Is the set of matrices with eigenvalues in an open set itself an open set? Let $U \subseteq \mathbb{C}$ be open. Is the set $\mathcal{U} := \{X \in \mathbb{M}_n(\mathbb{C}) : \sigma(X) \subseteq U\}$ open in (the Euclidean topology on) $\mathbb{M}_n(\mathbb{C})$?
 A: We start with a fixed $M$ with eigenvalues $a_1,\ldots,a_n$ such that the open balls $B(a_1,\epsilon),\ldots,B(a_n,\epsilon)$ are all contained in $U$.
Is it possible to find $\delta$ such that all matrices of distance less than $\delta$ from $M$ have eigenvalues in the union of the open balls $B(a_1,\epsilon)\cup\cdots\cup B(a_n,\epsilon)$?
Yes. Here's how.
Each coefficient of the characteristic polynomials of a matrix $T$ is a polynomial of the entries of $T$, and is therefore a continuous function of $T$. Remember that the characteristic polynomial is monic, i.e. has leading coefficient $1$.
We now just need to prove a result about the roots of polynomials.
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Fix $p(z)=(z-c_1)\cdots(z-c_n)=z^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0$.
For any $\epsilon>0$, there exists $\delta'>0$ such that the following holds:
whenever $|a_i-w_i|<\delta'$ for all $0\leq i\leq n-1$,
all the roots of the polynomial $z^n+w_{n-1}z^{n-1}+\cdots+w_1z+w_0$ are within a distance of $\epsilon$ from $\{c_1,\ldots,c_n\}$.
Proof:
For notational convenience, regard $w$ as a vector $(w_{n-1},\ldots,w_1,w_0)\in\mathbb{C}^n$.
Also, use the norm $||w||=\max\{|w_{n-1}|,\ldots,|w_1|,|w_0|\}$.
That way, we can write $q_w(z)=z^n+w_{n-1}z^{n-1}+\cdots+w_1z+w_0$, and the condition on the coefficients can be written as $||w-a||<\delta'$.
Consider $f(x)=x^n-(|a_{n-1}|+1)x^{n-1}-\cdots-(|a_1|+1)x-(|a_0|+1)$.
Let $r$ be either $0$ if $f(x)$ has no positive real roots, or else equal to the largest positive real root of $f(x)$.
Then for all $z\in\mathbb{C}$ with $|z|>r$, you find that $||w-a||<1$ implies $ |p_w(z)|>0$.
Therefore, we may as well only look for roots of $p_w(z)$ in $D=\overline{B(0,r)}-[B(a_1,\epsilon)\cup\cdots\cup B(a_n,\epsilon)]$.
Notice that $D$ is compact. Therefore, ${|f(D)|}$ has a minimum, which we can define to be $d$. Since $f(D)$ does not attain $0$, we know that $d>0$.
The task is now to choose $\delta'$ in such a way that:
$|q_w(z)-q_a(z)|<d$ for all $z\in \overline{B(0,r)}$, whenever $||w-a||<\delta'$.
This would ensure that for all $z\in D$, you have $|q_w(z)|=|q_a(z)+q_w(z)-q_a(z)|\geq|q_a(z)|-|q_w(z)-q_a(z)|>d-d=0$.
Here's one way to choose $\delta'$. We would ideally want $\delta' r^{n-1}+\cdots+\delta' r+\delta'=\delta'(r^{n-1}+\cdots+r+1)$ to be smaller than $d$. So choose $\delta'=\dfrac{d}{2(r^{n-1}+\cdots+r+1)}$.
A: Yes, it is an open set.
We know determinant is a continuous function on Mn(C). Det of all such matrix belongs to the set (-1) ^(n)U.U.... U(n times product). And U. U.... U(n times product) is an open set in C. So, det^(-1) (U.U.... U(n times product)) is an open set in Mn(C) . DetA is equal to (-1) ^n  times product of eigenvalues. So that's why det Of such matrix belong to the set (-1) ^nU.U.U.... U( n times product)
