# Is $V^+:=\{v\in V\space|\space\langle v,v\rangle>0\}\cup\{0\}$ a vector space?

Let $V$ be a finite dimensional vector space over $\mathbb{R}$ equipped with an inner product $\langle\cdot,\cdot\rangle$ and define: $$V^+:=\{v\in V\space|\space\langle v,v\rangle>0\}\cup\{0\}\\ V^-:=\{v\in V\space|\space\langle v,v\rangle<0\}\cup\{0\}$$ Are $V^+$ and $V^-$ vector spaces?

I'm quite clueless about this question. It seems to me that $V^+$ is a vector space if and only if the Cauchy-Schwarz inequality holds for every two vectors in $V^+$. But I cannot think of a proof of this inequality which doesn't use already the fact that $V^+$ is a vector space since in the standard proof, we consider the polynomial $p(\alpha)=\langle v+\alpha u,v+\alpha u\rangle$ which then has a negative discriminant. But for this, we already need $v+\alpha u\in V^+$. So how to prove or disprove it?

EDIT: $\langle\cdot,\cdot\rangle$ isn't necessarily positive definite i.e. $V^-\backslash \{0\}$ can be non empty.

EDIT: $V$ is finite dimensional.

Neither is a vector subspace, unless they are $V$ or $\{0\}$.

$\newcommand{\scal}[2]{\left\langle{#1},\,{#2}\right\rangle}$ Suppose $V^+$ is a proper non-zero subspace. Then, $\dim V>1$ and there is a finite-dimensional subspace $W$ such that $W\supsetneqq V^+\cap W\supsetneqq\{0\}$. (You can even assume $\dim W=2$)

So we can consider just the case $1\le\dim V^+<\dim V<\aleph_0$, by restricting $\scal\cdot\cdot$ to $W\times W$.

Let's select a basis of $V$ and pull back on $V$ the euclidean topology of $\Bbb R^n$.

So, we identify $V=\Bbb R^n$ and, in particular, $q\,:\,v\mapsto\scal vv$ is a continuous map $\Bbb R^n\to \Bbb R$

But $V^+=\{0\}\cup q^{-1}(0,+\infty)$. Hence $V^+\setminus\{0\}$ is an open euclidean set. But no proper non-zero subspace of $\Bbb R^n,\ n>1$ has this property.

If by "scalar product" you mean inner product, recall that inner products are usually required to be positive-definite:

"positive": $\langle v, v \rangle \geq 0$ for all $v \in V$,

definite: $\langle v, v \rangle = 0 \iff v = 0$.

Hence $V^+$ is the whole vector space $V$ and $V^-$ is the trivial vector space $\{0\}$.

• @G.Sassatelli, you're probably right. I thought I'd opt for the easy way out first, though – Jon Warneke Mar 16 '16 at 17:51
• I edited the question; there are no further assumptions on the inner product. Thanks anyway :) – Redundant Aunt Mar 16 '16 at 18:04
• If that's the case, then you should say that $\langle \cdot, \cdot \rangle$ is a (symmetric?) bilinear form, not an inner product. – Jon Warneke Mar 16 '16 at 18:09