# 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.

• So, the set of vectors you're working with is $W=\{w\in\mathbb R^k\mid v\perp w\}$. You need to prove it is a subspace of $\mathbb R^k$. How do you prove a subset of a vector space is a subspace? – user491874 Feb 11 '18 at 20:49
• You are supposed to know (and that includes to understand) the definition of "subspace". – Professor Vector Feb 11 '18 at 20:53

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.

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:

$$\vec v \cdot (a \vec x + b \vec y)= \vec v \cdot (a \vec x) + \vec v \cdot (b \vec y)= a(\vec v \cdot \vec x) + b (\vec v \cdot \vec y)= 0+0$$

• I don't really know how to go about proving this – Plz help me Feb 11 '18 at 21:05
• Added to my answer. – Emilio Novati Feb 12 '18 at 8:24