Maximal possible dimension of a subspace $U$ in $\Bbb{R}^n$ with property $\forall $ $x=(x_1,x_2...x_n)^T\in U$ \{$0$}$x_i\neq 0$ $ \forall i $

Question

Maximal possible dimension of a subspace $U$ in $\Bbb{R}^n$ with property $\forall$ $x=(x_1,x_2...x_n)^T\in U$ \{$0$}$x_i\neq 0$ $ \forall i $

I have a feeling that the maximal dimension should be n but i dont see how i could prove this. Because for every vector i add i can change one number so that they still are linear independent .

Can i get some tips ?

Answer 1

The maximal dimension is $1$. If there were two linearly independent vectors $x\in U$ and $y\in U$, and all their components were $\neq 0$, then $x_1\cdot y - y_1\cdot x$ would also have to be in $U$. This vector is not $0$ (because it is a non-trivial linear combination of two linearly independent vectors) but it has $0$ as its first component.

$\dim U =1$ works, simply take the subspace of all multiples of $(1,1,\ldots,1)^T$.