# Continuous eigenvalue of n by n matrix

1. May 7, 2006

### ImagineLab

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?

2. May 8, 2006

### J77

What do you mean, "continuous function"?

A general condition for an eigenvalue to be complex?

3. May 8, 2006

### Careful

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 rootsÂ´)
The latter are continuous functions (obviously, you have to be careful with the sqrt function).

Cheers,

Careful

Last edited: May 8, 2006
4. May 8, 2006

### mathwonk

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.