# The existence of continuous mapping

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 $\mathbb{C}$.

My question is : Does there exist a continuous function $f$ from $M_n(\mathbb{C})$ to $\mathbb{C}^n$ such that $f(A)=(\lambda_1,\lambda_2,...,\lambda_n)$ ? where all $\lambda_i$'s are eigenvalues of a matrix $A$.

I don't know how to proceed. If there exists such function, then how to prove its continuity ?

• how to order these $\lambda_i$'s? – user99914 Jan 4 '18 at 14:35
• @JohnMa Perhaps we can arrange $\lambda_i$'s into ascending order according to its absolute value, i.e., if $|\lambda_1|\leq|\lambda_2|\leq|\lambda_3|\leq ... \leq|\lambda_n|$ , then $f(A)=(\lambda_1,\lambda_2,...,\lambda_n)$ – UDAY PATEL Jan 4 '18 at 14:43
• But it is still not clear how to order, e.g. $\sqrt{-1}$ and $-\sqrt{-1}$. – user99914 Jan 4 '18 at 14:46
• @JohnMa Okay, I got the point. How about using dictionary order as order between $\lambda_i$'s ? – UDAY PATEL Jan 4 '18 at 14:49

$\newcommand{\F}{\mathscr F}$

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 $j$.