Is there a subspace $V \subset \mathbb{R}^n$ with $V \cap \mathbb{R}_{\geq0}^n = V^\perp \cap \mathbb{R}_{\geq0}^n = 0$? My geometric intuition tells me that there is no such $V$ but i'am completely stuck on proof.

  • $\begingroup$ What does $\mathbb R^n_{\geq 0}$ mean? Does it mean all vectors which consist of only non-negative entries, or something like that? $\endgroup$ Jan 18, 2018 at 22:47
  • $\begingroup$ @астонвіллаолофмэллбэрг $\mathbb{R}_{\geq 0}^n = \{(x_1, \dots, x_n) | x_i \in \mathbb{R} , x_i \geq 0\}$ $\endgroup$
    – qwenty
    Jan 18, 2018 at 22:50
  • $\begingroup$ Do you mean that both $V \cap \mathbb R^n$ and $V^\perp \cap \mathbb R^n$ should consist of only the zero vector? $\endgroup$ Jan 18, 2018 at 23:01
  • $\begingroup$ @астонвіллаолофмэллбэрг yes $\endgroup$
    – qwenty
    Jan 18, 2018 at 23:04
  • 1
    $\begingroup$ I see some easy parts: e.g. if $V$ or $V^\perp$ has dimension $1$. Now, if we could somehow extend $V$ to a hyperplane $\hat V$ with the same property $\hat V\cap\Bbb R^n_{\ge 0}=0$, then we're done. $\endgroup$
    – Berci
    Jan 19, 2018 at 0:23

3 Answers 3


Suppose $V\cap \mathbb{R}^n_{\geq 0} = \{0\}$, and let $P$ denote orthogonal projection onto $V$. A vector $w$ is in $V^\perp$ iff $P(w)=0$.

Let $H$ be the convex hull of $\{e_1,e_2,\dots,e_n\}$, where $e_i$ is a standard basis vector. Note that $H\subset \mathbb{R}^n_{\geq 0}$. The projection $P(H)$ is a convex subset of $V$. For it to not contain $0$, there must be a hyperplane separating $0$ and $P(H)$, which means there exists some non-zero $v\in V$ such that $\forall(w\in P(H))\: w\cdot v > 0$.

But the condition that $V\cap \mathbb{R}^n_{\geq 0} = 0 $ means that for any non-zero $v\in V$ the components of $v$ cannot all be non-negative (nor all non-positive, else $-v\in\mathbb{R}^n_{\geq 0}$). Hence for any $v\in V$ there is some $e_i$ such that $v\cdot e_i < 0$. Thus $v\cdot P(e_i) = v\cdot e_i < 0$, and since $P(e_i)\in P(H)$, so there cannot be a hyperplane separating $0$ and $P(H)$.

Hence $0\in P(H)$, and thus $H\subset \mathbb{R}^n_{\geq}$ intersects $V^\perp$.

  • $\begingroup$ Can you explain more detailed the existence of $v $? Thanks $\endgroup$ Jan 23, 2018 at 16:19
  • $\begingroup$ @FabioLucchini A hyperplane in $V$ is defined by $\{w\in V \: | \: v\cdot w = c\}$ for some vector $v\in V$ and constant $c$. $v$ is a vector normal to the hyperplane. For the hyperplane to separate $P(H)$ and $0$, we must have either $v\cdot 0 <c$ and $v\cdot w > c$ for all $w\in P(H)$, or the other way around. In the latter case, replace $v$ by $-v$ and $c$ by $-c$. So then $w\cdot v > c > 0$ for all $w\in P(H)$, as desired. $\endgroup$
    – Yly
    Jan 24, 2018 at 3:59

$\newcommand{\IR}{\mathbb{R}}$ Note that $C:=\IR_{\geq 0}^n$ is a cone, i.e. a subset that is closed under addition and multiplication with $\IR_{\geq 0}$. Following Paul Garrett's advice, we look at the dual cone, i.e. $C^\vee := \{v \mid \forall x\in C: \langle v,x\rangle \geq 0\}$. One can easily verify that $V^\vee = V^\perp$ for subspaces $V\leq\IR^n$ and $(\IR_{\geq 0}^n)^\vee = \IR_{\geq 0}^n$.

Now look at any basis $w_1,\ldots,w_k$ of $V^\perp$ and look at the linear map $\omega: \IR^n\to\IR^k, x\mapsto (\langle w_1,x\rangle, ..., \langle w_k,x\rangle)$. By assumption $\ker(\omega)=V$ so that $\omega(\IR_{\geq0}^n\setminus\{0\}) \subseteq \IR^k\setminus\{0\}$. This is now a cone in $\IR^k$ and by choosing a different basis of $\IR^k$ (or equivalently: choosing a different basis of $V^\perp$), we can assume that $\omega(\IR_{\geq0}^n) \subseteq \IR_{\geq0}^k$. But that means that $\langle w_i,x\rangle \geq 0$ for all $x\in C$, i.e. $w_i\in (\IR_{\geq 0}^n)^\vee = \IR_{\geq 0}^n$.

We have shown: If $V\cap\IR_{\geq 0}^n=0$, then $V^\perp\cap\IR_{\geq 0}^n$ contains a whole basis of $V^\perp$.

  • 1
    $\begingroup$ Why we can choose such ${w_i}$ that $x \in \mathbb{R}_{\geq 0}^n \implies \langle w_i, x\rangle \geq 0$? $\endgroup$
    – qwenty
    Jan 22, 2018 at 10:53

If $V$ has dimension strictly less than $n-1$ and it intersect trivially $\mathbb{R}^n_{n\ge0}$, it is easy to see that we can enlarge it keeping trivial the intersection with $\mathbb{R}^n_{n\ge0}$, hence it is enough to do the case in which $V$ has dimension $n-1$ and its orthogonal is a line $L$.

Now, if $L$ is not contained in $\mathbb{R}^n_{n\ge0}\cup\mathbb{R}^n_{n\le0}$, its points have two coordinates with opposite sign. But then it is easy to find a vector $v\in\mathbb{R}^n_{n\ge 0}$ orthogonal to $L$.


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