Timeline for Uniquely reconstruct a matrix $M$ from its inverse $M^{-1}$ if $n$ elements of $M^{-1}$ are unknown and $n$ elements of $M$ are given
Current License: CC BY-SA 4.0
25 events
| when toggle format | what | by | license | comment | |
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| Oct 13, 2023 at 9:03 | answer | added | Denis Serre | timeline score: 5 | |
| S Nov 2, 2021 at 9:15 | history | bounty ended | Carlo Beenakker | ||
| S Nov 2, 2021 at 9:15 | history | notice removed | Carlo Beenakker | ||
| Nov 2, 2021 at 9:15 | vote | accept | Carlo Beenakker | ||
| Nov 2, 2021 at 1:28 | answer | added | David E Speyer | timeline score: 32 | |
| Nov 1, 2021 at 7:57 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
conjecture
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| S Oct 31, 2021 at 18:08 | history | bounty started | Carlo Beenakker | ||
| S Oct 31, 2021 at 18:08 | history | notice added | Carlo Beenakker | Draw attention | |
| Oct 25, 2021 at 12:23 | comment | added | Steven Landsburg | Okay---I should indeed have realized what you were doing. I preferred to think of the $N^2$ missing variables from the two matrices combined as the variables, and to write the equation $M\times M^{−1}=I$. This gives $N^2$ equations (for the $N^2$ entries) that are at worst quadratic, which seemed easier to think about than fewer equations of higher degree. | |
| Oct 25, 2021 at 6:03 | comment | added | Carlo Beenakker | @StevenLandsburg --- I don't think so: take $M^{-1}$, label the $n$ unknowns $x_1,x_2,\ldots x_n$, then invert that matrix to obtain a matrix $\tilde{M}(x_1,x_2,\ldots x_n)$; let the known elements of $M$ be $M_{i_1,j_1}, M_{i_2,j_2}\ldots M_{i_n,j_n}$; then the $n$ equations with $n$ unknowns are $\tilde{M}_{i_k,j_k}(x_1,x_2,\ldots x_n) =M_{i_k,j_k}$, $k=1,2,\ldots n$. | |
| Oct 24, 2021 at 23:24 | comment | added | Steven Landsburg | Sorry if I'm being dense...but when you say "$n$ nonlinear equations in $n$ unknowns", should both instances of $n$ be replaced with $N^2$ (at least in the non-symmetric case)? | |
| Oct 24, 2021 at 20:52 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
also tested N=4,5
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| Oct 24, 2021 at 20:43 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
als tested N=4,5
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| Oct 24, 2021 at 16:29 | comment | added | Carlo Beenakker | @StevenLandsburg -- I have not thought (yet) about the case of a positive definite but non-symmetric matrix; in the context of a covariance matrix we can assume it is both p.d. and symmetric. | |
| Oct 24, 2021 at 16:15 | comment | added | Steven Landsburg | Are you claiming that the symmetric case suffices for the general case or that the symmetric case suffices for your purposes? | |
| Oct 24, 2021 at 14:10 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
link to Mathematica note book
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| Oct 24, 2021 at 6:01 | comment | added | Carlo Beenakker | it's true for $N=2$, I added the brief calculation for that case. | |
| Oct 24, 2021 at 6:00 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
N=2 case worked out
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| Oct 23, 2021 at 17:17 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
mentioned the $N=2$ case
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| Oct 23, 2021 at 15:30 | comment | added | Random | Actually, the conjecture is false for $N = 2$. | |
| Oct 23, 2021 at 15:28 | comment | added | Random | Perhaps you mean your conjecture to be "$n = N$ elements on the diagonal and $N^2 - n = N^2 - N$ off diagonal elements"? Otherwise already for $N = 2$, giving the two diagonal elements of $M^{-1}$ only determines the off diagonal element of $M$ up to sign. | |
| Oct 23, 2021 at 7:09 | comment | added | Carlo Beenakker | one can take $M$ real symmetric (in the linked post it is a covariance matrix). | |
| Oct 23, 2021 at 5:56 | comment | added | joro | Over what field are the entries? Are you asking about all fields? | |
| Oct 22, 2021 at 17:06 | comment | added | Carlo Beenakker | I hesitated whether to raise this as a follow-up of the cited MO post, but following this advice I ask it separately. | |
| Oct 22, 2021 at 17:03 | history | asked | Carlo Beenakker | CC BY-SA 4.0 |