1
$\begingroup$

Let $\mathbb{h} \in \mathbb{Q}_{>0}^n$ a vector of positive rational numbers. Can you, please, help me prove that the vector space \begin{equation} Q = \lbrace \mathbf{q} \in \mathbb{Q}^n \,|\, \forall i, j \in \lbrace 1, 2,\dots, n \rbrace : q_i\,h_i = q_j\,h_j\rbrace \end{equation} is at least one dimensional?

If $\mathbb{q} \in \mathbb{Q}^n$ satisfies the previous equations than $\alpha\mathbb{q} \in \mathbb{Q}^n$ also satisfies the equations. However, I have trouble proving that something other than $\mathbf{0}$ vector satisfies the solution. Is it enough to say that a vector $q_1 = 1$ satisfies the equations?

I think my intuition is correct, but I am really having trouble formalizing it.

$\endgroup$
6
  • 1
    $\begingroup$ The way you have set up the question, you have no way of knowing whether or not q is $\bar 0$, in which case the space generated by q would obviously be of $0$ dimension. Do you just want to know whether or not a non-zero possibility for q? $\endgroup$
    – CardioidAss22
    Aug 23, 2019 at 6:04
  • $\begingroup$ According to my understanding a zero-dimensional space is $\lbrace\mathbf{0}\rbrace$ (math.stackexchange.com/questions/664594/…). When I say at least one dimensional space, I would like to know if the space spanned by the set of vectors $\mathbf{q}$ is at least a line. I think this is shown by Chris Cutler below. There exists a non-zero vector that solves the equations which can be multiplied by a scalar and still satisfy the equations. This basically shows what I wanted. I hope I am answering you comment. Btw, I appreciate any discussion. $\endgroup$
    – Slaven Glumac
    Aug 23, 2019 at 8:02
  • $\begingroup$ The way you state the question (the very first word "Let"), the vector $\mathbf q$ is given to you (as is $\mathbf h$), and you need to prove that it must be nonzero (in which case it spans a $1$-dimensional space). There is no option of choosing a value for $\mathbf q$. However none of the information provided is inconsistent with $\mathbf q$ being the zero vector, so there is no way you could prove it is nonzero. What are you really asking here? $\endgroup$
    – Marc van Leeuwen
    Aug 23, 2019 at 16:15
  • $\begingroup$ @MarcvanLeeuwen If I understood correctly, your complaint is that words "Let $\mathbf{q} \in \mathbb{Q}^n$" fix the vector $\mathbf{q}$. I wanted to ask if for fixed $\mathbf{h} \in \mathbb{Q}_{>0}^n$ the vector space $Q = \lbrace \mathbf{q} | \mathbf{q} \in \mathbb{Q}^n, \, q_i\,h_i = q_j\,h_j\rbrace$ is at least one dimensional. It's easy to show it is a vector space by definition. Would this be an ok way to rephrase the question? I will rephrase it if you say this is ok. $\endgroup$
    – Slaven Glumac
    Oct 5, 2019 at 9:32
  • $\begingroup$ So why not just start "Let $\mathbf{h} \in \mathbb{Q}_{>0}^n$ be a vector..." and then ask whether the vector space $Q = \{\, \mathbf{q} \in \mathbb{Q}^n \mid \forall i,j\in\{1,2,\ldots, n\}:q_i\,h_i = q_j\,h_j\,\}$ is at least one dimensional? Or alternatively you could ask to show that there exists some nonzero $\mathbf q$ satisfying the given relations, as that is what it is really about. $\endgroup$
    – Marc van Leeuwen
    Oct 5, 2019 at 12:08

1 Answer 1

Reset to default
0
$\begingroup$

Given $\mathbb h$, the nonzero vector $q_i=h_i^{-1}$ solves the equation.

$\endgroup$
2
  • $\begingroup$ This (choosing $\mathbf q$) is not something you are allowed to do for the given problem statement; see my comment at OP. $\endgroup$
    – Marc van Leeuwen
    Aug 23, 2019 at 16:19
  • 1
    $\begingroup$ @MarcvanLeeuwen Yes, I assume that the OP wasn't sure how to phrase the question they really wanted to ask, so I extrapolated the real question from the rest of the post, and answered that. Given OP's comment "I would like to know if the space spanned by the set of vectors...", and the fact that they accepted this answer, I'm pretty confident that I extrapolated correctly. $\endgroup$
    – Chris Culter
    Aug 23, 2019 at 18:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .