# Prove that the vector space spanned by the presented system of equations is at least one dimensional

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.

• 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? Aug 23, 2019 at 6:04
• 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. Aug 23, 2019 at 8:02
• 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? Aug 23, 2019 at 16:15
• @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. Oct 5, 2019 at 9:32
• 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. Oct 5, 2019 at 12:08

Given $$\mathbb h$$, the nonzero vector $$q_i=h_i^{-1}$$ solves the equation.
• This (choosing $\mathbf q$) is not something you are allowed to do for the given problem statement; see my comment at OP. Aug 23, 2019 at 16:19