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Math Help - Reducing a matrix to its Jordan normal form

  1. #1
    MHF Contributor arbolis's Avatar
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    Reducing a matrix to its Jordan normal form

    I must reduce the following matrix to its Jordan normal form: 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 p(\lambda )= (2-\lambda)^5(-1-\lambda).
    So there are 2 different eigenvalues, namely \lamdba _1=2 and \lambda _2 =-1.
    Now, I think I have to find the solution of (M-\lambda _1I)^5 v_1=0.
    I did lots of arithmetics to reach that (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 v_1=\vec 0... Mathematica doesn't agree on (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 v_1 is still worth 0 which doesn't give me anything useful I believe.
    I don't know how to proceed.
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  2. #2
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    You don't want to solve (M- \lambda_1I)^5v=0. You want to start by solving (M- \lambda_1I)v= 0 (such solutions will, of course, also satisfy (M- \lambda_1I)^5v= 0). From the Jordan Normal form, it looks like you should be able to find two independent solutions, call them v_1 and v_2. Those correspond to the two " \lambda_1"s on the diagonal with "0" above them. Then look for "generalized eigenvectors" that satisfy (M- \lamba_1I)v= v_1 and (M- \lambda_1I)v= v_2 (those will also satisfy (M- \lambda_1I)^5v= 0).
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