Is there a polynomial time algorithm for creating a set of vectors in general position?
Let me give some details for the Cauchy matrix construction, which is simple. Let $x = (x_1, \ldots, x_n)$, $y = (y_1, \ldots, y_m)$ be sequences of pairwise distinct numbers. The corresponding Cauchy matrix is
$$
C = \begin{pmatrix}
\frac{1}{x_1 - y_1} &amp; \frac{1}{x_1 - y_2} &amp; \ldots &amp; \frac{1}{x_1 - y_m}\\
\frac{1}{x_2 - y_1} &amp; \frac{1}{x_2 - y_2} &amp; \ldots &amp; \frac{1}{x_2 - y_m}\\
\vdots &amp; \vdots &amp; \ddots &amp; \vdots\\
\frac{1}{x_n - y_1} &amp; \frac{1}{x_n - y_2} &amp; \ldots &amp; \frac{1}{x_n - y_m}\\
\end{pmatrix}
$$

Then any square submatrix of $C$ is an invertible square Cauchy matrix. In particular, the $n\times n$ submatrix $C_J$ with columns indexed by $J = \{j_1, \ldots, j_n\}$, $j_1 \leq \ldots \leq j_n$, has determinant
$$
\det C_J = \frac{\prod_{i = 2}^n\prod_{k = 1}^{i-1}{(x_i - x_k)(y_{j_i} - y_{j_k})}}{\prod_{i = 1}^n\prod_{k = 1}^n{(x_i - y_{j_k})}}.
$$

Because $x$ and $y$ are sequences of distinct numbers, the determinant is nonzero, and $C_J$ is invertible. 

If you prefer something even more concrete, you can take the Hilbert matrix: the Cauchy matrix with $x = (1, \ldots, n)$ and $y = (0, -1, -2, \ldots, -m)$. With $n = 3$ you get the points $(1, 1/2, 1/3),\ (1/2, 1/3, 1/4),\ (1/3, 1/4, 1/5),\ (1/4, 1/5, 1/6)$ and so on.