# Nonzero Projection onto Orthogonal Complement

Pardon my ignorance, but I don't know much about functional analysis and advanced linear algebra. Let $W$ be a closed convex cone in some Hilbert space (it can be finite, if we need to stipulate that for the following to be true), and let $v$ be some vector in the Hilbert space. Then

If the projection of $v$ onto $W^\perp$ (orthogonal complement) is not zero, then there exists a $w \in W$ such that $\langle v,w \rangle = 0$.

Is something like this true?

Is it necessary that $W$ be a closed convex cone, or can we weaken the conditions on $W$?

EDIT:

Let me try to reword my ideas. Suppose that I have some vector $v$, and I do not know whether it is in $W^\perp$ (in fact, it may not even be in $W^\perp$). What do I need to know in order to conclude there exists a $w \in W$ such that $\langle v,w \rangle = 0$. I was thinking that a nonzero projection of $v$ onto $W^\perp$ would suffice, but perhaps not. SO, what would suffice, given that $W$ is a closed convex cone?

• Sorry. I should have written $w \in W$, rather than $w \in W^\perp$. I fixed it. Yes, $\langle p(v), w \rangle = 0$ will be true, because $p(v) \in W^\perp$. I want to know something different. I want to know whether there exists a $w \in W$ such that $\langle v,w \rangle = 0$, if you assume $proj_{W^\perp} v \neq 0$. – user193319 Sep 18 '16 at 13:42
• Check what happens when $W$ is something like $\{\alpha \vec{w} : \alpha \in [1, \infty)\}$. In this case, we can't guarantee something like $\langle v, \alpha w \rangle = 0$ even if $v$ has a component in $W^\perp$. So at the very least we'll need that the span of $W$ is at least 2-dimensional. This might even be sufficient. – Josh Keneda Sep 18 '16 at 17:24
• You can also rephrase your question this way: As you've noted, the $p(v)$ part of $v$ is automatically orthogonal to everything in $W$. Since we can decompose $v$ as $p(v) + (1-p)(v)$, where $(1-p)v$ is the projection of $v$ onto the closed span of $W$, your question could be rephrased as "Given an element $v$ of the closed span of $W$, is there always a $w \in W$ with $\langle v, w \rangle = 0$?" – Josh Keneda Sep 18 '16 at 17:32