I have the following system of linear equations $$A\mathbf{v}=\mathbf{u}$$ where $$A=(a_{ij}),\quad a_{ij}=a^{p_iq_j},\ 0<a<1$$ where $p_i,q_j\forall 0\le i,j\le N-1$ are non-negative real numbers that I can choose. Also $\mathbf{u>0}$ is given i.e. each component of $\mathbf{u}$ is $>0$. Then my question is under what conditions upon $p_i,q_j$ I can have a solution for $\mathbf{v}$, and can I have a solution such that $\mathbf{v>0}$?
Update: Let me relax the conditions on $p_i, q_j$ a little bit more. Let, $q_j=j-1\quad 0\le j\le N-1$ and $p_i$ are in increasing order i.e. $p_0<p_1<\cdots \ p_{N-1}$. Also I have $u_1>u_2>\cdots \ >u_{N-1}$. Then we have the matrix equation as $$\mathbf{Mv}={u}$$ where $M=(m_{ij})$ is a Vandermonde matrix with $\displaystyle m_{ij}=a^{(i-1)p_{j-1}}$. Now the question is how to choose $p_j$'s such that $\mathbf{v>0}$ for the given $\mathbf{u>0}$?
If you want to find $p_i,q_j$ such that for any $u > 0$ (component-wise) you can always find a solution $v > 0$, then it is impossible. Note that for any choice of values for $p_i,q_j$, your linear transformation $A$ will carry the first orthant $S$ (set of all vectors with positive entries) to a cone which is a strict subset of $S$. So each $u \in S - AS$ will not have a solution $v > 0$. If you just have one given choice of $u > 0$ and you want to know when you can find $v > 0$ for that particular $u$, then it's more tricky.