Can Three Non-Collinear Points Always Define a Projective Plane?

[Euler2718]

Homework Statement

Let [itex]P(W)[/itex] be a projective space whose dimension is greater than or equal to 2 and let three non-colinear projective points, [itex][v_{1}],[v_{2}],[v_{3}]\in P(W) [/itex]. Prove that there is a projective plane in [itex]P(W)[/itex] containing all three points.

Homework Equations

The Attempt at a Solution

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For [itex]n=2[/itex] my reasoning was: WLOG assume [itex][v_{1}],[v_{2}],[v_{3}][/itex] are distinct. Then, define [itex]s = \{ v_{1},v_{2},v_{3} \}[/itex] which spans a vector space, S. The dimension here is 3, and since [itex]v_{1},v_{2},v_{3}\in S[/itex] it follows that [itex][v_{1}],[v_{2}],[v_{3}]\in P(S) [/itex], has dimension two and is thus a plane containing the points.

If this idea is correct, then I am at the point now where I am trying to prove it for all dimensions greater than 2. I was thinking that is [itex]P(W)[/itex] had dimension n (>2) , [itex]s = \{ v_{1},v_{2},v_{3} \}[/itex] still spans a three dimensional vector space which would be a subspace of W. So the projective space of S would be a linear subspace of the projective space of W?

 

[member 587159]

First, there is no need to ask that the points are distinct. They are non colinear, so projectively independent. It also follows that ##v_1, v_2, v_3## must be linearly independent in ##W##, so they span a three-dimensional vectorspace. Now, it is sufficient to define ##V = span\{v_1,v_2,v_3\}## and then say that ##P(V)## is the subspace of ##P(W)## you are looking for. Note that ##P(V)## is a plane, as it has dimension 2.