# Diagonalizable

### From Conservapedia

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 *A* is conjugate to a matrix *B* in Jordan canonical form. One can then perturb the diagonal elements *b*_{ii} of *B* by arbitrarily small numbers ε_{i} so that the diagonal elements *b*_{ii} + ε_{i}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 *A*. But since conjugation is a length-preserving operation on the inner product space of complex matrices, this shows that *A* is arbitrarily close to a diagonalizable matrix. This completes the proof.