3
$\begingroup$

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?

  • 0
    @ Pierre-Yves Gaillard: Sorry, you're right, it is a homeomorphism.2012-04-02

2 Answers 2

3

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 $t=\pi$.

  • 0
    Dear sir,I am not able to understand. Would it be possible to express elaborately?will be very pleased.2012-04-01
2

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 $\mathbb C^n\to S_n\backslash\mathbb C^n$.