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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)

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