Continuous eigenvalue of n by n matrix

[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?

 

[J77]

What do you mean, "continuous function"?

A general condition for an eigenvalue to be complex?

 

[Careful]

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

 

[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.