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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?

Pierre-Yves Gaillard
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Myshkin
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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. – Dylan Moreland Apr 01 '12 at 17:14
  • Dear Sir, I agree I must go to $\mathbb{C}$ but would you please tell me the notation $S_n\setminus\mathbb{R}^n$? – Myshkin Apr 01 '12 at 17:16
  • 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?" – Dylan Moreland Apr 01 '12 at 17:18
  • I don't get dear sir.would you please elaborate. – Myshkin Apr 01 '12 at 17:23
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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. – N. S. Apr 01 '12 at 17:24
  • If you find a "right" way (i.e. consistent way) of always ordering the eigenvalues, the question you ask is exactly what Dylan said. – N. S. Apr 01 '12 at 17:26
  • absolutely true,I got it. – Myshkin Apr 01 '12 at 17:30
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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. – Pierre-Yves Gaillard Apr 01 '12 at 17:37
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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). – Robert Israel Apr 01 '12 at 17:50
  • @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. Thanks in advance for telling me what's wrong. Please see also Statement (ii) at the top of page 208 here. – Pierre-Yves Gaillard Apr 01 '12 at 18:38
  • @RobertIsrael. Dear Robert: Please see also Theorem 4 p. 10 here. - 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? – Pierre-Yves Gaillard Apr 01 '12 at 19:27
  • Well frankly I am not able to understand any of last 4 comments :( – Myshkin Apr 01 '12 at 20:11
  • @ Pierre-Yves Gaillard: Sorry, you're right, it is a homeomorphism. – Robert Israel Apr 02 '12 at 07:19

2 Answers2

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There is no way to order the eigenvalues so that the function $f$ (as a function into $M_n({\mathbb C})$) is continuous. Consider the matrices $A(t) = \pmatrix{0 & 1\cr e^{2it} & 0\cr}$. Note that $A(0) = A(\pi)$. The eigenvalues are $\pm e^{it}$. But if you take the eigenvalue that is $1$ at $t=0$ and follow it continuously as $t$ goes from $0$ to $\pi$, it will be $-1$ at $t=\pi$.

Robert Israel
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  • Dear sir,I am not able to understand. Would it be possible to express elaborately?will be very pleased. – Myshkin Apr 01 '12 at 20:12
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I think what's behind the question is this:

Write $$ (X-z_1)\cdots(X-z_n)=X^n+c_1X^{n-1}+\cdots+c_n,\qquad(*) $$ where $X$ is an indeterminate and $$ z=(z_1,\dots,z_n),\quad c=(c_1,\dots,c_n)\in\mathbb C^n. $$ For any $z$ there is a unique $c$ satisfying $(*)$.

Moreover, the map $z\mapsto c$, $\mathbb C^n\to\mathbb C^n$, is polynomial, and induces a continuous map $S_n\backslash\mathbb C^n\to\mathbb C^n$, where $S_n\backslash\mathbb C^n$ is the space of orbits of the symmetric group $S_n$ acting on $\mathbb C^n$ by permuting the coordinates.

Conversely, let $c$ be given. By the Fundamental Theorem of Algebras, there is a $z$ satisfying $(*)$.

Moreover, the $S_n$-orbit $S_nz$ of $z$ is depends only on $c$, and by Rouché's Theorem, the map $c\mapsto S_nz$, $\mathbb C^n\to S_n\backslash\mathbb C^n$ is continuous.

Clearly the maps $S_nz\mapsto c$ and $c\mapsto S_nz$ are inverse.

In conclusion, we have a homeomorphism $$ S_n\backslash\mathbb C^n\simeq\mathbb C^n. $$ EDIT. What Robert Israel's answer shows is that there is no continuous section to the canonical projection $\mathbb C^n\to S_n\backslash\mathbb C^n$.

Pierre-Yves Gaillard
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