Continuous mapping from matrices to eigenvalues Let $n\in\mathbb{N}$, and let $M_n(\mathbb{C})$ be the set of all $n\times n$ matrices whose entries are taken from $\mathbb{C}$.
Now, is there a continuous function $f:M_n(\mathbb{C})\to\mathbb{C}^n$ that maps every matrix $A$ exactly to its eigenvalues $\lambda_1,...,\lambda_n$?
I don't know how to proceed. If there exists such a function, then how can I prove its continuity ?
Thanks in Advance.
 A: $\newcommand{\F}{\mathscr F}$ 
Let $\F$ denote the set of finite subsets of $\mathbb C$. Give $\F$ the Hausdorff metric: $$d(F_1,F_2)=\max_{z\in F_1}\min_{w\in F_2}|z-w|+\max_{z\in F_2}\min_{w\in F_1}|z-w|.$$
$\newcommand{\P}{\mathscr P}$ Let $\P$ denote the space of polynomials of degree $n+1$. Give $\P$ the obvious topology.
Let $p_A$ be the characteristic polynomial of $A$. Then the map $A\mapsto p_A$ is certainly continuous. For $p\in\P$ let $Z(p)$ be the zero set of $p$. It follows from Rouche's Theorem that $p:\P\to\F$ is continuous. So the map $A\mapsto Z(p_A)$ is continuous.
Which is at least analogous to what you want; seems possible that you can use this for the application  you have in mind.
Details regarding continuity of $Z$: Suppose $p_k\to p$ in $\P$. Then $p_k(z)\to p(z)$ uniformly on compact sets. Hence if $p(a)=0$ Rouche shows that there exist $z_k\to a$ with $p_k(z_k)=0$. 
We also need to show that for every $\epsilon>0$ there exists $K$ so that if $k>K$ and $p_k(a)=0$ then there is a zero of $p$ within $\epsilon$ of $a$. It's not so clear how this part follows from Rouche; the quantifiers don't work out right. Luckily this part is trivial: If not then we have $p_{k_j}(z_{k_j})=0$ and $|z_j-z|\ge\epsilon$ for every $j$ and every $z\in Z(p)$.. This is impossible:
Since $p_k\to p$ in $\P$ there is a uniform bound on all the coefficients, and the leading coefficient is bounded away from zero. This gives a uniform bound $|z|\le R$ on all the zeroes of $p_k$ and of $p$. So passing to a further subsequence we have $z_j\to z$, hence $p(z)=0$ and $|z_j-z|<\epsilon$ for large $j$.
A: I've actually been on the very last metres of writing my own question on this when I stumbled upon this post; as such, this is less of an actual anwser, and more a summary of what I have found out about this problem so far, as well as my thoughts on what remains to be answered here.
First of all, notation: let $\mathbb C_n[t]$ be the set of all normed polynomials of degree $n$, with its obvious topology. Also, let $\sim$ be the equivalence relation of two vectors in $\mathbb C^n$ being permutations of each other, $\mathbb C^n/\sim$ its corresponding quotient space with the quotient topology, and $\pi:\mathbb C^n\to\mathbb C^n/\sim,v\mapsto[v]_\sim$ its projection. This is useful because it gives us essentially a topology on the set $\mathbb C^n/\sim$ of unordered $n$-tupels of complex numbers, allowing us not to care about how values should be ordered for now.
Then, as David already noted, the mapping $\chi:\mathbb C^{n\times n}\to\mathbb C_n[t], A\mapsto p_A$ from matrices to their characteristic polynomials is continuous; and, as proven here (which is also where I burrowed much of this notation from), there exists a continuous function (actually even an homeomorphism) $\tau:\mathbb C_n[t]\to\mathbb C^n/\sim$ mapping those polynomials to their zeroes as those unordered tupels. Simply chaining those two mappings gives us a continuous function $\tau\circ\chi:\mathbb C^{n\times n}\to\mathbb C^n/\sim$ mapping matrices to their eigenvalues as unordered tupels.
Now the question is, can we use this to construct such a mapping to $\mathbb C^n$ too, instead of $\mathbb C^n/\sim$? Sadly, the answer is no - the projection $\pi:\mathbb C^n\to\mathbb C^n/\sim$ does not have a continuous right inverse $\pi^{-1}$, so we can't just write our wanted function $\varphi:\mathbb C^{n\times n}\to\mathbb C^n$ as $\pi^{-1}\circ\tau\circ\chi$. More generally, we can prove that for all $n\geq2$ no continuous function $\psi:\mathbb C_n[t]\to\mathbb C^n$ mapping polynomials to their zeroes can exist, so any attempt to construct our function $\varphi$ as a chain of continuous functions starting with $\chi$ (including David's one) will be futile.
To show this, let's assume we have a function $\psi:\mathbb C_2[t]\to\mathbb C^2$ mapping polynomials to their zeroes. Then, consider the following path $\gamma$ in $\mathbb C_2[t]$: $$\gamma:[0,1]\to\mathbb C_2[t],s\to(t-e^{s\pi i})(t+e^{s\pi i}).$$
The zeroes of $\gamma(0)$ are $-1$ and $1$, so $\psi(\gamma(0))$ would have to be either $(-1,1)$ or $(1,-1)$. Choosing one of those two, gradually increasing the parameter $s$ and using the assumption that $\psi$ is continuous along the way, we eventually end up at $\psi(\gamma(1))$ with the opposite of what we started with; however, since $\gamma(0)$ and $\gamma(1)$ are equal, $\psi(\gamma(0))$ and $\psi(\gamma(1))$ should be equal too. This is a contradiction, so our assumption that such a function $\psi$ exists must have been wrong.
To recap:

*

*a continuous function $\tau\circ\chi:\mathbb C^{n\times n}\to\mathbb C^n/\sim$ mapping matrices to their eigenvalues as unordered tupels does indeed exists, as proven in the paper linked above

*for $n\geq 2$, no continuous function $\psi:\mathbb C_n[t]\to\mathbb C^n$ mapping normed polynomials to their zeroes as ordered tupels can exists

*thus, if a continuous function $\varphi:\mathbb C^{n\times n}\to\mathbb C^n$ mapping matrices to their eigenvalues as ordered tupels exists, it can't be of the form $\psi\circ\chi$ with $\chi:\mathbb C^{n\times n}\to\mathbb C_n[t]$ mapping matrices to their characteristic polynomial

*this does not prove that such a function $\varphi$ can't exist; however, it shows that it would have to map matrices with the same eigenvalues to different permutations of those.

My personal suspicion is that there exist exactly $n!$ such functions $\varphi$, all just permutations of the same component functions but otherwise no different, and that exactly one of those maps all diagonal matrices to their diagonal entries in the correct order. However, I have not found any source / proof for this yet (which is why I was looking for this question in the first place), so a third answer detailing more of that would be greatly appreciated by me too.
