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Given a collection of vectors $x_1,\ldots,x_k$, which inner products $\langle x_i,x_j\rangle$ need to be known to uniquely determine all inner products? I'll begin with the specific case I am interested and then ask a more general question.

Suppose $x_1,\ldots,x_4$ are a collection of (unknown) unit vectors in $\mathbb{R}^2$. Given $\langle x_1,x_2\rangle,\langle x_2,x_3\rangle,\langle x_3,x_4\rangle,\langle x_4,x_1\rangle$ is the value of $\langle x_1,x_3\rangle$ uniquely determined? I'm essentially positive they are with some algebra, linear algebra, and a proof by image, but I'm looking for a more elegant way to solve this and/or results that yield a more straightforward way to show this.

The above can be considered like a square graph, with edges between two vectors indicating a known inner product. Given the same situation, with different dimension and/or different number of vectors, can we determine the uniqueness of all inner products given the graph structure?

Edit: I'm wondering if this can be nicely formalized as a graph rigidity problem... I'm also interested in the case where the norm of each $x_i$ is unknown, other than being nonzero.

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    $\begingroup$ ah, I thought about 3d sphere $\endgroup$
    – Fedor Petrov
    Commented Feb 13 at 13:33
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    $\begingroup$ on the plane the answer is still negative: take 4 points such that $x_1x_4$ is a diameter of the unit circle and replace $x_1$ to a point symmetric to $x_1$ with respect this diameter $\endgroup$
    – Fedor Petrov
    Commented Feb 13 at 13:39
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    $\begingroup$ If I understand correctly, $x_1=-x_4$? Then isn't $\left<x_1,x_3\right> = -\left<x_3,x_4\right>$ which is known.. $\endgroup$
    – RandomTensor
    Commented Feb 13 at 13:55
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    $\begingroup$ If you restrict to unit vectors, then your setting is equivalent to what people study in rigidity theory as (global) rigidity of spherical frameworks. Another keyword to look up in this context is "matrix completion". $\endgroup$
    – M. Winter
    Commented Feb 13 at 16:51
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    $\begingroup$ Another way to frame this question, at least for the standard inner product: if we take $X$ to be a matrix whose columns are $x_1,\dots,x_k$, then what you have is knowledge of some of the entries of the (positive semidefinite) matrix $X^TX$. For your case, we have $$ X^TX = \pmatrix{1 & \langle x_1,x_2 \rangle & ? & \langle x_1, x_4 \rangle\\ \langle x_2, x_1 \rangle & 1 & \langle x_2, x_3 \rangle & ?\\ ?&\langle x_3,x_2 \rangle & 1 & \langle x_3,x_4 \rangle\\ \langle x_4,x_1 \rangle & ? & \langle x_4, x_3 \rangle & 1}. $$ $\endgroup$
    – Ben Grossmann
    Commented Feb 13 at 17:19

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