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Consider a systemdxdt=Axwhere A is a22matrix withtrA<0. We are told that A has no real eigenvalue. What can you say about the stability of the system

Short Answer

Expert verified

The given system dxdt=Axis stable.

Step by step solution

01

Step 1:Explanation for the continuous dynamical systems with eigen values p±iq

Consider the linear systemdxdt=Ax, where A is the real 22matrix with complex eigenvaluespiqandq0

Consider an eigenvectorv+iw with eigenvaluepiq . Then:

xt=eptScosqt-sinqtsinqtcosqtS-1x0, whereS=wv

Recall that S-1x0is the coordinate vector of x0with respect to the basis w,v.

02

Explanation of the stability of a continuous dynamical system

For a system,dxdt=Ax where A is the matrix form.

The zero state is an asymptotically stable equilibrium solution if and only if the real parts of all eigen values of A are negative.

03

Solution for the stability of the system

Consider A has no real eigen value, which means the real part of all the eigen values are negative.

Thus, the system dxdt=Axis stable

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