[Math] A question about pairwise inner products

linear algebra

What is the largest $m$ such that there exist $v_1,\dots,v_m \in \mathbb{C}^n$ all with unit norm and all distinct such that for all $i$ and $j$, $v_i \cdot v_j = x$ for some fixed constant value $x$.

If $x=0$ we know the answer is $n$ as we are then requiring that all the vectors are orthogonal.

It was shown in the comments by Krassotkin that $m= n+1$ is possible at least for $n=1,2$.

Is this maximum $m = n + 1$ for all $n$?

There is a related question at https://mathoverflow.net/q/31436/106623 but I don't know if the proof technique carries over.

Best Answer

Yes, $n+1$ is the maximal value of $m$ for any $n$. Let the vectors $v_i$ span a $k$-dimensional subspace in $\mathbb{C}^n$. Then there are indices $a_1, \dots, a_k$ such that the vectors $v_{a_1}, \dots, v_{a_k}$ form a basis of this subspace. In particular, any $v_i$ can be represented as a linear combination of $v_{a_1}, \dots, v_{a_k}$. If $v_i$ and $v_j$ are both not among $v_{a_1}, \dots, v_{a_k}$, then $(v_i-v_j, v_{a_1}) = \dots = (v_i-v_j, v_{a_k}) = 0$. As $v_i-v_j$ is a linear combination of $v_{a_1}, \dots, v_{a_k}$, it means that $v_i - v_j = 0$, i.e., there are at most $k+1\leq n+1$ vectors.

And, as Ethan Bolker noted, the regular $n+1$-simplex gives an example of $m=n+1$.

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