Timeline for Given a collection of vectors $x_1,\ldots,x_k$, which inner products $\langle x_i,x_j\rangle$ are needed to uniquely determine all inner products
Current License: CC BY-SA 4.0
18 events
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| Feb 13 at 19:36 | comment | added | RandomTensor | Okay I get it now, thanks. Seems like the diameter, thing is necessary for this counterexample to work… | |
| Feb 13 at 19:23 | comment | added | Fedor Petrov | in my example the inner product of $x_1$ and $x_3$ also changes | |
| Feb 13 at 17:41 | comment | added | RandomTensor | @BenGrossmann this also looks good thank you. I guess I can attempt to solve this via a constrained optimization problem. | |
| Feb 13 at 17:20 | comment | added | Ben Grossmann | The possible ways of assigning the remaining inner products is equivalent to the possible ways of filling in the entries of $X^TX$ that maintain the positive semidefiniteness property. In other words, this is a positive semidefinite matrix completion problem. | |
| Feb 13 at 17:19 | comment | added | Ben Grossmann | 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 $$ X^TX = \pmatrix{1 & \langle x_1,x_2 \rangle & ? & \langle x_1, x_4 \rangle\\ \langle x_2, x_1 \rangle & 1 & \langle x_2, x_3 \rangle & ?\\ ?&\langle x_3,x_2 \rangle & 1 & \langle x_3,x_4 \rangle\\ \langle x_4,x_1 \rangle & ? & \langle x_4, x_3 \rangle & 1}. $$ | |
| Feb 13 at 17:18 | comment | added | RandomTensor | @M.Winter this looks like a very good starting point, thanks! | |
| Feb 13 at 16:51 | comment | added | M. Winter | 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". | |
| Feb 13 at 15:51 | comment | added | RandomTensor | The uniqueness of the inner products is all I am interested in not in vectors themselves. | |
| Feb 13 at 15:33 | comment | added | Fedor Petrov | Sorry, $x_2=-x_4$. In complex coordinates, $x_4=1,x_2=-1$, $x_3$ is arbitrary and $x_1$ may be replace to $\bar{x}_1$ without changing the four given scalar products | |
| Feb 13 at 13:55 | comment | added | RandomTensor | 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.. | |
| Feb 13 at 13:39 | comment | added | Fedor Petrov | 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 | |
| Feb 13 at 13:33 | comment | added | Fedor Petrov | ah, I thought about 3d sphere | |
| Feb 13 at 13:19 | comment | added | RandomTensor | I really don't think this is correct. These values don't change with transform by G in O(2). From this we can just assume x_1= [1,0], and x_2,1 x_4,1 are also determined then. We also know (x_2,2)^2 by (x_2,2)^2 = 1 -(x_2,1)^2, and likewise for x_4,2^2. Because its the general orthogonal group we can further assume x_2,2 is positive. Two degrees of freedom x_3 and one sign x_4,2 are all that are unknown. We have 2 equations involving x_3 from the inner products, and one equation since its norm is known. So we really have 2 unknown reals and 1 unknown sign. It gets messy from here. | |
| Feb 13 at 12:58 | history | edited | YCor | CC BY-SA 4.0 |
formatting, changed tag; edited tags
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| Feb 13 at 12:55 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
MathJax: \langle, \rangle
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| Feb 13 at 12:41 | history | edited | RandomTensor | CC BY-SA 4.0 |
added 85 characters in body; edited tags
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| Feb 13 at 12:34 | history | edited | RandomTensor | CC BY-SA 4.0 |
Added graph rigidity
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| Feb 13 at 12:28 | history | asked | RandomTensor | CC BY-SA 4.0 |