- #1

andrew410

- 59

- 0

dx/dt = x(1.5 - x - 0.5y)

dy/dt = y(2 - y -0.75x)

I got all the critical points. They are (0, 0), (1.5, 0), (0, 2), (4/5, 7/5).

I understand how to get the eigenvalues, but I can't seem to get the eigenvectors.

So first you need to do that jacobian thing to the dx/dt and dy/dt.

After getting that for the corresponding critical point, I find A-rI for the Jacobian matrix and then do det(A-rI) to find the eigenvalues.

Now, I try to get the eigenvectors by subbing in the first eigenvalue into the A-rI matrix to get a new matrix. I use this new matrix to get the eigenvector, but the answer I get is incorrect.

For example, at critical point (0, 0), the eigenvalues are 1.5 and 2. The book tells me that the eigenvector is [tex]\left(\begin{array}{cc}1\\0\end{array}\right) [/tex] for eigenvalue 1.5 and [tex]\left(\begin{array}{cc}0\\1\end{array}\right) [/tex] for eigenvalue 2.

What am I doing wrong? Please help me...Thanks in advance!