Prove that, for any vector $v$ in space $\mathbb{R}^k$ , the set $W$ of all vectors that are orthogonal to $V$ is a subspace. I'm not really sure what I am being asked in this question, I know that if the scalar product of two vectors is 0, then they are orthogonal complements, but I don't really know what I's supposed to do.
 A: Hint:
You have to prove that , given a vector $\vec v$, for any $\vec x$ and $\vec y$ such that:
$\vec v \cdot \vec x=0$  and $\vec v \cdot \vec y=0$ ( this means that the two vectors are orthogonal to $\vec v$)
and any scalars $a,b$  we have
$$
\vec v \cdot (a \vec x + b \vec y)=0
$$
(this means that any linear combination of the two vectors is orthogonal to $\vec v$)
Can you do this? 

It is a simple consequence of the linearity of the dot product:
$$
\vec v \cdot (a \vec x + b \vec y)= 
\vec v \cdot (a \vec x) + \vec v \cdot (b \vec y)=
a(\vec v \cdot  \vec x) + b (\vec v \cdot  \vec y)= 0+0
$$
A: Let $f(x) = v^T x$. Then $ w\bot v$ iff $v^T w = 0$ iff $f(w) = 0$ iff $\ker f = 0$.
Since the nullspace of a linear operator is a linear space we are finished.
More explicitly, suppose $W = \{ w | f(w) = 0 \}$ then we need to show
that if $w_1,w_2 \in W$ then $w_1+w_2 \in W$ and if $w \in W$ and $\lambda$ is
a scalar, then $\lambda w \in W$.
Since $f(w_1+w_2) = f(w_1)+f(w_2)$ and $f(\lambda w) = \lambda f(w)$ we can quickly check these conditions.
