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Let $V$ be a real vector space, and let be given $v \in V$ and $v_1, v_2, \dots, v_n \in V$($n \geq 1$). Is it true that the statements below are equivalent?

  1. $v$ writes as a positive linear combination of $v_1, v_2, \dots, v_n$, that is, there are $\alpha_1, \alpha_2, \dots, \alpha_n \geq 0$ such that

$$v = \sum_{1 \leq k \leq n}\alpha_kv_k$$

  1. Given any linear functional $L: V \rightarrow \mathbb{R}$, if $L(v_k) \geq 0$ for all $1 \leq k \leq n$, then $L(v) \geq 0$.

That 1) $\implies$ 2) is clear. But is the converse 2) $\implies$ 1) true? If not, is it true if we require $V$ to be finite dimensional?

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  • $\begingroup$ It certainly seems to hold in any inner-product space, whether finite-dimensional or not. $\endgroup$ – Ted Shifrin Jul 22 at 0:42
  • $\begingroup$ Finite-dimensional shouldn't matter, I would think, since one could just replace $V$ with the span of $\{v,v_1,\dots,v_n\}$. $\endgroup$ – Greg Martin Jul 22 at 0:42
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    $\begingroup$ Consider particular linear functionals. Think of simple ones. $\endgroup$ – Paulo Jul 22 at 0:45
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    $\begingroup$ Yes. Its a matter of writing $v$ conveniently. $\endgroup$ – Paulo Jul 22 at 1:02
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    $\begingroup$ @Paulo, yes, as you said, it is only a matter of writting $v$ conveniently; I have just figured out how to prove this theorem $\endgroup$ – Lucas Jul 22 at 1:29
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Yes, the converse is true. First, note that we may as well assume $V$ is finite-dimensional: just let $V_0$ be the subspace of $V$ spanned by $v_1,\dots,v_n$ and $v$ and then it suffices to prove the converse for $V_0$ in place of $V$ since every functional on $V_0$ can be extended to a functional on $V$.

So, we may assume that $V=\mathbb{R}^m$ for some $m$, with standard basis $e_1,\dots,e_m$. We can now prove that (2) implies (1) by induction on $m$. The base cases $m=0$ and $m=1$ are trivial.

Now suppose $m>1$ the result is known for $\mathbb{R}^{m-1}$, and suppose we have vectors $v_1,\dots,v_n\in\mathbb{R}^m$ and a vector $v\in\mathbb{R}^m$ which is not a positive linear combination of $v_1,\dots,v_n$. We wish to find a linear functional $L:\mathbb{R}^m\to\mathbb{R}$ such that $L(v)<0$ and $L(v_i)\geq 0$ for each $i$.

We may assume without loss of generality that $v$ lies in the subspace $\mathbb{R}^{m-1}$ of $\mathbb{R}^m$ (i.e., the last coordinate of $v$ is $0$). We would now like to apply the induction hypothesis on that subspace. To do so, note first that there is a finite set $S$ of elements of $\mathbb{R}^{m-1}$ such an element of $\mathbb{R}^{m-1}$ is a positive linear combination of $v_1,\dots,v_n$ iff it is a positive linear combination of elements of $S$. Namely, let $S$ consist of all the $v_i$ which are in $\mathbb{R}^{m-1}$, together with, for each pair $i,j$ such that $v_i$ and $v_j$ have last coordinates with opposite sign, some positive linear combination of $v_i$ and $v_j$ that cancels out their last coordinates. (Exercise: show that any positive linear combination of the $v_i$ which is in $\mathbb{R}^{m-1}$ can actually be written as a positive linear combination of elements of $S$. In fact, the proof below does not even need this exercise, since we will only actually use this fact for positive multiples of elements of $S$ themselves.)

So, applying the induction hypothesis to $v$ and this finite set $S$, we obtain a functional $L_0:\mathbb{R}^{m-1}\to\mathbb{R}$ such that $L_0(v)<0$ and $L_0(x)\geq 0$ for all $x\in\mathbb{R}^{m-1}$ which are positive linear combinations of the $v_i$. We now wish to extend $L_0$ to a functional $L$ on all of $\mathbb{R}^m$ such that $L(v_i)\geq 0$ for all $i$. Defining such a linear extension just amounts to picking a value for $L(e_m)$.

Write $v_i=w_i+a_ie_m$ where $w_i\in\mathbb{R}^{m-1}$ (so $a_i$ is the last coordinate of $v_i$). Without loss of generality, we may assume that $a_i>0$ for some $i$ (if $a_i=0$ for all $i$ our conclusion is immediate since each $v_i$ is actually in $\mathbb{R}^{m-1}$, and if $a_i<0$ for some $i$ we can flip the signs of all the last coordinates). Let $c$ be the maximum value of $-L_0(w_i)/a_i$ where $i$ ranges over all values such that $a_i>0$. I claim that if we set $L(e_m)=c$, then $L$ will satisfy $L(v_i)\geq 0$ for all $i$.

For $i$ such that $a_i>0$, this is immediate from the definition, since we have $L(v_i)=L_0(w_i)+a_ic$ and $c\geq -L_0(w_i)/a_i$. For $i$ such that $a_i=0$ we have $v_i=w_i\in\mathbb{R}^{m-1}$ so we already know that $L(v_i)=L_0(v_i)\geq 0$ by our choice of $L_0$. Finally, suppose $i$ is such that $a_i<0$. By our choice of $c$, there is some $j$ such that $a_j>0$ and $c=-L_0(w_j)/a_j$. For this $j$, we have $L(v_j)=L_0(w_j)+a_jc=0$. Now consider the vector $x=v_i-\frac{a_i}{a_j}v_j$. This is a positive linear combination of $v_i$ and $v_j$, and the coefficients are chosen such that the last coordinate cancels out and $x\in\mathbb{R}^{m-1}$. By our choice of $L_0$, we have $L(x)=L_0(x)\geq 0$. But since $L(v_j)=0$, we also have $L(v_i)=L(x)$, and thus $L(v_i)\geq 0$.

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(Partial answer to $2) \implies 1) $)

If $v_1, v_2, \dots, v_n$ are linearly independent, then $(v_1, v_2, \dots, v_n)$ can be completed to a basis $\mathcal{B}=(v_i)_{i\in I}$ of $V$ where $\{1,2,\dots,n\}\subset I$, then we can write every $v \in V$ uniquely as a linear combination of $(v_i)_{i\in I}$, $v= \sum\limits_{i\in I}x_iv_i \ , \ $ where only finitely many $x_i$'s are nonzero.

Let $P_i:V\rightarrow\mathbb{R}, \ P_i(x)=x_i $

For all $i \in \{1,2,\dots,n\}$ and $j \in I \setminus \{1,2,\dots,n\}$, we have $\pm P_j(v_i)=0 \ $ then $\pm P_j(v) \geq0 \ $ i.e $P_j(v)=x_j=0$.

For all $i \in \{1,2,\dots,n\}$ and $j \in\{1,2,\dots,n\}$, we have $P_j(v_i)=\delta_{ij} \geq0$ then $P_j(v)=x_j \geq0$.

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