Let $\{q_{k}>0;k\geq 1\}$ be a strictly decreasing sequence converging to zero. Let $\{t_{1}<t_{2}<\cdots<t_{M}\}$ be a set of $M$ fixed distinct real values greater than 1 (or 2 or 3 as needed). I wish to construct a $M\times M$ full-ranked square matrix by doing the following.
- Selecting $M$ index values, $k_{1},\cdots,k_{M}$, and
- constructing matrix $$ \begin{bmatrix} q_{k_{1}}^{t_{1}} & q_{k_{2}}^{t_{1}} & \cdots & q_{k_{M}}^{t_{1}} \\ q_{k_{1}}^{t_{2}} & q_{k_{2}}^{t_{2}} & \cdots & q_{k_{M}}^{t_{2}} \\ \vdots & \vdots &\cdots & \vdots\\ q_{k_{1}}^{t_{M}} & q_{k_{2}}^{t_{M}} & \cdots & q_{k_{M}}^{t_{M}} \\ \end{bmatrix}$$ My question is: does such a set $\{k_{1},\cdots,k_{M}\}$ always exist (to make the above matrix full-ranked)?
Or rephrasing the question:
For each chosen $k$, $$\begin{bmatrix} q_{k}^{t_{1}} \\ q_{k}^{t_{2}} \\ \vdots \\q_{k}^{t_{M}} \end{bmatrix} $$ is a $M$-dimensional vector. It is always possible to find $M$ linearly independent such vectors by choosing $M$ different $q_{k}$s from $\{q_{k};k\geq 1\}$?
Yes. In fact, any $M$ distinct indices $k_1,k_2,\ldots,k_M$ suffice. The resulting matrix, up to a permutation of columns, is a generalised Vandermonde matrix. Therefore the permuted matrix is totally positive (and in particular, nonsingular). For proofs of total positivity, see