How is the dual cone of a subspace its orthogonal complement?

Question

From Boyd and Vandenberghe's Convex Optimization:

A dual cone of a subspace $V \subseteq \Bbb R^n$ is it's orthogonal complement.

$V^{*} = \{y : v^Ty = 0, \forall v \in V\}$

but the dual cone is defined by: $V^{*} = \{y : v^Ty \ge 0, \forall v \in V\}$.

Why are there no vectors $v$ such that $v^Ty > 0$?

Answer 1

Let $y\in V^*$ and assume that there exists $v\in V$ such that $y^Tv >0.$ Since $V$ is a subspace, it follows that $-v \in V.$ But then

$$0\leq y^T(-v)= -y^Tv<0,$$ a contradiction.

Answer 2

If $y \in V^*$ and $v^Ty > 0$ then $(-v)^Ty < 0$ contradicting $y \in V^*$.