Understanding Matousek's proof of Equiangular lines

Question

In Miniature 9 in 33 Miniatures by Matousek, he proofs that:

The largest number of equiangular lines in $\mathbb R^3$ is 6, and in general, there cannot be more than $\binom{d+1}{2}$ equiangular lines in $\mathbb R^d$.

The proof starts with:

Let us consider a configuration of $n$ lines, where each pair has the same angle $\vartheta \in (0, \pi/2]$. Let $v_i$ be a unit vector in the direction of the $i$th line (we choose one of the two possible orientations of $v_i$ arbitrarily). The condition of equal angles is equivalent to $$ |\langle v_i, v_j\rangle| = \cos \vartheta, \quad \text{for all } i\neq j.$$ Let us regard $v_i$ as a column vector, or a $d\times 1$ matrix. Then $v_i^Tv_j$ is the scalar product $\langle v_i, v_j \rangle$. On the other hand, $v_iv_j^T$ is a $d\times d$ matrix.

We show that the matrices $v_iv_i^T, i = 1,2,\dots, n$ are linearly independent.

And I do not see how showing the independence of these matrices shows that we can not have more than the claimed number of equiangular lines?

Thanks!

Answer 1

The key may be the first sentence "Let us consider a configuration of lines..."

You can think of it as proof by contradiction. If we had more than 6 equiangular lines, we would construct more than 6 such independent symmetric $3\times 3$ matrices (by the outlined procedure), which is impossible.

Answer 2

All those matrices are symmetric, so they are uniqely determined by an upper triangular part. So if we choose a matrices with one 1 in an upper triangular part it and all other 0 (in upper part) say (for $d=3$): $$\pmatrix{\color{\red}{0}&\color{\red}{1}&\color{\red}{0}\\1&\color{\red}{0}&\color{\red}{0}\\0&0&\color{\red}{0}}$$

you will get ${d+1\choose 2}$ inedpendent matrices and clearly you can not have more than that in a vector space $S$ of symetric matrices, so $\dim S ={d+1\choose 2}$.

Since all matrices $v_iv_i^T, i = 1,2,\dots, n$ are linearly independent (and they are symmetric), we have $n\leq {d+1\choose 2}$