In an expository paper on isometries of $\Bbb R^n$ Keith Conrad proves the following corollary:

Corollary 2.7. Let $P_0,...,P_n$ be $n+1$ points in $\Bbb R^n$ in “general position”, i.e., they don’t all lie in a hyperplane. Two isometries of $\Bbb R^n$ that are equal at $P_0,...,P_n$ are the same.

So what exactly does "don't all lie in a hyperplane" mean? That these are $n+1$ linearly independent points? How is that possible in a vector space of dimension $n$? The final step is also something I have trouble with:

Upon subtracting $P_0$ from $P_0,...,P_n$, the points $0,P_1−P_0,...,P_n−P_0$ are in general position. That means no hyperplane can contain them all, so there is no nontrivial linear relation among $P_1−P_0,...,P_n−P_0$ (a nontrivial linear relation would place these $n$ points,along with $0$, in a common hyperplane), and thus $P_1−P_0,...,P_n−P_0$ is a basis of $\Bbb R^n$.

Why do the points remain "in general position" after one was substracted from the rest and why does that lead to their being a basis? If there were a nontrivial linear relation between these points does that mean that they'd be linearly dependent? I guess my issue is understanding the definition of "general position" and how it relates to linear dependence, so any help on that count is appreciated.


A hyperplane in $\mathbb{R}^n$ is the solution set of a single linear equation. Equivalently, it is the translation of a linear subspace of dimension $n-1$. For example, $x - y = 1$ defines a hyperplane in $\mathbb{R}^2$. It is a translation of the one-dimensional linear subspace given by $x - y = 0$.

Points $P_0, \ldots, P_n$ in $\mathbb{R}^n$ are in general position if there is no hyperplane that contains them all. If a collection of points is in general position, any translation of this set of points is in general position (do you see why?). Therefore, if $P_0, P_1, \ldots, P_n$ are in general position, then $0, P_1 - P_0, \ldots, P_n - P_0$ are in general position as well. Now if $P_1 - P_0, \ldots, P_n - P_0$ would be linearly dependent (in other words, if there is a non-trivial linear relation between them), they would be contained in an $n-1$-dimensional linear subspace. Since a linear subspace always contains $0$, this would be a contradiction.

  • $\begingroup$ Is a hyperplane necessarily of dimension $n-1$? Can't a line in $\Bbb R^3$ be one? A translation of points in general position keeps them that way because the original set was already a translation of a subspace and a composition of two translation is a translation by the sum, making the outcome also points in general position, right? $\endgroup$ – V.Ch. Aug 17 at 22:50
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    $\begingroup$ Yes, a hyperplane must be of dimension $n-1$. But note that if four points in $\mathbb{R}^3$ are contained in a line, they are also contained in a hyperplane (which in the case $n=3$ is just called a plane). Your idea about the translations is in the right direction, but be careful: a set of points in general position is not contained in a translation of a subspace. $\endgroup$ – merle Aug 17 at 23:16

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