix} → \begin{pmatrix}1 & 1 & 0 & |~~ 0\\0 & 0 & 1 & | ~~-1\\0 & 0 & -1 & |~~ \end{pmatrix} → \begin{pmatrix}1 & 1 & 0 & |~~ 0\\0 & 0 & 1 & | ~~-1\\0 & 0 & 0 & | ~~-1\end{pmatrix} [/itex]

    Therefore, there does not exist an invertible matrix X because the last matrix can not be solved to get [itex]v_{3}[/itex] so that X can be created.

    I don't think this is right as my teacher generally doesn't give us 'unsolvable' questions. I'm not sure what I've done wrong, and I've tried it a few different ways and I can never get it to work. Thanks!
     
  2. jcsd
  3. Jan 18, 2012 #2
    I've only just finished my linear algebra course. This looks like you're being given the Jordan Normal form of the matrix, and asked to find the decomposition. Every square matrix has a Jordan decomposition.
     
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