We have a hyperplane through the origin in N dimensional space and T vectors in this hyperplane.
I want to proof that if the T vectors are linearly independent (maximal N-1) than I can tilt the hyperplane such that all vectors are on the same side of the hyperplane (and not in the hyperplane anymore or one vector is on the other side). Here an example in two dimensions:
If the vectors are linearly dependent (N or more vectors are now in the hyperplane) this is not true in general:
If I tilt this hyperplane, the vectors are on two different sides.
Can anyone help to prove that in N Dimensions?
edit:
Let $\vec{x}$ denote any non-zero vector orthogonal to that hyperplane. Than I want to show: $$\exists \vec{x}: \vec{x} \cdot \vec{r}_k \geq 0 \quad \forall k=1...T \quad \Leftrightarrow \quad \exists \vec{x}': \vec{x}' \cdot \vec{r}_k > 0 \quad \forall k=1...T$$
And this is only true if the vectors $\vec{r}_k$ are linearly independent in the hyperplane. I want to show it for general vectors $\vec{r}_k$ with the only property that in N dimensions N or fewer vectors are linearly independent.
Hint: Let $A$ denote a matrix whose columns are the vectors $r_1,\dots,r_T$. Let $x$ denote any non-zero vector. We find that the hyperplane orthogonal to $x$ satisfies the requirement if and only if $r_k^\top x$ has the same sign for all $k = 1,\dots,T$.
In other words, it suffices to show that there exists a vector $x$ for which the vector $A^\top x$ has either all positive or all negative entries.
Further Hint: Show more specifically that there exists an $x$ such that we have $r_k^\top x = 1$ for all $k$.