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  "body": "\n## Definition\n\nLet $K$ be a field and let $m \\ge 1$.\nLet $A(n)$, for $n \\in \\N$, be $m \\times m$ [[matrices|0VD9ZSWWJG]] over $K$, and let $b(n)$, for $n \\in \\N$, be vectors in $K^m$.\nA *matrix difference equation* (of first order) is the equation\n$$x(n+1) = A(n)\\, x(n) + b(n) \\qquad (n \\in \\N)$$\nfor a [[sequence|0G9EKM0E15]] of vectors $x \\colon \\N \\to K^m$.\n\n1. It is *homogeneous* if $b(n) = 0$ for all $n$.\n2. It has *constant coefficients* if $A(n) = A$ does not depend on $n$.\n\nA *solution* is a sequence $x$ that satisfies the equation for all $n \\in \\N$.\n\n## Remarks\n\nExample: with $m = 2$, $A = \\begin{pmatrix} 1 \u0026 1 \\\\ 1 \u0026 0 \\end{pmatrix}$ and $b = 0$, the sequence $x(n) = (F_{n+1}, F_n)^T$ of [[Fibonacci numbers|0GA9K57A39]] is a solution, since $F_{n+2} = F_{n+1} + F_n$ and the second entry just shifts.\nNon-example: $x(n+1) = x(n)^2$ for $x(n) \\in K$ is not a matrix difference equation, since the right side is not of the form $A(n) x(n) + b(n)$.\nIn the homogeneous constant-coefficient case the solutions are $x(n) = A^n x(0)$.\nEquations of higher order, such as $x(n+1) = A_0 x(n) + A_1 x(n-1)$, reduce to first order by stacking $(x(n), x(n-1))$ into one vector of size $2m$.\nIn this way a scalar linear recurrence of order $d$ becomes a first-order system with a [[companion matrix|0ZCWPNZTQ3]].\n",
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