Question

http://imageshack.us/photo/my-images/856/screenshot20120613at102.png/
For this theres a typo it should be a+d <0 is stable and a+d >0 is unstable

How do you start with these proofs? I know that if the eigenvalues are both negative then its stable and if there both positive its unstable. And if one is negative and the other is positive then its unstable saddle. How do you put that in proof form?

Answer 1

The characteristic equation of that system is going to be:\[\lambda^2-(a+d)\lambda+(ad-bc)=0\] which is the same as:\[\lambda^2-tr(A)+\det(A)=0\]where tr(A) is the trace, and det(A) is the determinant. Now in each case apply the conditions. For example, if det(A) = 0 and a+d is positive, then you equation would be:\[\lambda^2-(a+d)\lambda-\det(A)=\lambda^2-(a+d)\lambda=\lambda(\lambda-(a+d))=0\]\[\Longrightarrow \lambda = 0,\lambda = a+d>0\]Your eigenvalues are 0 and a positive number

Answer 2

Come up with a generic form of the eigenvalues and eigenvectors in terms of \(a\), \(b\), \(c\), and \(d\), and work it from there.

Answer 3

I get everything there but how do you plug in the first and last ones?

Answer 4

This may help:\[\lambda^2-\text{tr}(A)\lambda +\det(A)=0\]implies\[\lambda = \frac{\text{tr}(A)}{2}\pm \sqrt{\frac{\left( \text{tr}(A)\right)^2}{4}-\det (A) }\]

Answer 5

Yes it does thanks