# There exist an infinite subset $S\subseteq\mathbb{R}^3$ such that any three vectors in $S$ are linearly independent.

Could anyone just give me hint for this one?

There exist an infinite subset $S\subseteq\mathbb{R}^3$ such that any three vectors in $S$ are linearly independent. True or false?

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 Do you know any facts about linearly independent sets? – Chris Eagle Dec 12 '12 at 22:27

Try $\left(\begin{matrix}1\\t\\t^2\end{matrix}\right)$ with $t\in\mathbb R$. Do you know to compute $$\left\vert\begin{matrix}1&1&1\\r&s&t\\r^2&s^2&t^2\end{matrix}\right\vert?$$

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 good one!!. Please clear my doubt, consider the random gaussian vector $[x,y,z]$, where each entries are independent gaussian random variables. So shouldn't the set of all realizations of this random vector be a set of the kind he is looking for? – dineshdileep Dec 13 '12 at 5:44 the determinant is $(st^2-s^2t)-(rt^2-tr^2)+(rs^2-sr^2)$ – Taxi Driver Dec 14 '12 at 5:08 @Kuttus ... and that equals $(r-s)(s-t)(t-r)$ and is non-zero unless two of the numbers are equal. :) Actually, the fact that the functions $x\mapsto 1$, $x\mapsto x$ and $x\mapsto x^2$ are linearly independent lurks behind this as a shortcut: If a linear combinaton of the rows is the zero vector, then the correspondnig quadratic polynomial has three distinct roots $r,s,t$, hence is the zero polynomal, hence the linear combination was in fact the trivial combination, hence the rows are linearly independent, hence so are the columns (which is what we were actually after). – Hagen von Eitzen Dec 14 '12 at 11:39

Three vectors in $\mathbb{R}^3$ are linearly dependent if and only if they lie in a plane.

Consider the following process for building $S$. We can start with the empty set, and choose any two vectors $v_1, v_2 \in \mathbb{R}^3$ and add them to $S$. Then to choose a third vector $v_3$ to add to $S$, we must make sure it is not in the unique plane containing (i.e. spanned by) $v_1$ and $v_2$. Thus $v_3$ can be any vector in $\mathbb{R}^3 \backslash span(v_1, v_2)$.

Similarly, if at some stage $S = \{v_1, \ldots, v_k\}$, we can add to $S$ any vector $v_{k+1}$ in $\mathbb{R}^3 \backslash \bigcup_{x_i, x_j} span(x_i, x_j)$. Note that $\bigcup_{x_i, x_j} span(x_i, x_j)$ is a finite union of planes, so it can never be all of $\mathbb{R}^3$. In this way we can choose an infinite set with the desired property.

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The key issue with this approach is to prove that a finite union of planes is not all of $\mathbb R^3$. For a more general result: mathoverflow.net/questions/26/… – Andres Caicedo Dec 12 '12 at 22:47

The parametric curve $t\mapsto(1,t,t^2)$ has the property that any three distinct points on it are linearly independent (without the "distinct" one clearly cannot have a solution). To check, just compute the determinant, which is Vandermonde.

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Inspired by the Vandermonde matrix example, let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a strictly convex function. Then $$\left\vert\begin{matrix}1&1&1\\x&y&z\\f(x)&f(y)&f(z)\end{matrix}\right\vert\not=0$$ for any three distinct numbers $x,y,z$.

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