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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?

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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 $$\prod_{1\leq s<r\leq k}(j_{r}-j_{s})\neq0.$$

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  • $\begingroup$ I was thinking about something like that.. But I wasn't sure that it will be always independent.. Thank you for your answer and mentioning it name $\endgroup$ – Kareem Metwaly Feb 8 '18 at 22:58
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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?

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  • $\begingroup$ I am not sure that I understand your answer, what do you mean by extending the definition using the recursive definition.. Could you please elaborate by an example? $\endgroup$ – Kareem Metwaly Feb 8 '18 at 23:00

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