All Eigenvalues Lie on the Unit Circle

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Discussion Overview

The discussion revolves around the eigenvalues of a 2x2 partitioned matrix and whether they lie on the unit circle. Participants explore the conditions under which this might be true, including the properties of the matrix and its components.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant asserts that all eigenvalues of the matrix A are on the unit circle but is unsure how to prove this, asking if there is a relevant theorem.
  • Another participant questions whether A is a unitary matrix, suggesting that this property might be relevant.
  • A participant provides a counterexample with a specific matrix that has an eigenvalue of 2, indicating that not all eigenvalues lie on the unit circle.
  • One participant notes that there are no special conditions on A and describes the components B1, B2, B3, and B4 as constructed from several matrices, some of which are symmetric and positive definite.
  • A later reply emphasizes the importance of the interrelations of the block matrices and questions whether the matrix is normal, suggesting that understanding these relationships is key to the discussion.

Areas of Agreement / Disagreement

Participants express disagreement regarding the claim that all eigenvalues lie on the unit circle. While some propose conditions that might lead to this property, others provide counterexamples and challenge the initial assertion.

Contextual Notes

The discussion highlights the need for clarity regarding the properties of the block matrices and their interrelations, as well as the implications of matrix normality on the eigenvalue distribution.

ali987
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Hi everyone

Consider a 2x2 partitioned matrix as follow:

A = [ B1 B2 ; B3 B4 ]

I'm sure that all eigenvalues of A are on the unit circle (i.e., abs
(all eig) = 1 ). but, I don't know how to prove it. Is there any
theorem?
 
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Is it, by chance, a unitary matrix: [tex]AA^\dag=I[/tex]?
 
[tex]\begin{pmatrix}2 & 0\\ 0 & 1\end{pmatrix}\begin{pmatrix}1 \\ 0\end{pmatrix}=2\begin{pmatrix}1\\ 0\end{pmatrix}[/tex]
 
ali987 said:
Hi everyone

Consider a 2x2 partitioned matrix as follow:

A = [ B1 B2 ; B3 B4 ]

I'm sure that all eigenvalues of A are on the unit circle (i.e., abs
(all eig) = 1 ). but, I don't know how to prove it. Is there any
theorem?
You can't prove it- it's not true! Fredrik gave an example in which the eigenvalue is 2, not on the unit circle.

Now, is there some condition on A, such as [tex]AA^\dag=I[/tex], as arkajad suggests, that you haven't told us?
 
Unfortunately, there is no special condition on A.
B1, B2, B3 and B4 are constructed from several matrices themselves, some of those matrices are symmetric and positive definite.
Is there any theorem which relate orthogonality of the A to orthogonality (or sth like that) of B1...B4?
 
ali987 said:
Is there any theorem which relate orthogonality of the A to orthogonality (or sth like that) of B1...B4?

No. What is important are the interrelations of your block matrices. Write [tex]AA^\dag[/tex] in the block matrix form and you will see. Is your matrix normal: [tex]AA^\dag=A^\dag A[/tex]? If you do not know even this - then what do you know about your B matrices and relations between them? The key to your question is there.
 

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