Geometric independence

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Let us have a set of points, [ilmath]\{a_0,a_1,\ldots,a_N\}\subseteq\mathbb{R}^n [/ilmath][Note 1], we say they are geometrically independent if[1]:

  • [math]\forall \{t_0,t_1,\ldots,t_N\}\subset\mathbb{R}\left[\left(\sum^N_{i=1}t_i=0\wedge\sum^N_{i=1}t_ia_i=0\right)\implies\left(t_0=t_1=\ldots=t_N=0\right)\right][/math]

The reader should note that this is very similar to linear independence; in fact, [ilmath]\{a_0,\ldots,a_N\} [/ilmath] is geometrically independent iff the set [ilmath]\{a_1-a_0,\ldots,a_N-a_0\} [/ilmath] is linearly independent


  1. Consider [ilmath]n=0[/ilmath], then set equality is possible, hence [ilmath]\subseteq[/ilmath] rather than [ilmath]\subset[/ilmath] - see Importance of being pedantic about strict-subset and subset relations


  1. Elements of Algebraic Topology - James R. Munkres

TODO: Is this in the right category?