Diagonalizable

An operator T on a finite-dimensional vector space V is diagonalizable if V has a basis of eigenvectors for T.

Denseness of diagonalizable operators

The space of complex linear operators on <math>\mathbb{C}^n</math> may be identified with the vector space <math>M_n(\mathbb{C})</math> of nxn matrices with complex coefficients. As such, it inherits a natural structure as a topological space.

Given this topology, the set of diagonalizable functions is a dense subset of <math>M_n(\mathbb{C}^n)</math>.

We can prove this as follows: Every complex matrix <math>A</math> is conjugate to a matrix <math>B</math> in Jordan canonical form. One can then perturb the diagonal elements <math>b_{ii}</math> of <math>B</math> by arbitrarily small numbers <math>\epsilon_i</math> so that the diagonal elements <math>b_{ii}+\epsilon_i</math>of the perturbed matrix are distinct. But this implies that the perturbed matrix is diagonalizable. Thus, we can find a diagonalizable matrix arbitrarily close to a conjugate of <math>A</math>. But since conjugation is a length-preserving operation on the inner product space of complex matrices, this shows that <math>A</math> is arbitrarily close to a diagonalizable matrix. This completes the proof.