Let $T_{nm}$ be the set of all possible binary rectangular matrices of dimension $n\times m$. The cardinality of $T_{nm}$ would be $2^{nm}$.

Let f be a map from $T_{nm}$ to itself. Consider a dynamical system $C_{t+1}=f(C_t)$ where $C_t$ $\in$ $T_{nm}$.

Let us think that $C_{t+1}=A*(C_t)$ where $A$ is a square matrix of dimension $m\times m$. Therefore the map $f$ precisely is linear. There are essentially $2^{n^2}$ such $A$.

Question: Do we have any non-linear maps and if there is, then how can we construct those map?

  • $\begingroup$ Just to clarify: When you say "binary", do you mean "with entries taken modulo 2"? (If $1+1=2$ rather than $1+1=0$, then those matrices wouldn't form a vector space, so "linear map" wouldn't be a well-defined concept.) $\endgroup$ Feb 10, 2014 at 6:55
  • $\begingroup$ Yes, the entries of the matrices are either 0 or 1. Yes modulo 2 operation is to be done. $\endgroup$
    – Fukuzita
    Feb 10, 2014 at 7:06
  • $\begingroup$ How about a constant map, say $f(C)$ equal to the matrix with $1$ in every position, regardless of what $C$ is? That would be nonlinear (but perhaps not very interesting). $\endgroup$ Feb 10, 2014 at 8:51
  • $\begingroup$ Yes, but it is not interesting in the context of dynamical system study. $\endgroup$
    – Fukuzita
    Feb 10, 2014 at 9:06


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