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Math Help - jordan basis problem

  1. #1
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    jordan basis problem

    find th jordan form of A and find P for which J=P^{-1}AP
    A=\left(\begin{array}{ccc}1 & -3 & 3\\-2 & -6 & 13\\-1 & -4 & 8\end{array}\right)
    |kI-A|=\left|\begin{array}{ccc}k-1 & +3 & -3\\2 & k+6 & -13\\+1 & +4 & k-8\end{array}\right|=(k-1)[(k+6)(k-8)+52]-3[2(k-8)+13]-3[8-k-6]
    =(k-1)[k^{2}-8k+6k-48+52]-3[2k-16+13]-[2-k]=(k-1)[k^{2}-2k+4]-3[2k-3]-[2-k]=k^{3}-2k^{2}+4k-k^{2}+2k-4-6k+9-2+k=k^{3}-3k^{2}+k+3

    the answer in the book is that the caracteristic polinomial is (x-1)^3
    i didnt get this answer
    why??


    then regarding the finding of the basis problem
    dont know what they have done here
    any guidence
    ??
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  2. #2
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    Re: jordan basis problem

    check your arithmetic! you dropped a factor of 3 from the term -3[8-k-6] (you should have -3[2-k] at the end, not -[2-k]).

    as far as finding P goes, i believe that dim(ker(A-I)) = 1 (prove this!), so the Jordan form of A is:

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

    now A-I has nilpotency 3, right? so we can pick any vector in R^3 for an element in ker((A-I)^3).

    it appears your text has chosen v1 = e1, and then chosen

    v2= (A-I)v1, which is in ker((A-I)^2), but not in ker((A-I)^3), so...
    Last edited by Deveno; October 7th 2011 at 05:30 PM.
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  3. #3
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    Re: jordan basis problem

    how did you know that A-I has nilpotency of 3?
    its a lot of calulations to do
    (A-I)(A-I)=A'
    A'A'=0
    did you actually calculated these 2 multiplications
    ??
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  4. #4
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    Re: jordan basis problem

    no, i deduced that it has nilpotency 3 from the fact that there is only 1 Jordan block.
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  5. #5
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    Re: jordan basis problem

    the number of jordan blocks=order of matrices-the rank of A

    number of jordan blocks=3-the rank of A

    for row reduction i got 3 independant lines
    so the rank is 3
    so the rank of A=3-3=0
    which is wrong?
    where is my mistake
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  6. #6
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    Re: jordan basis problem

    you agree that A-I is nilpotent, right? since A is not I, it's nilpotency must be 2 or 3. now we could verify (A-I)^2 is non-zero, or...

    we could find the dimension of the eigenspace E1, which is ker(A-I), which will tell us the number of Jordan blocks.

    you are checking the rank of the wrong matrix, try checking the rank of A-I.
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  7. #7
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    Re: jordan basis problem

    you must multiply it by itself and get zero and some point in order to establish that its nilpotent
    how do you see that its nilpotent just from looking at a matrices?
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  8. #8
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    Re: jordan basis problem

    what is dim(ker(A-I))? if we row-reduce A-I, we find it has rank 2. therefore dim(R^3) = rank(A-I) + dim(ker(A-I)):

    3 = 2 + dim(ker(A-I)), so dim(ker(A-I)) = 1.

    OR...let's find ker(A-I) explicitly:

    \begin{bmatrix}0&-3&3\\-2&-7&13\\-1&-4&7 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix}0\\0\\0 \end{bmatrix}

    right away, we see that x2 = x3.

    from either 2 of the remaining equations, we get x1 = 3x2.

    so {(3,1,1)} is a basis for A-I. so the Jordan form for A has just one Jordan block, in other words the entire superdiagonal is 1's.

    now...you agree that if J is the jordan form for A, that there is some matrix P with:

    A = PJP^{-1} right? so

    A - I = PJP^{-1} - I = PJP^{-1} - PIP^{-1} = P(J-I)P^{-1}

    what is the nilpotency of J-I? just look at it...it's obviously 2! J-I =

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

    (J-I)^2 will raise the superdiagonal to the upper-right-hand corner, so (J-I)^2 won't be 0, so (A-I)^2 won't be 0.
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  9. #9
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    Re: jordan basis problem

    Quote Originally Posted by Deveno View Post
    what is dim(ker(A-I))? if we row-reduce A-I, we find it has rank 2. therefore dim(R^3) = rank(A-I) + dim(ker(A-I)):

    3 = 2 + dim(ker(A-I)), so dim(ker(A-I)) = 1.

    OR...let's find ker(A-I) explicitly:

    \begin{bmatrix}0&-3&3\\-2&-7&13\\-1&-4&7 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix}0\\0\\0 \end{bmatrix}

    right away, we see that x2 = x3.

    from either 2 of the remaining equations, we get x1 = 3x2.

    so {(3,1,1)} is a basis for A-I. so the Jordan form for A has just one Jordan block, in other words the entire superdiagonal is 1's.

    now...you agree that if J is the jordan form for A, that there is some matrix P with:

    A = PJP^{-1} right? so

    A - I = PJP^{-1} - I = PJP^{-1} - PIP^{-1} = P(J-I)P^{-1}

    what is the nilpotency of J-I? just look at it...it's obviously 2! J-I =

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

    (J-I)^2 will raise the superdiagonal to the upper-right-hand corner, so (J-I)^2 won't be 0, so (A-I)^2 won't be 0.
    ok so because the rank of A-I is two its joudan form can have only one block
    and its
    \begin{bmatrix}0&1&0\\0&0&1\\0&0&0 \end{bmatrix}
    but its not a proper prof because we have to espablish the minimal polinomial to find the smallest block.

    our caracteristic polinomial is p(t)=(x-1)^3
    and the minimal could be M(t)=(x-1)^3 or M(t)=(x-1)^2
    or M(t)=(x-1)

    so again i must reprlace x with A and make very long calculations to see if i get the zero matrices in one of them
    ?

    is there another way to find the minimal polinomial

    ?
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  10. #10
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    Re: jordan basis problem

    the minimal polynomial could only be (x-1)^2 or (x-1)^3.

    if (x-1)^2 was the minimal polynomial, (A-I)^2 would be 0, so (J-I)^2 would be 0.

    but (J-I)^2 is not 0, it's just not...try it, you'll see (very easy calculation).

    suppose x-1 was the minimal polynomial. then A-I = 0 --> A = I. but A is NOT I. so x -1 is not the minimal polynomial. no. isn't. never will be.

    that only leaves (x-1)^3. it's our only hope, obi-wan kenobi....
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  11. #11
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    Re: jordan basis problem

    if A-I nilpotent 3 then its largest jordan block is 3
    so we dont need minimal polinomial

    then why do we need minimal polinomial
    in findind jordan form
    ?
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  12. #12
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    Re: jordan basis problem

    there are different ways of solving linear algebra problems. surely you know that by now.
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  13. #13
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    Re: jordan basis problem

    Quote Originally Posted by Deveno View Post
    there are different ways of solving linear algebra problems. surely you know that by now.
    ok coming beack to the originak question

    we want to find P:
    you said
    "now A-I has nilpotency 3, right? so we can pick any vector in R^3 for an element in ker((A-I)^3).

    it appears your text has chosen v1 = e1, and then chosen

    v2= (A-I)v1, which is in ker((A-I)^2), but not in ker((A-I)^3), so... "

    why we can pick any vector in R^3 for ker((A-I)^3)?
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  14. #14
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    Re: jordan basis problem

    Quote Originally Posted by transgalactic View Post
    number of jordan blocks=3-the rank of A
    No, the number of Jordan blocks associated to an eigenvalue \lambda of A\in\mathbb{K}^{n\times n} is n-\textrm{rank}(A-\lambda I) . In our case 3-\textrm{rank}(A- I)=3-2=1 .
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  15. #15
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    Re: jordan basis problem

    so from the jourdan form
    if our basis B={v1,v2,v3}
    Tv1=1v1+0v2+0v3
    Tv2=1v1+1v2+0v3
    Tv3=0v1+1v2+1v3

    Tv1=Av1


    but in the solution they take A-I insteat of A

    and they choose e1 e2 e3 in the standart basis

    why?
    Last edited by transgalactic; October 9th 2011 at 02:43 AM.
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