Consider the map which sends an $n \times n$ matrix $A$ to its eigenvalues. Mathematically, let $\phi:M(n,\mathbb{C}) \to \mathbb{C}^n$ given by $\phi(A)=(\lambda_1(A),\cdots,\lambda_n(A))$, where $\lambda_i(A)$ are the eigenvalues of the matrix $A$. For concreteness one can fix an increasing or decreasing order for the eigen values. I can only guess that the map should be measurable. Is there a concrete logic of some kind of proof to show it. The complete random matrix theory is based upon this thing. It will be very helpful to have an answer to this question. Thank you. I tried as these are solutions of the characteristic polynomial, but failed to conclude measurablility.
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I don't understand the definition of $\phi$. We are talking about complex numbers here. So, what do you mean when you say “increasing or decreasing order”? Besides, what is $\phi(\operatorname{Id}_n)$? Is it $(1)$ or is it $(1,1,\ldots,1)$ ($n$ times)? – José Carlos Santos Aug 10 '17 at 11:33
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I wanted to fix a convention about the well-definedness of the map. For complex numbers the ordering is not possible. and $\phi(Id_n)=(1,\cdots,1)$. – TRUSKI Aug 10 '17 at 14:29
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The roots of the characteristic polynomial are "continuous in the coefficients" (Continuity of the roots of a polynomial in terms of its coefficients). The coefficients are continuous in the matrix entries. So your map (under the right interpretation) is continuous.
Tim kinsella
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