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I want to know whether this is absurd question or reasonable to ask: Let $f:M_n(\mathbb{C})\to M_n(\mathbb{C})$ be given by $f(A)= B$, where $B$ is a diagonal matrix having the same eigenvalues as $A$. Is $f$ continuous?

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    It might be better to go to $\mathbb C$, for what will you do if eigenvalues are outside of $\mathbb R$? It also might be better to map into something like $S_n\backslash\mathbb R^n$, where the symmetric group permutes coordinates, since the order shouldn't really matter.2012-04-01
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    Dear Sir, I agree I must go to $\mathbb{C}$ but would you please tell me the notation $S_n\setminus\mathbb{R}^n$?2012-04-01
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    Maybe it isn't so important. I would recommend searching around here for questions involving the words "do the roots of a polynomial depend continuously on its coefficients?"2012-04-01
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    I don't get dear sir.would you please elaborate.2012-04-01
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    The main problem is that this is not really a function. There are actually up to $n!$ different matrices $B$. To understand the problem, if $A$ has eigenvalues $1,2$ which is the matrix $B$? $1,2$ on the diagonal or $2,1$ on the diagonal? The choice you make in the order of the eigenvalues matters.2012-04-01
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    If you find a "right" way (i.e. consistent way) of always ordering the eigenvalues, the question you ask is exactly what Dylan said.2012-04-01
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    absolutely true,I got it.2012-04-01
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    The ordering of the roots isn't a issue. The natural setting is to view the set of roots of a degree $n$ polynomial with complex coefficients as an element of the quotient on $\mathbb C^n$ by the group of permutation of the coordinates, quotient which is a well-defined topological space (homeomorphic to $\mathbb C^n$). Then the continuity follows from Rouché's Theorem.2012-04-01
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    Note also that eigenvalues must be counted by algebraic multiplicity so that there are always $n$ of them. The quotient is certainly not homeomorphic to ${\mathbb C}^n$, in fact it is not a manifold (think of what a neighbourhood of $(0,\ldots,0)$ looks like).2012-04-01
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    @RobertIsrael. Dear Robert: I think your last comment refers to my previous comment. I suppose you forgot to ping me. I was assuming implicitly that the multiplicities must indeed be taken into account. I tried to prove the homeomorphism $S_n\backslash\mathbb C^n\simeq\mathbb C^n$ [here](http://math.stackexchange.com/a/121607/660). Thanks in advance for telling me what's wrong. Please see also Statement (ii) at the top of page 208 [here](http://books.google.fr/books?id=nDgBsOurnAIC&lpg=PA210&hl=fr&pg=PA208#v=onepage&q&f=false).2012-04-01
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    @RobertIsrael. Dear Robert: Please see also Theorem 4 p. 10 [here](http://www.numdam.org/item?id=SHC_1953-1954__6__A12_0). - Even if you don't believe in the homeomorphism $S_n\backslash\mathbb C^n\simeq\mathbb C^n$, do you agree that: (a) $X:=S_n\backslash\mathbb C^n$ is a topological space, (b) the multiset of roots of a degree $n$ polynomial defines a point of $X$, (c) this point depends continuously on the polynomial?2012-04-01
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    Well frankly I am not able to understand any of last 4 comments :(2012-04-01
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    @ Pierre-Yves Gaillard: Sorry, you're right, it is a homeomorphism.2012-04-02

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