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If we have a linear subspace of projective $P^n_k$ where $k=\bar k$ that contains all the points $[0:\ldots:0:x:0:\ldots:0]$ with the nonzero $x$ in the $i$th slot for all $i$, how can we see that this is actually the whole space.

E.g. in $P^2$, suppose $L$ is a linear subspace which contains the points $[x:0:0]$ and $[0:x:0]$ and $[0:0:x]$. How do we see that $L$ is in fact equal to $P^2$.


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Replacing everywhere $x$ by $1$ doesn't change the problem but makes the question clearer by eliminating an irrelevant letter. Also, $k=\bar k$ is irrelevant. – Georges Elencwajg Mar 29 '13 at 13:01
up vote 2 down vote accepted

No hyperplane $l(z)=\sum c_jz^j=0$ contains all your points $P_i$: substitution yields $l(P_i)=c_i=0$ .

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how do we know there cannot be more than one hyperplane? is the span of two distinct hyperplanes the whole space necessarily? – Steven-Owen Mar 29 '13 at 13:03
Is it easy to show that the "biggest" linear proper linear subspace is a hyperplane? And thus, by what you've shown since it cannot be contained in that, it must be the whole thing? – Steven-Owen Mar 29 '13 at 13:05
Dear Ricky, if some linear subspace given by a bunch of linear forms contained your points, you would obtain a hyperplane through these points by throwing away all the equations in the bunch save one. You must remember that a strict linear subspace of projective space is given by linear equations. – Georges Elencwajg Mar 29 '13 at 14:15

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