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Math Help - jordan basis problem

  1. #16
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    Re: jordan basis problem

    why we look at A-I and not A
    ?
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  2. #17
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    Re: jordan basis problem

    because all of A's eigenvalues are 1?
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  3. #18
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    Re: jordan basis problem

    why did they desided that one of the vectors is e1
    ?
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  4. #19
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    Re: jordan basis problem

    for generalized eigenvectors, you pick the first eigenvector v1 like you normally do, and continue until you run out (in this case, one is all there is).

    ker(A - I) has basis {(3,1,1)} now we want to pick our next generalized eigenvector so that:

    \begin{bmatrix}0&-3&3\\-2&-7&13\\-1&-4&7 \end{bmatrix} \begin{bmatrix}x\\y\\z \end{bmatrix} = \begin{bmatrix}3\\1\\1 \end{bmatrix}

    solve this for (x,y,z), which should give you (0,-2,-1), which is in ker((A-I)^2), but not in ker(A-I).

    do it again, and you should get e1 as the 3rd generalized eigenbasis vector.
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