Geometric independence
From Maths
(Redirected from Geometrically independent)
Stub grade: C
This page is a stub
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Suggest why this definition is useful
Definition
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
Notes
- ↑ 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
References
TODO: Is this in the right category?
