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 diagonalizeble 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 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 diagonalizeable 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.
