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Thread: Jordan Canonical Form

  1. #1 Jordan Canonical Form 
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    Good morning guys,

    I've got a matrix with multiplicity 3, where all of the eigenvalues must be 0. (Ie: |sI-A| = s^3)

    I understand that in order to get the Q to put this in JCF I need to chain generalized Eigenvectors, but I'm having issues. The way the professor explained it I start with:

    (A-sI)^2v=0, and solve that. (I did, and I get [0 -4 5]' (col vector))

    Then solve (A-sI)v=0 (which gives [-1 0 0]' (col vector))

    and then his instructions get blurry, and the book also does not explain what happens at this point. It glosses over it and says the final v is [...]' where they give a col vector that is a solution to their problem.

    My A matrix is [0 4 3; 0 20 16; 0 -25 -20].

    So my question is what do I do to get this last eigenvector?

    Thanks in advance,


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    safety deserve neither liberty nor safety."

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  3. #2 Re: Jordan Canonical Form 
    . DrRocket's Avatar
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    Quote Originally Posted by alienmindsinc
    Good morning guys,

    I've got a matrix with multiplicity 3, where all of the eigenvalues must be 0. (Ie: |sI-A| = s^3)

    I understand that in order to get the Q to put this in JCF I need to chain generalized Eigenvectors, but I'm having issues. The way the professor explained it I start with:

    (A-sI)^2v=0, and solve that. (I did, and I get [0 -4 5]' (col vector))

    Then solve (A-sI)v=0 (which gives [-1 0 0]' (col vector))

    and then his instructions get blurry, and the book also does not explain what happens at this point. It glosses over it and says the final v is [...]' where they give a col vector that is a solution to their problem.

    My A matrix is [0 4 3; 0 20 16; 0 -25 -20].

    So my question is what do I do to get this last eigenvector?

    Thanks in advance,
    Your eigenvalues are all 0, so your matrix is nilopotent and in particular so you see immediately that you must be looking for generalized eigenvectors. If there were three linearly independent true eigenvectors then your matrix would represent a linear transformation in a 3-dimensional space spanned by 3 null bectors and hence would in fact be 0. But it is not 0.

    So you need three vectors such that

    [0 3 -4] should work.


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  4. #3  
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    Okay, I can verify that that does work, but why?

    When I do A*v = 0 (v = [v1;v2;v3]), I get two incompatible solutions.

    There is no v1 component because that column is 0. I would normally take this to mean that v1 can be anything, and assume you assigned 0 to it because it is easiest.

    Then I get v2 = -3/4*v3, followed by v2 = -4/5*v3 and v2 = -4/5*v3. If I assume the second 2 equations don't matter, and choose -4 as v3, that gives me a 3 as v2, and I get your answer, but then it makes the other equations incorrect. So why would I do this?

    Thanks again.
    --
    -M

    "Those that would give up essential liberty to obtain a little temporary
    safety deserve neither liberty nor safety."

    -Benjamin Franklin, An Historical Review of Pennsilvanya, 1759
    Reply With Quote  
     

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