Consider a set $S$ of unit vectors in $\mathbb R^2$ such that $\left<x,y\right>=-\frac12$ if $x,y\in S,x\ne y$.

Question

This is a question from an entrance exam paper.

Consider a set $S$ of unit vectors in $\mathbb R^2$ such that $\left<x,y\right>=-\frac12$ if $x,y\in S,x\ne y$.Then it is necessarily true that

1. the set S is linearly independent
2. the set S generates $\mathbb R^2$
3. the set S is either linearly independent or generates $\mathbb R^2$
4. if the set S is linearly independent then $S$ generates $\mathbb R^2$

(Choose correct option(s))

My attempt so far:

I took $\{x_1,x_2,...x_n\}$ as a finite subset of $S$ and considered $\displaystyle \sum_{i=1}^nc_ix_i=0$ Now taking inner product with each $c_j\forall j\in \{1,2,..,n\}$ we have $c_j-\frac12\displaystyle \sum_{i\ne j}c_i=0\Rightarrow -2c_j+\displaystyle \sum_{i\ne j}c_i=0 \forall j\in \{1,2,..,n\}$

So in matrix form we have$Ac=O$

where $A=\begin{pmatrix} -2 & 1 &\cdots &1 &1 \\ 1 & -2 & \cdots &1 &1 \\ \vdots & \vdots &\ddots &\vdots&\vdots\\1&1&\cdots&1&-2\end{pmatrix}$, $c=\begin{pmatrix} c_1 \\ c_2 \\ \vdots\\c_n\end{pmatrix}$,$O=\begin{pmatrix} 0\\ 0\\ \vdots\\0\end{pmatrix}$

Thus if $A$ is invertible then clearly 1. is true.But I don't know how to show it and going for a row reduction seems very tedious for me.So I can't proceed anymore. Also I have no idea about rest of the options.

I apologise because I can't show much effort,but I am totally stuck here.

Please help me. Thnx in advance.

Answer 1

An empty set of vectors is linearly dependent (there is no non-trivial combination of vectors = 0 because there are no vectors), but doesn't span (generate) $R^2$.

A set with one (non-zero) vector is linearly independent, but doesn't span $R^2$.

A set with 2 vectors is linearly independent (else inner product is +/- 1) and therefore spans $R^2$.

A set with more than 2 vectors in $R^2$ cannot be independent, but spans $R^2$.

So looking at these four cases it is true that S is linearly independent and/or spans $R^2$. Check the wording of option (3) in your question: if this can be an inclusive or (i.e. and/or) , then it is true.

Answer 2

If $|S|\le 1$, the inner product condition is vacuously true. This set does not span $\mathbb{R}^2$.

If $|S|>1$, let $x,y\in S$. These two cannot be linearly dependent, because then $x=-y$ and $\langle x,y\rangle=-\langle x,x\rangle=-1\neq 1/2$. Now $S$ spans $\mathbb{R}^2$ by a dimensionality argument.

Answer 3

Hint: Using the geometric interpretation of the inner product, you have that $$\langle x,y \rangle=\|x\|\|y\| \cos \theta=\cos \theta=-\frac{1}{2} $$ for all $x\neq y$. Thus all distinct pairs of vectors of $S$ have an angle of $\arccos \left( -\frac{1}{2}\right)=\frac{2 \pi}{3}$ between them. This means that $S$ can't have more than two vectors.

The options for $S$ are thus:

1. an empty set
2. one unit vector
3. Two unit vectors with the appropriate angle ($2 \pi/3$) between them.
4. Three unit vectors which form an equilateral triangle.