# How to form a genral matrix $n \times k, k \leq n$ such that any $k$ rows are linearly independent?

Let's say that I have $n \ge k$.

I want a general formula for generating an $n \times k$ matrix $A_{n\times k}$ such that selecting any $k$ rows will be linearly independent and can span a space of dimension $k$.

Example: $n=3$ and $k=2$

$$A_{3\times 2} = \begin{bmatrix} 1&0\\0&1\\1&1 \end{bmatrix}$$ selecting any 2 rows will be linearly independent and, thus, span the sapce of dimension 2.

Is this generally possible?

Consider the matrix $$A=\begin{bmatrix}1&1&1&\cdots&1\\1&2&2^{2}&\cdots&2^{k}\\1&3&3^2&\cdots&3^{k}\\\vdots&\vdots&\vdots&\ddots&\vdots\\1&n&n^{2}&\cdots&n^{k}\end{bmatrix}.$$
Any $k$ rows of $A$ form a Vandermonde matrix with values $\{j_{1},\ldots,j_{k}\}\subset\{1,2,\ldots,n\},$ which has determinant $$\prod_{1\leq s<r\leq k}(j_{r}-j_{s})\neq0.$$
Does $\pmatrix{1 & 1\\ 1 & 2\\ 1 & 3\\ 1 & 4 \\ \cdots & \cdots}$ help you for the case $N\times 2$? Can you extend this to arbitrary $k$ by using the recursive definition of the determinant?