Stability of a System of Linear Difference Equations¶

Proposition¶

Let $A$ be a real $n\times n$ matrix and consider the system $\boldsymbol{x}_{t+1}=A\boldsymbol{x}_t+\boldsymbol{b}$ with constant $\boldsymbol{b}\in\mathbb{R}^n$ $R^{n}$ . The following are equivalent:

every eigenvalue $\lambda$ of $A$ , real or complex, satisfies $\lvert\lambda\rvert<1$ $1$ ;
$A^t\to O$ as $t\to\infty$ $\infty$ ;
for every $\boldsymbol{x}_0$ the solution of the homogeneous system satisfies $\boldsymbol{x}_t=A^t\boldsymbol{x}_0\to\boldsymbol{0}$ $0$ .

In that case $1$ is not an eigenvalue, the equilibrium $\boldsymbol{x}^*=(I-A)^{-1}\boldsymbol{b}$ is unique, and every solution converges to it: the equilibrium is (globally) asymptotically stable.

Interpretation¶

The deviation from equilibrium evolves as $\boldsymbol{x}_t-\boldsymbol{x}^*=A^t(\boldsymbol{x}_0-\boldsymbol{x}^*)$ $x^{*})$ . For a diagonalizable matrix it is a sum of terms $c_i\lambda_i^t\boldsymbol{p}_i$ , and each dies out exactly when the modulus $\lvert\lambda_i\rvert$ is below $1$ . In general the Jordan normal form adds factors $t^j$ , which do not change the conclusion. This is the matrix version of the scalar criterion $\lvert a\rvert<1$ $1$ .

The unit circle in the complex plane is drawn with its interior shaded; the eigenvalues 0.5, −0.5 and 0.7 lie inside, 1.5 lies outside, and 1 lies exactly on the circle.

The figure plots the eigenvalues of the three examples below. The first matrix has the single eigenvalue $0.5$ (green), inside the unit circle, so the system is asymptotically stable. The second has $-0.5$ inside but $1.5$ outside (red), and one eigenvalue outside is enough for instability. The third has $0.7$ inside and $1$ on the circle (orange): solutions stay bounded and converge, but not to a single equilibrium.

Change the entries of $A$ and watch its eigenvalues: while every modulus is below $1$ the iterates converge to $\boldsymbol{x}^*=(I-A)^{-1}\boldsymbol{b}$ from any $\boldsymbol{x}_0$ ; push one eigenvalue outside the circle and they diverge.

Complex eigenvalues must be included: a conjugate pair $\rho \mathrm{e}^{\pm\mathrm{i}\theta}$ produces rotations whose size is multiplied by $\rho$ each period. A common mistake is to inspect the entries of $A$ instead of its eigenvalues; small diagonal entries do not imply stability unless $A$ is triangular.

In Math V, asymptotic stability means that all solutions converge, which needs all eigenvalues strictly inside the unit circle. The weaker notion, stable, means that all solutions stay bounded; it holds if and only if all $\lvert\lambda\rvert\le1$ and every eigenvalue with $\lvert\lambda\rvert=1$ has equal algebraic and geometric multiplicity.

Key Results¶

If some eigenvalue has $\lvert\lambda\rvert>1$ $1$ , solutions starting on its eigenvector (or its real invariant plane, for a complex pair) grow without bound; solutions starting in the span of the generalized eigenvectors with $\lvert\lambda\rvert<1$ still converge.
Eigenvalues with $\lvert\lambda\rvert=1$ exclude asymptotic stability: $\lambda=1$ gives constant modes, $\lambda=-1$ a two-cycle, a nonreal pair a rotation of constant size.
For $n=2$ $2$ : both eigenvalues lie inside the unit circle if and only if $\lvert\det A\rvert<1$ and $\lvert\operatorname{tr}A\rvert<1+\det A$ (the determinant and trace are the product and sum of the eigenvalues).
Necessary condition for any $n$ : $\lvert\det A\rvert<1$ $1$ , since $\lvert\det A\rvert=\prod_i\lvert\lambda_i\rvert$ $\prod_{i} ∣ λ_{i} ∣$ .
A scalar equation of order $k$ is (asymptotically) stable if and only if its companion matrix is, since the eigenvalues are the characteristic roots (higher-order equation).

Examples¶

Asymptotically stable, not diagonalizable. $A=\begin{pmatrix}0.5&0\\0.25&0.5\end{pmatrix}$ $(0.5 0.25 0 0.5)$ , $\boldsymbol{b}=(1,1)^{\top}$ $(1, 1)^{⊤}$ . The only eigenvalue is $0.5$ , so the system is asymptotically stable. Solving $(I-A)\boldsymbol{x}^*=\boldsymbol{b}$ gives $\boldsymbol{x}^*=(2,3)^{\top}$ $(2, 3)^{⊤}$ ; check: $A\boldsymbol{x}^*+\boldsymbol{b}=(1+1,\,0.5+1.5+1)^{\top}=(2,3)^{\top}$ $(2, 3)^{⊤}$ . Here $A^t=0.5^t\begin{pmatrix}1&0\\0.5t&1\end{pmatrix}$ $0. 5^{t} (1 0.5 t 01)$ , and $0.5^{t}\cdot0.5t\to0$ despite the factor $t$ .

Small diagonal, unstable. $A=\begin{pmatrix}0.5&1\\1&0.5\end{pmatrix}$ has $\operatorname{tr}A=1$ $1$ , $\det A=-0.75$ and eigenvalues $1.5$ and $-0.5$ . Unstable; the $2\times2$ test agrees, since $\lvert\operatorname{tr}A\rvert=1\not<0.25=1+\det A$ $det A$ .

Boundary case. For $\boldsymbol{x}_{t+1}=A\boldsymbol{x}_t$ with the column-stochastic matrix $\begin{pmatrix}0.9&0.2\\0.1&0.8\end{pmatrix}$ of a two-state Markov chain the eigenvalues are $1$ and $0.7$ . Solutions converge, but to a limit that depends on $\boldsymbol{x}_0$ (a multiple of $(2,1)^{\top}$ ), so the zero solution is not asymptotically stable. It is stable: both eigenvalues are simple and $\lvert\lambda\rvert\le1$ $1$ , so all solutions stay bounded.

Prerequisites¶

Sources¶

Math V: Advanced Linear Algebra · Week 2

Aliases: stability of a linear discrete-time system, asymptotic stability of x_{t+1} = A x_t Keywords: all eigenvalues of modulus less than 1, spectral radius below 1, A^t -> 0, |det A| < 1 and |tr A| < 1 + det A, unit circle Symbols: element of, infinity, lambda, product, real numbers, rho, theta

Last updated: 16 October 2026 22:28