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Neutrinos lead to unexpected discovery in basic math

Anonymous Fri 15 Nov 03:27:46 2019 No.11147914 View Reply Original Report

Quoted By: >>11148013 >>11148147 >>11148553 >>11148670 >>11148679

https://www.quantamagazine.org/neutrinos-lead-to-unexpected-discovery-in-basic-math-20191113/
Apparently you can calculate eigenvectors from just the eigenvalues and you don't need to know anything about the original matrix. Apparently this has just been discovered.

Anonymous

Anonymous Fri 15 Nov 2019 03:59:16 No.11148013 Report

Quoted By: >>11148042

>>11147914
post the math

Anonymous

Anonymous Fri 15 Nov 2019 04:12:23 No.11148042 Report

Quoted By: >>11148051 >>11148670 >>11148679

>>11148013
https://www.google.com/amp/s/terrytao.wordpress.com/2019/08/13/eigenvectors-from-eigenvalues/amp/

Anonymous

Anonymous Fri 15 Nov 2019 04:15:59 No.11148051 Report

Quoted By:

>>11148042
the theorem requires the eigenvalue of a matrix based on the original matrix. im assuming youd need the original matrix to compute the eigenvalues of the new one

Anonymous

Anonymous Fri 15 Nov 2019 04:30:16 No.11148081 Report

Quoted By: >>11150030

Yeah, only if the matrix is Hermitian which is a pretty big assumption. One that is generally taken for granted in the context of QM calculations, though.

Anonymous

Anonymous Fri 15 Nov 2019 04:57:10 No.11148147 Report

Quoted By: >>11148155

>>11147914
Is this trolley meme a reference to the usage of medical knowledge acquired from unethical Japanese human experimentation?

Anonymous

Anonymous Fri 15 Nov 2019 04:59:01 No.11148155 Report

Quoted By: >>11148162

>>11148147
No, it's about healthcare reform

Anonymous

Anonymous Fri 15 Nov 2019 05:01:15 No.11148162 Report

Quoted By: >>11148675

>>11148155
Or forgiving student debt.

Anonymous

Anonymous Fri 15 Nov 2019 08:45:28 No.11148553 Report

Quoted By:

>>11147914
>hermitean matrices are diagonalizable
>diagonal contains all the eigenvalues
That's all this means, if I'm not mistaken. You cannot reconstruct the original matrix from eigenvalues only since there is no unique correspondence, infinitely many hermitean matrices have the same eigenvalues.

Anonymous

Anonymous Fri 15 Nov 2019 10:10:51 No.11148670 Report

Quoted By:

>>11147914
>>11148042
Actually very cute :3
I wonder if it's related to the cofactor expansion of the determinant
And I wonder if it has any usable equivalent in functional analysis (spectrum of an operator)

Anonymous

Anonymous Fri 15 Nov 2019 10:16:07 No.11148675 Report

Quoted By:

>>11148162
the dumbest policy. It should be illegal to campaign on the promise of forgiveness of debt or amnesty.

Anonymous

Anonymous Fri 15 Nov 2019 10:18:48 No.11148679 Report

Quoted By: >>11149983

>>11147914
>Apparently you can calculate eigenvectors from just the eigenvalues and you don't need to know anything about the original matrix

>>11148042
According to this you need to know the eigenvalues for all submatrices by deleting a row and column too. That's not just the eigenvalues of the original matrix. That's a lot of extra information about the matrix.

El Arcón

El Arcón Fri 15 Nov 2019 22:30:52 No.11149916 Report

Quoted By: >>11149959

Does God judge according to any standard of fairness other than what is right in his own eyes?

Anonymous

Anonymous Fri 15 Nov 2019 22:42:43 No.11149959 Report

Quoted By: >>11150103

>>11149916
Why are you asking us? Do you?

Anonymous

Anonymous Fri 15 Nov 2019 22:50:21 No.11149983 Report

Quoted By:

>>11148679
This.

I don't see how this is supposed to be ground-breaking.

Anonymous

Anonymous Fri 15 Nov 2019 23:03:56 No.11150030 Report

Quoted By:

>>11148081
$ A=\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \\ \\ \\ \bar{A} = \begin{pmatrix} \overline{a}_{11} & \overline{a}_{12} & \overline{a}_{13} \\ \overline{a}_{21} & \overline{a}_{22} & \overline{a}_{23} \\ \overline{a}_{31} & \overline{a}_{32} & \overline{a}_{33} \end{pmatrix} \\ \\ \\ A^T = \begin{pmatrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \end{pmatrix} \\ \\ \\ \bar{A}^T=\overline{A^T}= A^*=A^\dagger = \begin{pmatrix} \overline{a}_{11} & \overline{a}_{21} & \overline{a}_{31} \\ \overline{a}_{12} & \overline{a}_{22} & \overline{a}_{32} \\ \overline{a}_{13} & \overline{a}_{23} & \overline{a}_{33} \end{pmatrix} \\ \\ \\ \text{if } A = A^\dagger \text{, it's called a HERMITIAN} $

El Arcón

El Arcón Fri 15 Nov 2019 23:35:55 No.11150103 Report

Quoted By: >>11150113

>>11149959
I don't, no.

Anonymous

Anonymous Fri 15 Nov 2019 23:40:40 No.11150113 Report

Quoted By:

>>11150103
Toker, I suggest you buy a gun and then feel what your intuition tells you about judgment.

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