Stable Matrices

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Consider the system of ordinary differential equations x｢ (t) = Ax(t) + b. Let x(t) = F (t)c + x_p where x_p 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:

Stable Matrix: The matrix A is stable if all the solutions x｢ (t) = Ax(t) converge toward zero as t converges toward infinity; that is, F (t)c ｮ 0 when t ｮ ･ .

We now prove a quite standard theorem:

Theorem: (Negative Eigenvalues) The matrix A is stable if and only if all its eigenvalues have negative real parts.

Proof: Suppose A is stable and let l _i = u_i + iw_i be an eigenvalue (complex or real) of A. Let v_i be the associated eigenvector, thus Av_i = l _iv_i. Now, we know that F(t) = [v₁e^l1t, v₂e^l2t, .., v_ne^lnt] thus, defining j _i(t) = v_ie^lit, then:

F (t) = [j ₁(t), j ₂(t), .., j _n(t)]

Now, as j _i(t) = v_ie^lit, then taking the first derivative j_i｢ (t) = l _iv_ie^lit = Av_ie^lit = Aj _i(t) by definition. Thus, j_i(t) is a solution to the homogeneous system, x｢ = Ax. Now, by definition, A is stable iff F(t)c ｮ 0 when t ｮ ･ , thus it must be that j _i(t) ｮ 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. u_i < 0 for all i. Consider the kth coordinate of v_i and thus of j _i(t), i.e. j _ik(t) = v_ike^lit. Since lim_{tｮ ･} j _i(t) = 0, then it must be that lim_{tｮ ･} j _ik(t) = 0, or lim_{tｮ ･} |j _ik(t)| = 0 or lim_{tｮ ･} |v_ike^lit| = 0. Dropping the subscript i for cleaner notation, so that we now are consider l = u + iw as a representative eigenvalue, then lim_{tｮ ･} |v_ke^lit| = lim_{tｮ ･} |v_k||e^lit| = lim_{tｮ ･} |v_k||e^{(u + iw)t}| = lim_{tｮ ･} |v_k||e^ut||e ^iwt| = 0. As |e^iwt| has a finite upper bound, i.e. as |e^iwt| ｣ M for some M < ･ and e^ut > 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) = u_i < 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 = u_i + iw_i so Re(l _i) = u_i < 0 for all i = 1, .., n. Let x(t) be a solution to x｢ = Ax and let x_k(t) be the kth coordinate of x(t). Then, x_k(t) = ・/font> _j=1ⁿ a _kj e^ljt where a _kj are coefficients and l _j = u_j + iw_j. Thus, |x_k(t)| = |・/font> _j=1ⁿ a _kje^ljt| = |・/font> _j=1ⁿ a _kje^{(uj + iwj) t}| =|・/font> _j=1ⁿ a _kje^ujte^iwjt|. This implies, by triangular inequality that |x_k(t)| ｣ ・/font> _j=1ⁿ|a _kje^ujte^iwjt| = ・/font> _j=1ⁿ|a _kj||e^ujt||e^iwjt|. Now, again, by upper boundedness |e^iwt| ｣ M and e^ut > 0, then |x_k(t)| ｣ ・/font> _j=1ⁿ|a _kj|e^ujtM. Thus, lim_{tｮ ･} |x_k(t)| ｣ lim_{tｮ ･} ・/font> _j=1ⁿ|a _kj|e^ujtM. Since u_j < 0 for all j = 1, 2, .., n, then lim_{tｮ ･} ・/font> _j=1ⁿ|a _kj|e^ujtM = 0, thus it follows that lim_{tｮ ･} |x_k(t)| ｣ 0, i.e. lim_{tｮ ･} |x_k(t)| = lim_{tｮ ･} x_k(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.ｧ

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:

Theorem: (Symmetric Case) The symmetric matrix A is stable iff (-1)^k|A_k| > 0 for all k = 1, 2, .., n, where |A_k| is the kth order principal leading minor of A (or the determinant of a submatrix of A obtained by deleting the last n - k rows and columns).

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|A_k| > 0.ｧ

Theorem: (Necessity) If a n ｴ n matrix A with real elements is stable, then its trace is negative and its determinant has the same sign as (-1)ⁿ, that is A stable implies trA = ・/font> _i=1ⁿ a_ii < 0 and (-1)ⁿ|A| > 0.

Proof: This is elementary. Recall that trA = ・/font> _i=1ⁿ l _i and |A| = ﾕ _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.ｧ

The above theorem is actually sufficient if A is a 2 ｴ 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.

Theorem: (Lyapunov) A real n ｴ n matrix A is a stable matrix if and only if there exists a positive definite matrix H such that A｢ H + HA is negative definite.

Proof: Given elsewhere via Lyapunov's method.ｧ

Theorem: (Quasi-Negative Case) A real n ｴ n matrix A is stable if A is quasi-negative definite.

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.ｧ

Theorem: (Symmetric Negative Case) A real n ｴ n matrix A is stable if A is symmetric and negative definite.

Proof: If A is symmetric and negative definite, then A + A｢ is negative definite, and Lyapunov's theorem applies.ｧ

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:

D-Stability: A matrix A is "D-Stable" if BA is stable for any positive diagonal matrix B.

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:

Theorem: (Arrow-McManus) matrix A is D-stable if there exists a positive diagonal matrix C such that A｢ C + CA is negative definite.

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.ｧ

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).

Diagonal Dominance: a n ｴ n matrix A with real elements is dominant diagonal (dd) if there are n real numbers d_j > 0, j = 1, 2, .., n such that

d_j|a_jj| > ・/font> _{iｹ j} d_i|a_ij|

for j = 1, 2, .., n.

This is also known as "column" diagonal dominance. "Row" diagonal dominance is also available and defined as the existence of d_i > 0 such that d_i|a_ii| > ・/font> _{jｹ i} d_j|a_ij| 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 ｹ j and |a_ii| is the ii entry, then y > 0 such that yA > 0 implies x > 0 such that Ax > 0.

Hadamard: If n ｴ n matrix A is dominant diagonal and d_j = 1 for all j = 1,2, .., n, or:

|a_jj| > ・/font> _{iｹ j} |a_ij|

for j = 1, 2, .., n, then A is a "Hadamard" matrix.

Alternatively, if we consider a matrix D with diagonal elements d_j as defined in the dd case, and A is dd, then obviously DA is a Hadamard matrix.

Theorem: (McKenzie) If A is dominant diagonal, then |A| ｹ 0.

Proof: (McKenzie, 1960) Suppose not. Suppose |A| = 0. Then there is an x ｹ 0 such that DAx = 0, thus d_ja_jjx_j + ・/font> _{iｹ j} d_ia_ijx_i = 0 for all j = 1, 2, .., n. In absolute value terms, then, using the triangular inequality, this implies for all j:

|d_ja_jj| |x_j| = |・/font> _{iｹ j} d_ia_ijx_i| ｣ ・/font> _{iｹ j} |d_ia_ij ||x_i|

Let J be an index set such that if j ﾎ J, then |x_j| ｳ |x_i| for all i = 1, .., n. Then, for j ﾎ J:

|d_ja_jj| |x_j| ｣ ・/font> _{iｹ j} |d_ia_ij ||x_i| ｣ ・/font> _{iｹ j} |d_ia_ij ||x_j|

which implies:

|d_ja_jj| ｣ ・/font> _{iｹ j} |d_ia_ij |

or, as all d_j > 0, then:

d_j|a_jj| ｣ ・/font> _{iｹ j} d_i |a_ij |

for all j ﾎ J, which contradicts diagonal dominance.ｧ

The following theorem, due to McKenzie (1960), establishes that if A has a negative dominant diagonal, then A is stable:

Theorem: (Sufficiency) If an n ｴ n matrix A is dominant diagonal and the diagonal is composed of negative elements (a_ii < 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 ｳ 0. As a_ii < 0, then:

|l - a_ii| = ﾖ ((u_i - a_ii)² + v_i²) ｳ u_i - a_ii ｳ |a_ii|

Since A has a dominant diagonal, there exist d_j > 0 (j = 1, .., n) such that d_j|a_jj| ｳ ・/font> _{iｹ j} d_i |a_ij |, thus:

d_j|l - a_jj| ｳ d_j|a_ii| ｳ ・/font> _{iｹ j} d_i |a_ij |

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| ｹ 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 (u_i < 0 for all i) , thus A is stable.ｧ

Finally, note the following corollary:

Corollary: If A has negative diagonal dominance, then it is D-stable.

Proof: We have to prove that BA is stable, where B is a positive diagonal matrix (thus, b_ii > 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 d_j > 0 (j = 1, .., n) such that d_j|a_jj| ｳ ・/font> _{iｹ j} d_i |a_ij |. Consequently, dividing and multiplying by the relevant b_jj > 0, we obtain:

(d_j/b_jj)|b_jja_jj| ｳ ・/font> _{iｹ j} (d_i /b_ii)|b_iia_ij | for all j = 1, 2, .., n.

Thus, d_j/b_jj > 0 will satisfy diagonal dominance for matrix BA. Finally, as a_jj < 0 and b_jj > 0 for all j, then b_jja_jj < 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.ｧ

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