A1. Horn’s conjecture = Theorem

Answer 1, posted on March 31, 2019

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The following article deals with the case of Hermitian matrices:

254A, Notes 3a: Eigenvalues and sums of Hermitian matrices

Terence Tao

January 12, 2010

https://terrytao.wordpress.com/tag/schur-horn-inequalities/

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“A basic question in linear algebra asks the extent to which

the eigenvalues of two Hermitian matrices

constrains the eigenvalues of their sum.”

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“The complete answer to this problem is a fascinating one,

requiring a strangely recursive description

(once known as Horn’s conjecture, which is now solved),

and connected to a large number of other fields of mathematics,

such as geometric invariant theory, intersection theory,

and the combinatorics of a certain gadget known as a “honeycomb”.”

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With best regards, Jean-Claude

A2. The original Horn’s conjecture

Answer 2, posted on March 31, 2019

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The following article contains the original Horn’s conjecture:

Eigenvalues of sums of Hermitian matrices

Alfred Horn

Pacific Journal of Mathematics

Volume 12, number 1, pages 225—241, 1962

https://projecteuclid.org/euclid.pjm/1103036720

https://projecteuclid.org/download/pdf_1/euclid.pjm/1103036720

“It is therefore natural to conjecture that the set E

of all possible gammas forms a convex subset of the hyperplane (1).”

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With best regards, Jean-Claude

Dear Jean-Claude Evard ,

Yes, I mentioned the details and the (Honeycomb paper) in the introduction of the question. I wonder if any updates were published.

Best regards

A4. Perturbation bounds for eigenvalues

Answer 4, posted on April 1, 2019

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The following copy of a talk gives a beginning of answer

to the question of this thread:

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On Perturbation Bounds for Eigenvalues,

Eigenvectors and Invariant Subspaces

Michael Karow (TU-Berlin)

Joint work with Daniel Kressner

Talk is based on the paper:

On a Perturbation Bound for Invariant Subspaces of Matrices

SIMAX 35-2 (2014), pp. 599-618

http://page.math.tu-berlin.de/~karow/talks/gamm_anla_14.pdf

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With best regards, Jean-Claude

Dear Jean-Claude Evard ,

The previously mentioned paper related to the perturbations and the invariance of eigenvalues and eigenvectors of the same matrix, which is beyond the requirements of the comparison of the eigenspaces of three matrices

A, B and A + B.

Best regards

A long time ago, I worked in array processing. As by-products of my work, I developped recursive algorithms for matrices updated by adding dyads or joining columns (or) rows. In the eigendecomposition case, the new eigenvalues are located in well defined intervals defined by the ancient eigenvalues. The same happens with the singular values. I attach two papers.

Thank you professor Manuel Duarte Ortigueira , for the valuable attached papers.

One needs time to read and understand the results. So, I have the following questions:

You know the results gave good relations for the sequence of matrices that connected by a recurrence equation.

1. Are the matrices generated by the recurrence commute?

2. Do you think that we can apply the results to the posted problem?

At least partially.

best regards

A8. Eigenvalues of the sum A + B

Answer 8, posted on April 4, 2019

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In the copy of the slides of the talk that I mentioned in answer 4,

in the perturbation theorem for simple eigenvalues,

on pages 15—16, the authors give an upper bound

for the distance between a simple eigenvalue of A

and the corresponding eigenvalue of the sum A + E.

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With best regards, Jean-Claude

Assume that we have two matrices A = UdV and B. We can consider te matrix

d + U B U^{-1} = V D V^{-1}

where d and D are diagonal and U and V are eigenvector matrices. If B is already decomposed B=TST^{-1} we can put d + U B U^{-1} as a sum of a diagonal matrix and several dyads. Then the algorithm applies directly.

I do not remember all the details because I left the theme ober 25 years.

Mark Allien Roble ,

In the case of A and B are Hermitian matrices, I mentioned

the work ( If A and B are Hermitian matrices, Allen Knutson and Terence Tao, show excellent relations about the eigenvalues of A + B).

So, you need to read and understand the posted question and all previous answers, then show your contribution.

We have nothing to do with Sylvester's law of inertia.

Best

A-13. The case of simultaneous triangularizability

Answer 13, posted on July 22, 2019.

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When the given matrices are simultaneously triangularizable,

we can change both A and B to the common basis

where they are in triangular form,

then in this basis A, B, and A + B are in triangular form

and the eigenvalues of A + B are on the diagonal,

as sums of eigenvalues of A and eigenvalues of B.

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Simultaneous triangularisability:

https://en.wikipedia.org/wiki/Triangular_matrix#Simultaneous_triangularisability

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Theorem:

A 1 , … , A k is simultaneously triangularisable

if and only if

for every polynomials p in k non-commuting variables,

the matrix p ( A 1 , … , A k ) [ A i , A j ] is nilpotent

where [ A i , A j ] denotes the commutator.

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This was proven in (Drazin, Dungey & Gruenberg 1951);

a brief proof is given in (Prasolov 1994, pp. 178–179).

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Problems and Theorems in Linear Algebra

V. V. Prasolov

Translations of Mathematical Monographs

225 pages, softcover, American Mathematical Society, 1994

ISBN-13: 9780821802366

https://bookstore.ams.org/mmono-134/

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With best regards, Jean-Claude