I came across the following problem while studying geometry.
Given $v_1, ..., v_7 \in \mathbb{R}^3$ with no three vectors in the same plane. Show that among the dot products $v_i \cdot v_j, i \neq j$ there are at least three different values.
I managed to find a counterexample for six vectors but I'm not sure how to prove it for 7 vectors.
Any help is appreciated.
On
First I will give a counterexample for 6. The figures of it are shown here.
There are two dot product values in it. All of these six vectors have unit length, and the dot product of two of them is either $1/\sqrt 5$ or $-1/\sqrt5$. (In the figures the length has been scaled so you can ignore the scale of the axes.) You may reconstruct these vectors and verify the dot products on your own.
Proof for the case 7:
Consider the Gram matrix of these 7 vectors, that is, the matrix whose $(i,j)$-th element is $v_i^\mathsf T v_j$. We choose 3 rows from this matrix first and there are $\binom{7}{3}=35$ ways to choose. Then we choose 3 columns such that any index of them is different from those indexes of the chosen rows. Hence we have $\binom{4}{3}=4$ ways to choose the columns. In total we have $35\times 4=140$ ways to choose rows and columns and therefore the corresponding 140 $3\times3$ matrices. For each matrix with rows $r_1,r_2,r_3$ and columns $c_1,c_2,c_3$, the determinant of it is
$$ \begin{aligned} \det((v_{r_i}^{\mathsf T} v_{c_j}) _ {1\leqslant i,j\leqslant 3})&=\det( \begin{pmatrix} v_{r_1}^\mathsf T \\\\ v_{r_2}^\mathsf T \\\\ v_{r_3}^\mathsf T \end{pmatrix}\begin{pmatrix} v_{c_1} & v_{c_2} & v_{c_3} \end{pmatrix})\\\\ &=\det( \begin{pmatrix} v_{r_1}^\mathsf T \\\\ v_{r_2}^\mathsf T \\\\ v_{r_3}^\mathsf T \end{pmatrix})\det(\begin{pmatrix} v_{c_1} & v_{c_2} & v_{c_3} \end{pmatrix})\\\\ &\neq0. \end{aligned} $$
Both of the two determinants in the product are not zero since $v_{r_1},v_{r_2},v_{r_3}$ are not coplanar and neither are $v_{c_1},v_{c_2},v_{c_3}$.
Now suppose there are only two different dot product values between different vectors. Then we can assert that at least one of the 140 chosen matrices has zero determinant, since there are $2^3=8$ possibilities for one column and we have at most $\binom{8}{3}=56$ matrices that have no repeated columns. This contradicts with our previous result of nonzero determinant.
Update: Another answer by @Marco has been improved and it turns out our ideas essentially coincide.
Updata 2: Here is my proof on the existence of a singular matrix.
Case 1: For any vector $v_i$ ($1\leqslant i\leqslant6$), there are three other vectors having dot product $a$ with it and three having dot product $b$ ($\neq a$). Then half of $v_i^\mathsf Tv_j$ ($i\neq j$) takes value $a$ and half takes value $b$. There are $7\times6=42$ combinations of $(i,j)$ so there are 21 $a$ and 21 $b$. But this is impossible because $v_i^\mathsf T v_j=v_j^\mathsf T v_i$ implies the number of $a$ is even and so is the number of $b$.
Case 2:
There exists one vector, say $v_1$, such that at least 4 other vectors have the same dot product with it. Suppose $\color{blue}{v_1^\mathsf Tv_2=v_1^\mathsf Tv_3=v_1^\mathsf Tv_4=v_1^\mathsf Tv_5=a}$.
Consider the dot products of $v_6$ and $v_2,v_3,v_4,v_5$. Then two of them are $a$ and the other two are $b$, because if there are three having the same dot product, then it is not hard to see we have obtained a $3\times3$ submatrix in the Gram matrix with two identical columns/rows. Suppose $\color{blue}{v_6^\mathsf Tv_2=v_6^\mathsf Tv_3=a}$ and $\color{blue}{v_6^\mathsf Tv_4=v_6^\mathsf Tv_5=b}$.
Consider the dot products of $v_2,v_3$ and $v_4,v_5$. We assert that $v_2^\mathsf Tv_4\neq v_3^\mathsf T v_4$. Otherwise, together with $v_2^\mathsf Tv_1= v_3^\mathsf T v_1=a$ and $v_2^\mathsf Tv_6= v_3^\mathsf T v_6=a$ we obtain a $3\times3$ submatrix in the Gram matrix with two identical columns/rows. Suppose $\color{blue}{v_2^\mathsf Tv_4=a}$ and $\color{blue}{v_3^\mathsf T v_4=b}$. Similarly, $v_2^\mathsf Tv_5\neq v_3^\mathsf T v_5$. We assert that $\color{blue}{v_2^\mathsf Tv_5=b}$ and $\color{blue}{v_3^\mathsf T v_5=a}$. Otherwise, $v_2^\mathsf Tv_5=a$ and $v_3^\mathsf T v_5=b$, so $v_4^\mathsf Tv_1= v_5^\mathsf T v_1=a$, $v_4^\mathsf Tv_2= v_5^\mathsf T v_2=a$, $v_4^\mathsf Tv_3= v_5^\mathsf T v_3=b$, indicating we obtain a $3\times3$ submatrix in the Gram matrix with two identical columns/rows.
Consider $v_1^\mathsf Tv_7$ and $v_6^\mathsf Tv_7$. Since $v_1^\mathsf Tv_2=v_6^\mathsf Tv_2=a$ and $v_1^\mathsf Tv_3=v_6^\mathsf Tv_3=a$, we have $v_1^\mathsf Tv_7\neq v_6^\mathsf Tv_7$. Since $v_1^\mathsf Tv_4=a$ and $v_6^\mathsf Tv_4=b$, we have $\color{blue}{v_1^\mathsf Tv_7=b}$ and $\color{blue}{v_6^\mathsf Tv_7=a}$.
Consider $v_4^\mathsf Tv_7$ and $v_5^\mathsf Tv_7$. Since $v_4^\mathsf Tv_1=v_5^\mathsf Tv_1=a$ and $v_4^\mathsf Tv_6=v_5^\mathsf Tv_6=b$, we have $v_4^\mathsf Tv_7\neq v_5^\mathsf Tv_7$. If $v_4^\mathsf Tv_7=a$ and $v_5^\mathsf Tv_7=b$, then $v_7^\mathsf Tv_4=v_2^\mathsf Tv_4=a$ and $v_7^\mathsf Tv_5=v_2^\mathsf Tv_5=b$ and $v_7^\mathsf Tv_6=v_2^\mathsf Tv_6=a$. If $v_4^\mathsf Tv_7=b$ and $v_5^\mathsf Tv_7=a$, then $v_7^\mathsf Tv_4=v_3^\mathsf Tv_4=b$ and $v_7^\mathsf Tv_5=v_3^\mathsf Tv_5=a$ and $v_7^\mathsf Tv_6=v_3^\mathsf Tv_6=a$. In both cases we obtain a $3\times3$ submatrix in the Gram matrix with two identical columns/rows.
The following Gram matrix is for illustration.
$$ \begin{pmatrix} . & a & a & a & a & & b\\ a & . & & a & b & a & \\ a & & . & b & a & a & \\ a & a & b & . & & b & \\ a & b & a & & . & b & \\ & a & a & b & b & . & a\\ b & & & & & a &. \end{pmatrix} $$
If three vectors are not coplanar their scalar triple product is not zero. In the Wikipedia article is also shown that the product of two scalar triple products can be expressed in terms of dot product i.e. if we call $\mathbf{v_{ij}}:=\mathbf{v_i} \cdot \mathbf{v_j}$ $$ ((\mathbf{v_{i_1}}\times \mathbf{v_{i_2}}) \cdot \mathbf{v_{i_3}})\;((\mathbf{v_{j_1}} \times \mathbf{v_{j_2}}) \cdot \mathbf{v_{j_3}}) = \det \begin{bmatrix} \mathbf{v_{i_1j_1}} & \mathbf{v_{i_1j_2}} & \mathbf{v_{i_1j_3}}\\ \mathbf{v_{i_2j_1}} & \mathbf{v_{i_2j_2}} & \mathbf{v_{i_2j_3}} \\ \mathbf{v_{i_3 j_1}} & \mathbf{v_{i_3j_2}}& \mathbf{v_{i_3j_3}}\\ \end{bmatrix} \ne 0 $$ We are interested in the product $\mathbf{v_{ij}}$ with $i \ne j$, so we can suppose that the six indeces are all distinct. I will moreover suppose that $i_1<i_2<i_3$ and $j_1<j_2<j_3$
Given $7$ numbers the number of triplets in ascending order is $35$ (found by using a Python script). If we fix a triplet $(i_1,i_2,i_3)$ the number of triplets in ascending order of the remaining $4$ numbers are $4$. So there are $35 \cdot 4=140$ distinct matrices. if there are only 2 possible values for the dot products the number of possible row vectors is $8$. So there are $\binom{8}{3}=56$ possible matrices with distinct rows. So some of this matrices are singular, that means that at least three vectors have to be complanar. (Notice that the number of distinct matrices for the case with $6$ vectors is $20$, consistently with the fact that you found a counterexample)