Clockwise or counterclockwise? (linear system phase portrait)

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SUMMARY

The discussion focuses on determining the rotational direction (clockwise or counterclockwise) of solutions to a linear system represented by a 2x2 matrix A with complex eigenvalues. The key finding is that the sign of the derivatives x1' and x2' can indicate the rotation direction, specifically that clockwise rotation corresponds to x1' > 0 and x2' < 0 for positive x1 and x2. The example matrices provided, such as A = [[0, -1], [1, 0]] for counterclockwise and A = [[0, 1], [-1, 0]] for clockwise, illustrate this concept effectively.

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Homework Statement



Given a 2x2 matrix A with entries a,b,c,d (real) with complex eigenvalues I would like to know how to find out whether the solutions to the linear system are clockwise or counterclockwise. (Some kind of inequality between a,b,c,d).


Homework Equations





The Attempt at a Solution



I tried looking at the signs of each component of the derivative. It seems to me that clockwise means x1' > 0 and x2' < 0 for x1,x2 > 0 as long as x1 or x2 are not too small. Then I am not sure...
 
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The simplest way is to see what the matrix does to (1, 0) and (0, 1).

For example, if the problem is
[tex]\frac{dX}{dt}= \left( \begin{array}{cc}0 & -1\\ 1 & 0\end{array}\right)X[/tex]
Then
[tex]\left( \begin{array}{cc}0 & -1\\ 1 & 0\end{array}\right)\left(\begin{array}{c} 1 \\ 0\end{array}\right)= \left(\begin{array}{c}0 \\ 1\end{array}\right)[/tex]
That's counter-clockwise rotation. The matrix
[tex]\left(\begin{array}{cc}0 & 1\\ -1 & 0\end{array}\right)[/tex]
has exactly the same eigenvalues but is clockwise rotation.
 

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