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Math Help - Eigenvectors for the Phase Plane of a Two-Dimensional Linear System

  1. #1
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    Eigenvectors for the Phase Plane of a Two-Dimensional Linear System

    I'm having trouble seeing how eigenvectors help in picturing the phase plane for a two-dimensional linear system in the general form

    \frac{dx}{dt}=ax+by
    \frac{dy}{dt}=cx+dy

    The eigenvectors of the matrix
    \[<br />
M =<br />
\left[ {\begin{array}{cc}<br />
 a & b  \\<br />
 c & d  \\<br />
 \end{array} } \right]<br />
\]

    tells us the direction of the straight-line trajectories (assuming the eigenvalues are real and distinct). I'm having trouble seeing how the eigenvalues determine the straight-line trajectories.
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  2. #2
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    If you write this system in vector/matrix form, you have

    \dot{\mathbf{x}}=M\mathbf{x},

    where M is as you've defined it, and

    \mathbf{x}=\begin{bmatrix}x\\y\end{bmatrix}.

    The eigenvalues of the matrix M are those numbers \lambda such that M\mathbf{x}=\lambda\mathbf{x}, for a nonzero eigenvector \mathbf{x}. That is, if you apply the matrix M to an eigenvector, you don't change its direction (except possibly an exact reversal in the case of a negative eigenvalue). Hence, if you have an eigenvector \mathbf{x}, with corresponding eigenvalue \lambda, then

    \dot{\mathbf{x}}=M\mathbf{x}=\lambda\mathbf{x}. That is,

    \dot{\mathbf{x}}=\lambda\mathbf{x}.

    What this DE is saying, geometrically, is that any change in the vector \mathbf{x} can only be an elongation or a reversal. I can't have any rotations. That is, it must be straight-line motion.

    In summary, I would partially agree/partially disagree with the statement. It's the eigenvectors that determine the straight-line motion. But in order to find the eigenvectors, of course, you have to find the eigenvalues.

    Hope this helps.
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    If you write this system in vector/matrix form, you have

    \dot{\mathbf{x}}=M\mathbf{x},

    where M is as you've defined it, and

    \mathbf{x}=\begin{bmatrix}x\\y\end{bmatrix}.

    The eigenvalues of the matrix M are those numbers \lambda such that M\mathbf{x}=\lambda\mathbf{x}, for a nonzero eigenvector \mathbf{x}. That is, if you apply the matrix M to an eigenvector, you don't change its direction (except possibly an exact reversal in the case of a negative eigenvalue). Hence, if you have an eigenvector \mathbf{x}, with corresponding eigenvalue \lambda, then

    \dot{\mathbf{x}}=M\mathbf{x}=\lambda\mathbf{x}. That is,

    \dot{\mathbf{x}}=\lambda\mathbf{x}.

    What this DE is saying, geometrically, is that any change in the vector \mathbf{x} can only be an elongation or a reversal. I can't have any rotations. That is, it must be straight-line motion.

    In summary, I would partially agree/partially disagree with the statement. It's the eigenvectors that determine the straight-line motion. But in order to find the eigenvectors, of course, you have to find the eigenvalues.

    Hope this helps.
    Thanks!

    This did really help and I now understand how the eigenvectors determine the straight-line motion. I actually did make a mistake in saying that the eigenvalues determine the straight-line trajectory; I meant to say eigenvectors.
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  4. #4
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    Great! Any other questions?
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