Consider the system of ordinary differential equations x¢ (t) = Ax(t) + b. Let x(t) = F (t)c + xp where xp is a particular solution to the system and F(t)c is the general solution of the homogeneous case x¢ = Ax. We define a matrix A as "stable" iff x(t) is stable, or:
We now prove a quite standard theorem:
Proof: Suppose A is stable and let l i = ui + iwi be an eigenvalue (complex or real) of A. Let vi be the associated eigenvector, thus Avi = l ivi. Now, we know that F(t) = [v1el1t, v2el2t, .., vnelnt] thus, defining j i(t) = vielit, then:
Now, as j i(t) = vielit, then taking the first derivative ji¢ (t) = l ivielit = Avielit = Aj i(t) by definition. Thus, ji(t) is a solution to the homogeneous system, x¢ = Ax. Now, by definition, A is stable iff F(t)c ® 0 when t ® ¥ , thus it must be that j i(t) ® 0 for all i = 1, .., n for A to be stable. We now show that this implies that the real parts of the l i must be negative for all i, i.e. ui < 0 for all i. Consider the kth coordinate of vi and thus of j i(t), i.e. j ik(t) = vikelit. Since limt® ¥ j i(t) = 0, then it must be that limt® ¥ j ik(t) = 0, or limt® ¥ |j ik(t)| = 0 or limt® ¥ |vikelit| = 0. Dropping the subscript i for cleaner notation, so that we now are consider l = u + iw as a representative eigenvalue, then limt® ¥ |vkelit| = limt® ¥ |vk||elit| = limt® ¥ |vk||e(u + iw)t| = limt® ¥ |vk||eut||e iwt| = 0. As |eiwt| has a finite upper bound, i.e. as |eiwt| £ M for some M < ¥ and eut > 0, then necessarily u < 0, i.e. Re(l ) < 0. Since this is true for any l (i.e. for all l i, i = 1, .., n), it follows that if A is stable, then Re(l i) = ui < 0 for all i = 1, .., n, i.e. the real parts of all eigenvalues must be negative. Conversely, suppose that the real parts of all eigenvalues l i = ui + iwi so Re(l i) = ui < 0 for all i = 1, .., n. Let x(t) be a solution to x¢ = Ax and let xk(t) be the kth coordinate of x(t). Then, xk(t) = å j=1n a kj eljt where a kj are coefficients and l j = uj + iwj. Thus, |xk(t)| = |å j=1n a kjeljt| = |å j=1n a kje(uj + iwj) t| =|å j=1n a kjeujteiwjt|. This implies, by triangular inequality that |xk(t)| £ å j=1n|a kjeujteiwjt| = å j=1n|a kj||eujt||eiwjt|. Now, again, by upper boundedness |eiwt| £ M and eut > 0, then |xk(t)| £ å j=1n|a kj|eujtM. Thus, limt® ¥ |xk(t)| £ limt® ¥ å j=1n|a kj|eujtM. Since uj < 0 for all j = 1, 2, .., n, then limt® ¥ å j=1n|a kj|eujtM = 0, thus it follows that limt® ¥ |xk(t)| £ 0, i.e. limt® ¥ |xk(t)| = limt® ¥ xk(t) = 0. Since this is true for any k = 1, 2, .., n, then the solution x(t) converges towards to zero as t converges towards infinity. Thus, the matrix A is stable.§ Under what conditions are all eigenvalues negative? The Routh-Hurwitz conditions on n-degree polynomials are necessary and sufficient for negativity of all eigenvalues. Murata's (1977: p.92) "modified Routh-Hurwitz" conditions can be simpler to use. Nonetheless, we can still look for properties of A that will yield a stable A. The following are straightforward:
Proof: From quadratic forms on a real symmetric matrix, we know that A is negative definite if and only if all its eigenvalues are negative. We also know for symmetric matrices that A is negative definite if and only if (-1)k|Ak| > 0.§
Proof: This is elementary. Recall that trA = å i=1n l i and |A| = Õ i=1n l i. Thus, if all l i < 0, then necessarily, tr A < 0. Similarly, if all l i < 0, then |A| > 0 if n is even and |A| < 0 if n is odd, thus it must be that (-1)n|A| > 0.§ The above theorem is actually sufficient if A is a 2 ´ 2 matrix. However, for higher dimensions, this is no longer the case. As a result we must look elsewhere for sufficient conditions. The following are a few.
Proof: Given elsewhere via Lyapunov's method.§
Proof: This follows from Lyapunov's theorem. Namely, if A is quasi-negative definite, then A + A¢ is negative definite. Thus, A¢ I + IA is negative definite. Obviously, letting I = H, then we can see this is merely a special case of the previous theorem.§
Proof: If A is symmetric and negative definite, then A + A¢ is negative definite, and Lyapunov's theorem applies.§ A more interesting set of sufficiency conditions has been derived in the course of the analysis of the local stability of a Walrasian tatonnement process in general equilibrium. In short, we have the following:
It is obvious that if A is D-stable, then A is stable (just set B = I). Sufficient conditions for D-stability are the following:
Proof: (Arrow and McManus, 1958). To see this, we must prove that BA is stable. Thus, from the earlier theorem, we need (BA)¢ H + H(BA) to be negative definite. Let H = CB-1. As B is a positive diagonal matrix, then (B-1)¢ = B-1 and B = B¢ . If C is a positive diagonal matrix, then H = CB-1 is also a positive diagonal matrix. Substituting in for H, then the condition becomes (A¢ B¢ )CB-1 + CB-1(BA) = A¢ C + CA. Since we assumed that A¢ C + CA is negative definite, then it also must be that (BA)¢ H + H(BA) is negative definite, thus BA is stable and thus A is D-stable.§ It is easy to notice that if we let C = I, then we obtain the condition that A¢ + A is negative definite. We know that if (i) A is quasi-negative definite or if (ii) A is symmetric and negative definite, then A¢ + A is negative definite. Thus (i) and (ii) are also sufficient conditions for stability. A more interesting set of results is derived from the property of "diagonal dominance" introduced into economics by Lionel McKenzie (1960).
This is also known as "column" diagonal dominance. "Row" diagonal dominance is also available and defined as the existence of di > 0 such that di|aii| > å j¹ i dj|aij| for all i = 1, 2, ...., n. It is easy to show that a matrix which is column dd will also be row dd. This is from the Hawkins-Simon condition: namely, suppose -|aij| is the ij entry of A for i ¹ j and |aii| is the ii entry, then y > 0 such that yA > 0 implies x > 0 such that Ax > 0.
Alternatively, if we consider a matrix D with diagonal elements dj as defined in the dd case, and A is dd, then obviously DA is a Hadamard matrix.
Proof: (McKenzie, 1960) Suppose not. Suppose |A| = 0. Then there is an x ¹ 0 such that DAx = 0, thus djajjxj + å i¹ j diaijxi = 0 for all j = 1, 2, .., n. In absolute value terms, then, using the triangular inequality, this implies for all j:
Let J be an index set such that if j Î J, then |xj| ³ |xi| for all i = 1, .., n. Then, for j Î J:
which implies:
or, as all dj > 0, then:
for all j Î J, which contradicts diagonal dominance.§ The following theorem, due to McKenzie (1960), establishes that if A has a negative dominant diagonal, then A is stable: : (Sufficiency) If an n ´ n matrix A is dominant diagonal and the diagonal is composed of negative elements (aii < 0 for all i = 1, .., n), then the real parts of all its eigenvalues are negative, i.e. A is stable. Proof: (McKenzie, 1960) Suppose that A has an eigenvalue l = u + iv with a non-negative real part, i.e. u ³ 0. As aii < 0, then:
Since A has a dominant diagonal, there exist dj > 0 (j = 1, .., n) such that dj|ajj| ³ å i¹ j di |aij |, thus:
But notice that this implies that the system (l I - A) has a dominant diagonal as well. By our previous theorem, we know that if a matrix has dd, then it is non-singular, i.e. it must be that |l I - A| ¹ 0. But then that contradicts the postulate that l is an eigenvalue of A. Thus, there is no eigenvalue l with a non-negative real part, i.e. the real parts of all eigenvalues of A are negative (ui < 0 for all i) , thus A is stable.§ Finally, note the following corollary:
Proof: We have to prove that BA is stable, where B is a positive diagonal matrix (thus, bii > 0 for all i). If A has negative diagonal dominance, then we know all its eigenvalues have negative real parts. Recall that diagonal dominance implies that there are dj > 0 (j = 1, .., n) such that dj|ajj| ³ å i¹ j di |aij |. Consequently, dividing and multiplying by the relevant bjj > 0, we obtain:
Thus, dj/bjj > 0 will satisfy diagonal dominance for matrix BA. Finally, as ajj < 0 and bjj > 0 for all j, then bjjajj < 0, thus BA has a negative diagonal. Thus, BA is negative dominant diagonal and thus, by our previous theorem, is stable. Thus, A is D-stable.§
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