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

[Euler2718]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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For $n=2$ my reasoning was: WLOG assume $[v_{1}],[v_{2}],[v_{3}]$ are distinct. Then, define $s = \{ v_{1},v_{2},v_{3} \}$ which spans a vector space, S. The dimension here is 3, and since $v_{1},v_{2},v_{3}\in S$ it follows that $[v_{1}],[v_{2}],[v_{3}]\in P(S)$, 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 $P(W)$ had dimension n (>2) , $s = \{ v_{1},v_{2},v_{3} \}$ 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.