1
$\begingroup$

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 $n=2$.

  • 0
    darij - I see the subspaces, and now the theorem is more intuitive, but I haven't solved it yet (thinking about it right now). I think I did encounter your theorem in some abstract algebra course, I wonder where exactly. Also, I found your counterexample. Generalization: all the vectors in $\mathbb{F}_2^{2n+1}$ perpendicular to $(1,1,\cdots,1)$.2012-03-24

2 Answers 2

1

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 $V$.

0

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).