# Norm closure of diagonalizable operator on Hilbert space

Problem: Prove the norm closure of diagonalizable operator in $$\mathcal{B(H)}$$ (bounded operators on Hilbert space which is not necessarily finite dimensional) is the set of normal operators.

A bounded operator $$\mathbb{T}$$ is said to be Diagonalizable if there exists an orthonormal basis consisting of eigenvectors of $$\mathbb{T}$$.

What I am thinking: Since I already known that a normal operator can be approximated by a linear combination of pairwise commuting projections, that is, for $$\mathbb{T}$$ is bounded normal operator, then there exists a sequence of projections $$\{P_n\}$$ summing to identity operator $${I}$$ and $$\{\lambda_n\}$$ such that $$|| \mathbb{T} - \sum_{n}\lambda_nP_n||<\epsilon$$.

And we know that for diagonalizable operator $$\mathbb{T}$$ in general Hilbert space $$\mathcal{H}$$, we have another equivalent definition: There exists an orthonormal basis $$\{e_i\}_{i\in J}$$ of $$\mathcal{H}$$ and values $$\{\lambda_i\}_{i \in J}$$ such that $$\mathbb{T}(x) = \sum_{i \in J}\lambda_i e_i$$. So my idea is to express a diagonalizable operator as a linear combination of projections then combined with the fact that a normal operator can be approximate by a linear combination of projections to finish the proof.

I am not sure whether expressing a diagonalizable operator as linear combination of projections is correct in general Hilbert space. In finite dimension, which should be true. Any idea will be appreciated.

Yes, that's how it works. You need to check that a diagonalizable operator $$T=\sum_j\lambda_jP_j$$ is normal, which is easy to do.