# 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 $\mathbb{C}^n$ may be identified with the vector space $M_n(\mathbb{C})$ 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 $M_n(\mathbb{C}^n)$.

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 bii of B by arbitrarily small numbers εi so that the diagonal elements bii + εiof 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.