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A vector $v=(a_1, ... , a_n) \in \Bbb{R}^n$ is called strongly positive if $a_i > 0$ for all $i=1, ..., n$.

a) Suppose that $v$ is strongly positive. Show that any vector that is "close enough" to $v$ is also strongly positive. (Formulate carefully what "close enough" should mean.)

We can define a "close enough" vector $w$ to mean $(a_1, ..., a_{i-1}, a_i + 1, a_{i+1} , ..., a_n)$. So $w$ is different from $v$ in only one entry. Of course, $w$ must also be strongly positive, because we are adding one to the $i$-th entry of $v$.

b) Prove that if a subspace $S$ of $\Bbb{R}^n$ contains a strongly positive vector, then $S$ has a basis of strongly positive vectors.

If $S$ contains $v$, then it must also contain all vectors that are close enough to $v$. So we have $n+1$ total vectors. Now for any $i$ (say $1$), we have $(a_1 + 1, a_2, ... ,a_n) - (a_1 , a_2 , ... , a_n) = (1 , 0, ..., 0)$. So we can generate the standard basis from these vectors, meaning that they must span all of $\Bbb{R}^n$. However, since the standard basis has only $n$ vectors and we have $n+1$ vectors, we must remove one of the vectors in order for them to be linearly independent. But which vector we remove from the list doesn't really matter, because the other vectors will still be strongly positive and that's what we were looking for.

Do you think my answer is correct?

Thank you in advance

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3  
How do you prove your claim that if a subspace contains a vector then it must contain all the vectors close enough to it, in particular with your definition of "close enough"? I honestly can't see it... –  DonAntonio Jun 19 '13 at 18:33
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No, we can't: the subspace $\;S\;$ may be such that not even one single vector from the standard basis is contained in it. –  DonAntonio Jun 19 '13 at 18:37
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@Artus Your (a) is true, but it is a strange definition. My impression is that the question would like you to say that a strongly positive vector has an open neighborhood of strongly positive vectors. –  Cocopuffs Jun 19 '13 at 18:39
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@Artus The formal definition would be that for some $\epsilon > 0$, any vector within $\epsilon$ of $v$, according to some notion of distance, is strongly positive. An example of a "notion of distance" would be the maximum absolute value of a difference between corresponding coordinates of the two vectors. –  Jack M Jun 19 '13 at 18:57
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@Artus Of course, you are indeed meant to make up your own, and yours does work for (a), but the way you apply it for (b) is incorrect, see DonAntonio's comments. –  Jack M Jun 19 '13 at 18:58

1 Answer 1

up vote 1 down vote accepted

In the comments, it's already been indicated that you should pick a definition of "close enough" that helps you with the second question.

So for a),

Theorem. Let $v \in \mathbb R^n$ be strongly positive. Then there exists an $\varepsilon > 0$ such that for all $w \in \mathbb R^n$ with $||v - w|| < \varepsilon$, $w$ is strongly positive.

(Can you prove this yourself? Given $v$, pick a concrete $\varepsilon$ that works.)

This can be used to answer b) as follows:

Theorem. Let $S$ be a subspace of $\mathbb R^n$ containing at least one strongly positive vector. Then $S$ has a basis of strongly positive vectors.

Proof. Let $v$ be a strongly positive vector in $S$ and let $e_1, \dots, e_m$ be a basis of $S$ (not necessarily consisting of strongly positive vectors). Now note that for every $\lambda \neq 0$, $v + \lambda e_1, \dots, v + \lambda e_m$ is also a basis of $S$. Using the previous theorem, pick $\lambda$ small enough that all $v + \lambda e_i$ are close enough to $v$ to be strongly positive as well. (With $\varepsilon$ as in the previous theorem, $\lambda = \varepsilon / \max(||e_1||, \dots, ||e_m||)$ will do).

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Does the first theorem require some topology? –  user58289 Jun 19 '13 at 22:04
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No. You could reformulate that theorem in topological terms (as Cocopuffs did in an earlier comment): every strongly positive vector has an open neighbourhood of strongly positive vectors. Its proof, however, is by elementary means. (Picking $\varepsilon = \min(v_1, \dots, v_n)$ works). –  Magdiragdag Jun 19 '13 at 23:10
    
Actually, $v+\lambda e_1,\ldots,v+\lambda e_m$ is not necessarily a basis. Consider $e_1=(-1,0), e_2=(0,-1)$ and $v=(1/2,1/2)$. Then $v$ is strictly positive, but $v+e_1=(-1/2,1/2)=-(1/2,-1/2)=-(v+e_2)$ –  Marcin Łoś Jun 20 at 9:20

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