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Math Help - System of equations solution description

  1. #1
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    System of equations solution description

    If I have

    x' = Ax

    where A is the matrix below

    (1, -2
    3, -4)

    then the direction field of the solutions would have everything moving towards (0, 0) as t --> infinity. My question is: how would I describe the behavior as t --> infinity in words? Is it just "stable"?

    EDIT:

    A much more important question is: if I have a 2X2 matrix A, and the eigenvalues of that A are both greater than zero, then what would the direction field look like?

    Edit:

    Again, an even MORE important question is: if I have A as a 3X3 matrix, and so have three eigenvalues for it, and two of them are the same.... writing the general solution wouldn't be just

    x = (constant1)(eigenvector1)(e^eigenvalue1t) + (constant2)(eigenvector2)(e^eigenvalue2t) + (constant3)(eigenvector3)(e^eigenvalue3t)

    would it? That just doesn't seem right, and I can't get my answer doing it that way to match the book's answer.

    Thanks.

    ANOTHER EDIT:

    While I'm at it, let's put in another question here. I'm sure it's a simple question, but my class is at eight in the morning, and lasts forever, so... I drift.

    Anyway, how could I write an eigenvector in terms of its eigenvalue? What I'm trying to do is show that, considering the 2X2 system x' = Ax,
    (A - r1I)E1 = (r1 - r2)E1
    where r1 and r2 are the eigenvalues and E1 is the eigenvector corresponding to r1.
    Last edited by Oijl; April 8th 2009 at 02:02 PM.
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  2. #2
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722
    Ill throw in a bit of help.

    Here's how you classify linear systems...

    Eigenvalues ------------------- Condition ------------------------Type of system
    real, opposite sign ------------ det A < 0 ------------------------ saddle

    real, positive ----------------- 0 < det A < \frac{(tr A)^2}{4}, tr A > 0 ----- unstable node

    real, negative ---------------- 0 < det A < \frac{(tr A)^2}{4}, tr A < 0 ------- stable node

    complex, positive real part ----- det A > \frac{(tr A)^2}{4}, tr A > 0 -------- unstable focus

    complex, negative real part ----- det A > \frac{(tr A)^2}{4}, tr A < 0 -------- stable focus

    complex, zero real part --------- det A > 0, tr A = 0 -------------- centre

    Where det = determinant and tr = trace of a matrix

    Sorry im no good at doing latex tables!
    Attached Thumbnails Attached Thumbnails System of equations solution description-classify.jpg  
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