# 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 may be identified with the vector space 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 .

We can prove this as follows: Every complex matrix is conjugate to a matrix in Jordan canonical form. One can then perturb the diagonal elements of by arbitrarily small numbers so that the diagonal elements 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 . But since conjugation is a length-preserving operation on the inner product space of complex matrices, this shows that is arbitrarily close to a diagonalizable matrix. This completes the proof.