I have the following problem:
Consider two points, $v_1,v_2 ∈ R^n$. Show that there exist $c ∈ R^n and d ∈ R$ (and find them!) such that $${x : ||x−v1||_2 ≤||x−v2||_2}={x : c^Tx≤d}$$ Thus, you are showing that the set of points in Rn that are closer to point $v_1$ than to point $v_2$, form a half-space.
I am just getting back into this sort of math so I am having some trouble understanding what the equation is asking. From the statement underneath I can infer that I need to find the half-way point (hyperplane) between the two points, and then show the points that are on the $v_1$ side of the hyperplane. So thinking it should be something like:
$$x ∈ R^n : a^Tx < b$$
Also if someone could explain the equation version of the question that would be awesome, specifically the 2 subscript and why x - 2. I've also been wondering why we use a transform when discussing a hyperplane instead of dot product
Think about the geometric interpretation in $3$-dimensional space. For a concrete example, suppose that the two initial points were $\vec v_1 = (-3, 1, 2)$ and $\vec v_2 = (1, -5, 4)$. Then intuitively the boundary for the points closer to $\vec v_1$ than $\vec v_2$ (i.e.: the locus of points equidistant from both points) should form a $2$-dimensional plane, perpendicular to the line connecting the two points, and halfway between them.
Finding the normal vector of the plane (in the direction of $\vec v_2$), we have: $$ \vec c = \vec v_2 - \vec v_1 = (4, -6, 2) $$ So the plane has the form $4x - 6y + 2z = d$. This plane must pass through the midpoint, namely: $$ \vec m = \frac{1}{2}(\vec v_1 + \vec v_2) = (-1, -2, 3) $$ so that $d = 4(-1) - 6(-2) + 2(3) = 14$. Thus, the half-space is: $$ S = \{(x, y, z) \in \mathbb R^3 \mid (4, -6, 2)^\top(x, y, z) \leq 14 \} $$ We can verify that $\vec v_1, \vec m \in S$ while $\vec v_2 \notin S$. Now generalize!