Beyond characterizing the dynamical system in question,considering stability is extremely impor- tant for any real-world applications.Indeed it is virtually impossible for any object to be precisely at an actual equilibrium point.Thus it is crucial to know what an object will do when close to an equilibrium for these are the actual dynamics that the object will necessarily follow. Before starting,we need a theorem and a few corollaries. Theorem 1.1 Given a system X'(t)=AX(t)where A has distinct paired complex eigenvalues, a1+iB1,a1-iB1...,ak+iBk:ak-iBx.Let T be the matrix such that T-AT is in canonical form: B T where Then the general solution of X'(t)=AX(t)is TY(t)where aleat cos Bit+breait sin Bit -aleait sin Bit+bieait cos Bit Y(t)= (1.1) akeaxt cos Bkt+bkeakt sin Bkt -akeakt sin Brt bkeakt cos Bxt The proof of this theorem is beyond the scope of this paper,but can be found in chp.6 of Differential Equations,Dynamical Systems,and an Introduction to Chaos 4]. Corollary 1.2 A point in a dynamical system whose matrir of equations of motion,A,has an eigenvalue with positive real part is unstable. Proof Let m be the index s.t.am >0.To show the point is stable,it suffices to show that the point is unstable in only one direction.To do so we let ai=bi=0Vim and assume that either am0 or bm0(or both).This is effectively considering small perturbations of the equilibrium point in only one direction.We denote ci;=(T)i.i(where T is the matrix described in theorem 1.1). We begin by noting that T is invertible so every column must have at least one non-zero entry and no column is the multiple of another column.This means that,for every 1 <j<k,there exists an i such that c,2j-1≠0orc,2i≠0andc,2-1≠±ci,2j(if this second condition were not the case then we would have(T)2j-1=+(T)2j which is impossible). Suppose the point is stable.Then we have |TY(0)-TY(t)<s for some s and all t>0.For any row,n,we get: E>am(Cn.2m-1-Cn.2m) -cmt[Cn.2m-1(am cos Bmt+om sin Bmt)+cn,2m (-am cos Bmt+om sin Bmt)] (1.2) There are two cases to consider: 3Beyond characterizing the dynamical system in question, considering stability is extremely impor￾tant for any real-world applications. Indeed it is virtually impossible for any object to be precisely at an actual equilibrium point. Thus it is crucial to know what an object will do when close to an equilibrium for these are the actual dynamics that the object will necessarily follow. Before starting, we need a theorem and a few corollaries. Theorem 1.1 Given a system X0 (t) = AX(t) where A has distinct paired complex eigenvalues, α1 +iβ1, α1 −iβ1 . . . , αk +iβk, αk −iβk. Let T be the matrix such that T −1AT is in canonical form: T −1AT =   B1 . . . Bk   where Bi =  αi βi −βi αi  Then the general solution of X0 (t) = AX(t) is T Y (t) where Y (t) =   a1e α1t cos β1t + b1e α1t sin β1t −a1e α1t sin β1t + b1e α1t cos β1t . . . ake αkt cos βkt + bke αkt sin βkt −ake αkt sin βkt + bke αkt cos βkt   (1.1) The proof of this theorem is beyond the scope of this paper, but can be found in chp. 6 of Differential Equations, Dynamical Systems, and an Introduction to Chaos [4]. Corollary 1.2 A point in a dynamical system whose matrix of equations of motion, A, has an eigenvalue with positive real part is unstable. Proof Let m be the index s.t. αm > 0. To show the point is stable, it suffices to show that the point is unstable in only one direction. To do so we let ai = bi = 0 ∀i 6= m and assume that either am 6= 0 or bm 6= 0 (or both). This is effectively considering small perturbations of the equilibrium point in only one direction. We denote ci,j = (T)i,j (where T is the matrix described in theorem 1.1). We begin by noting that T is invertible so every column must have at least one non-zero entry and no column is the multiple of another column. This means that, for every 1 ≤ j ≤ k, there exists an i such that ci,2j−1 6= 0 or ci,2j 6= 0 and ci,2j−1 6= ±ci,2j (if this second condition were not the case then we would have (T)2j−1 = ±(T)2j which is impossible). Suppose the point is stable. Then we have ||T Y (0) − T Y (t)|| < ε for some ε and all t > 0. For any row, n, we get: ε >      am(cn,2m−1 − cn,2m) − e αmt [cn,2m−1 (am cos βmt + bm sin βmt) + cn,2m (−am cos βmt + bm sin βmt)]      (1.2) There are two cases to consider: 3