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## Properties of distinct eigenvalues operators

Let $E$ be a complex $n$-dimensional vector space. Denote by $L(E,E)$ the set of linear and bounded operators $A:E \to E$. Prove that the set $\{A \in L(E,E) : A\text{ has }n\text{ distinct eigenvalues}\}$ is open and dense in $L(E,E)$.
Source: Marsden, Ratiu, Manifolds, Tensor Analysis

Proof: Let $p$ be the characteristic polynomial of $A$, and $\mu_1,...,\mu_{n-1}$ be the roots of $p^\prime$. Then $A$ has multiple eigenvalues if and only if $p(\mu_1)...p(\mu_{n-1})=0$. The last expression is a symmetric polynomial in $\mu_1,...,\mu_{n-1}$. We know that any symmetric polynomial can be expressed as a polynomial of the elementary symmetric polynomials (for example see this link). Therefore, $q=p(\mu_1)...p(\mu_{n-1})$ can be regarded as a polynomial in the coefficients of $p^\prime$ (since $\mu_i$ are the roots of this polynomial and its coefficients are exactly the elementary symmetric polynomials in variables $\mu_i$), coefficients which can be expressed as polynomials of entries of the matrix of the operator $A$ in a given basis. Therefore $q((a_{ij}))=0$ is equivalent to $A=(a_{ij})$ has multiple eigenvalues. This means that the set of matrices which have multiple eigenvalues is $q^{-1}(0)$. Its complement is closed, and therefore the set of operators with distinct eigenvalues is open in $L(E,E)$.
On the other hand, suppose $q^{-1}(0)$ vanishes on an open set. Because $q$ is a polynomial, all its partial derivatives vanish on that open set, and therefore, all its coefficients are $0$, which is not the case. Therefore, $q_{-1}(0)$ does not contain any open set. This means that the closure of its complement is the whole space $L(E,E)$.

The ideas presented here might be useful. Another proof can be seen in the following article: density.  (original source)