# Thread: Eigenvectors for the Phase Plane of a Two-Dimensional Linear System

1. ## 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

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

The eigenvectors of the matrix
$\displaystyle $M = \left[ {\begin{array}{cc} a & b \\ c & d \\ \end{array} } \right]$$

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.

2. If you write this system in vector/matrix form, you have

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

where $\displaystyle M$ is as you've defined it, and

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

The eigenvalues of the matrix $\displaystyle M$ are those numbers $\displaystyle \lambda$ such that $\displaystyle M\mathbf{x}=\lambda\mathbf{x},$ for a nonzero eigenvector $\displaystyle \mathbf{x}.$ That is, if you apply the matrix $\displaystyle M$ 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 $\displaystyle \mathbf{x},$ with corresponding eigenvalue $\displaystyle \lambda,$ then

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

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

What this DE is saying, geometrically, is that any change in the vector $\displaystyle \mathbf{x}$ 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.

3. Originally Posted by Ackbeet
If you write this system in vector/matrix form, you have

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

where $\displaystyle M$ is as you've defined it, and

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

The eigenvalues of the matrix $\displaystyle M$ are those numbers $\displaystyle \lambda$ such that $\displaystyle M\mathbf{x}=\lambda\mathbf{x},$ for a nonzero eigenvector $\displaystyle \mathbf{x}.$ That is, if you apply the matrix $\displaystyle M$ 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 $\displaystyle \mathbf{x},$ with corresponding eigenvalue $\displaystyle \lambda,$ then

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

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

What this DE is saying, geometrically, is that any change in the vector $\displaystyle \mathbf{x}$ 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.

4. Great! Any other questions?