# A vector space such that all its vectors have a zero coordinate

Say I have a finite dimensional vector space $F^{d}$, and I have subspace of it, $V$, such that for each vector in it there is some coordinate which is zero, i.e. each vector is perpendicular to some basis vector.

What can be said about this space? Does it have some structure? Perhaps, all vectors are perpendicular to a common basis vector?

I tried to prove the last claim by Gauss-eliminating the basis matrix, but I didn't succeed. I did prove it for $n=2$.

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They must be orthogonal to some basis vector, if not then if one can take d vectors in the subspace such that for each basis element one of the d vectors is not orthogonal to it, then consider the sum of the vectors. – user21725 Mar 24 '12 at 22:21
That's exactly my question - does it always happen that a specific coordinate is 0? For $n=2$, you're correct and the subspace is either Span(0,1), Span(1,0) or {$\vec{0}$}. – Ofir Mar 24 '12 at 22:23
@EricGregor - consider the sum, and then what? Why can't the sum be perpendicular to some vector? What about $v_1=(1,0,0), v_2=(-1,1,0), v_3=(0,0,1)$? $v_i$ is perpendicular to $e_i$ but the sum is $(0,1,1)$, which is perpendicular to $e_1$. – Ofir Mar 24 '12 at 22:27
Hint for the case when $F$ is an infinite field: If a vector space over an infinite field is the union of finitely many subspaces, then it must equal to one of these subspaces. (You should apply this to the space $V$ here, not to $F^d$. Do you see what the subspaces are?) -- I have no idea what happens if $F$ is finite. – darij grinberg Mar 24 '12 at 22:28
Multiply the first by a scalar $((2,0,0)$. You can always do this in such a way to obtain a vector with nontrivial projection on each of the coordinate bases. – user21725 Mar 24 '12 at 22:29

## 2 Answers

In $F^d$ with $F=\mathbb F_2$ and $d$ odd, consider the set $V$ of $(x_k)_{1\leqslant k\leqslant d}$ in $F^d$ such that $x_1+x_2+\cdots+x_d=0$.

Every $(x_k)_{1\leqslant k\leqslant d}$ in $V$ is such that $x_k=0$ for an odd number of $k$, hence $x_k=0$ for at least one $k$, but, if $d\geqslant3$, no vector of the canonical basis of $F^d$ is orthogonal to $V$, that is, there exists no index $1\leqslant k\leqslant d$ such that $x_k=0$ for every $(x_k)_{1\leqslant k\leqslant d}$ in $V$.

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What is right in this case that there exists a coordinate $1\leq i\leq d$ such that all vectors in $V$ have $0$ on the $i$-th coordinate.
This is true since if for all $1\leq i\leq d$ there exist $v_i$ with non-zero $i$-th coordinate, you can find $v=\alpha_1v_1+...+\alpha_dv_d\in V$ which has all non-zero coordinates.
In this case, $V$ is orthogonal (w.r.t the natural inner-product) to $e_i$ (the vector which has $1$ on the $i$-th coordinate and $0$ on all other).

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