Talk:Matrix difference equation
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Higher-order can be written as first-order[edit]
I'm afraid I don't know how to write it in Wikipedia's mark-up language, so if someone else could, that'd be great:
It should be noted that x_t = A*x_t-1 + B*x_t-2 can be written as
X_t = [ x_t ; x_t-1 ] (i.e. x_t stacked on top of x_t-1)
X_t = C*X_t-1
where C = [ A , B ; I , 0 ] (i.e. the top "row" is A:B and the bottom "row" is I:0)
Generalisation to any higher order should be obvious.