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Math Help - Jordan Normal Form quick help!!!

  1. #1
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    Jordan Normal Form quick help!!!

    I have two matricies A and B:

    A=\left[\begin{array}{ccc}<br />
 0 & 0 & 1\\<br />
 1 & 0 & -3\\<br />
 0 & 1 & 3\end{array}\right] B=\left[\begin{array}{ccccc} 3 & 1 & 0 & 1 & -2\\<br />
 1 & 3 & -1 & 0 & 1\\<br />
 -1 & -1 & 4 & 3 & -3\\<br />
 1 & 1 & -1 & 2 & 1\\<br />
 -2 & -2 & 2 & 2 & 1\end{array}\right]<br />

    which I need to put in Jordan Normal Form:

    J_{A}=\left[\begin{array}{ccc}<br />
 -1 & 0 & 0\\<br />
 0 & 1 & 1\\<br />
 0 & 0 & 1\end{array}\right] J_{B}=\left[\begin{array}{ccccc}<br />
 2 & 1 & 0 & 0 & 0\\<br />
 0 & 2 & 0 & 0 & 0\\<br />
 0 & 0 & 3 & 1 & 0\\<br />
 0 & 0 & 0 & 3 & 1\\<br />
 0 & 0 & 0 & 0 & 3\end{array}\right]

    (Not sure if I got this right)

    and then (my biggest problem) construct matrices S an T so that
    S^{-1}AS=J_{A} and T^{-1}BT=J_{B} (I know th significance, but I'm not sure how to construct these matricies.)
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by realpart1/2 View Post
    I have two matricies A and B:

    A=\left[\begin{array}{ccc}<br />
 0 & 0 & 1\\<br />
 1 & 0 & -3\\<br />
 0 & 1 & 3\end{array}\right] B=\left[\begin{array}{ccccc} 3 & 1 & 0 & 1 & -2\\<br />
 1 & 3 & -1 & 0 & 1\\<br />
 -1 & -1 & 4 & 3 & -3\\<br />
 1 & 1 & -1 & 2 & 1\\<br />
 -2 & -2 & 2 & 2 & 1\end{array}\right]<br />

    which I need to put in Jordan Normal Form:

    J_{A}=\left[\begin{array}{ccc}<br />
 -1 & 0 & 0\\<br />
 0 & 1 & 1\\<br />
 0 & 0 & 1\end{array}\right] J_{B}=\left[\begin{array}{ccccc}<br />
 2 & 1 & 0 & 0 & 0\\<br />
 0 & 2 & 0 & 0 & 0\\<br />
 0 & 0 & 3 & 1 & 0\\<br />
 0 & 0 & 0 & 3 & 1\\<br />
 0 & 0 & 0 & 0 & 3\end{array}\right]

    (Not sure if I got this right)

    and then (my biggest problem) construct matrices S an T so that
    S^{-1}AS=J_{A} and T^{-1}BT=J_{B} (I know th significance, but I'm not sure how to construct these matricies.)
    Can you do A? From what I can understand, you manage to get to the Jordan form but you can't find the base change matrices?

    What are you able to do:

    Can you find the eigenvalues?
    Do you know the definition of generalized eigenspaces?
    From the definition you should be able to find a basis for the generalized eigenspaces corresponding to the eigenvalues -1 and 1.

    You have to say where you are stuck precisely.
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  3. #3
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    I got the Eigenvalues by using the respective characteristic polynomials of A and B, and as far as I understood, the Jordan form is made by putting the eigenvalues on the diagonal, keeping the identical values together and adding 1's in-between. The rest was never fully explained. I know what the inverse matrices around A and B are supposed to do but not how to construct them (If I'm right, their existence somehow defines the Jordan form).
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    The Jordan form you have for A is wrong.

    <br />
A=\left[\begin{array}{ccc}<br />
 0 & 0 & 1\\<br />
 1 & 0 & -3\\<br />
 0 & 1 & 3\end{array}\right]<br />

    Characteristic polynomial :

    <br />
f(t)=\mbox{det }\left[\begin{array}{ccc}<br />
 -t& 0 & 1\\<br />
 1 & -t & -3\\<br />
 0 & 1 & 3-t\end{array}\right] = t^2(3-t)+1-3t = -(t-1)^3<br />

    So 1 is the only eigenvalue. You will have

    K_1 = \mbox{Kernel }(A-I)^3 = \mbox{Kernel }\left[\begin{array}{ccc}<br />
 -1& 0 & 1\\<br />
 1 & -1 & -3\\<br />
 0 & 1 & 2\end{array}\right]^3=\mbox{Kernel }\left[<br />
\begin{array}{lll}<br />
 -1 & 0 & 1 \\<br />
 1 & -1 & -27 \\<br />
 0 & 1 & -1<br />
\end{array}<br />
\right]

    Now find a vector v in the kernel such that (A-I)^2v, (A-I)v are nonzero. Your basis for K_1 will be given by \{(A-I)^2v, (A-I)v, v\}.
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