Discussion
Let \(\langle \cdot, \cdot \rangle\) denote the standard inner product on \(\mathbb{R}^{7}\). Let \(\Sigma = \{v_{1}, \ldots, v_{5}\} \subseteq \mathbb{R}^{7}\) be a set of unit vectors such that \(\langle v_{i}, v_{j} \rangle\) is a non-positive integer for all \(1 \le i \neq j \le 5\). Define \(N(\Sigma)\) to be the number of pairs \((r, s)\), \(1 \le r, s \le 5\), such that \(\langle v_{r}, v_{s} \rangle \neq 0\). The maximum possible value of \(N(\Sigma)\) is equal to
Answer: A
The maximum value is 9.
For unit vectors, \(|\langle v_{i}, v_{j}\rangle| \le 1\), so a non-positive integer inner product is 0 or -1, and -1 happens only when \(v_{j} = -v_{i}\).
Each vector can have at most one antipode in \(\Sigma\): if \(v_{2} = -v_{1}\) and \(v_{3} = -v_{1}\), then \(v_{2} = v_{3}\), which is not allowed in a set of five distinct elements (and would give inner product 1). So the five vectors form at most two antipodal pairs plus one more vector.
The ordered pairs \((r, s)\) with non-zero inner product are the five diagonal pairs \((r, r)\), with value 1, plus 2 ordered pairs for each antipodal pair: \(5 + 2 \times 2 = 9\). This is attained in \(\mathbb{R}^{7}\) by \(\pm e_{1}, \pm e_{2}, e_{3}\). 10 and 14 are not attainable, and 5 ignores the antipodal pairs.
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