Continuous eigenvalue of n by n matrix

Click For Summary

Discussion Overview

The discussion revolves around the conditions under which the eigenvalues of a complex n by n matrix are continuous functions of the matrix itself. Participants explore the implications of continuity in the context of matrix properties and eigenvalue behavior.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions the meaning of "continuous function" in relation to eigenvalues.
  • Another participant asserts that there are no restrictions on the continuity of eigenvalues, citing the continuity of trace and determinant as examples.
  • A later reply challenges the previous assertion, stating that while the eigenvalues themselves may not be functions of the matrix due to their unordered nature, the elementary symmetric functions of the eigenvalues, such as trace and determinant, are continuous functions of the matrix.
  • This participant also draws a parallel to the continuity of polynomial roots with respect to their coefficients.

Areas of Agreement / Disagreement

Participants express differing views on the nature of eigenvalues as functions of the matrix. There is no consensus on whether eigenvalues can be considered continuous functions in the traditional sense, leading to an unresolved discussion.

Contextual Notes

Participants highlight the importance of considering the ordering of eigenvalues and the implications of using functions like the square root in the context of continuity.

ImagineLab
Messages
3
Reaction score
0
Hello, all,
This is my fisrt time post on this forum, I have this question for long time but people around me couldn't really answer it, hopes I can get the answer from you guys...

Given a complex n by n matrix A, Under what restriction, its eigenvalue(s) is the continuous function of A?
 
Physics news on Phys.org
What do you mean, "continuous function"?

A general condition for an eigenvalue to be complex?
 
ImagineLab said:
Hello, all,
This is my fisrt time post on this forum, I have this question for long time but people around me couldn't really answer it, hopes I can get the answer from you guys...

Given a complex n by n matrix A, Under what restriction, its eigenvalue(s) is the continuous function of A?

There are no restrictions whatsoever. For example, in two dimensions the trace and determinant are continuous functions in the matrix topology and the eigenvalue polynomial is:
x^2 - tr(A)x + det(A) with roots (tr(A) + sqrt(tr(A)^2 - 4 det(A) ) )/2 (here sqrt means `all possible roots´)
The latter are continuous functions (obviously, you have to be careful with the sqrt function).

Cheers,

Careful
 
Last edited:
well this is correct except the answer may be misleading, i.e. since their ordering is undetermined, the eigenvalues are not functions of the matrix at all ["be careful of the square root"], but the elementary symmetric functions of the eigenvalues, e.g. the trace and determinant, are.

it is the unordered set of eigenvalues that is a continuous function of the matrix, and this is equivalent to saying the elementary symmetric functions of them are continuous.

this is the same question as asking whether the roots of a polynomial are continuous functions of the coefficients.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 18 ·
Replies
18
Views
4K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 17 ·
Replies
17
Views
7K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
1
Views
3K
  • · Replies 69 ·
3
Replies
69
Views
9K