A normal operator $T$ in some inner product space usually has spectral decomposition $T=\sum_j\lambda_j P_j$ where $P_j$ are orthogonal projections and $\lambda_j$ are usually distinct values often converging to 0.
What happens if $\lambda_1=\lambda_2\dots$ (all eigenvalues are fixed constants)? is $T$ still a normal operator? Are this kind of operators defined as a specific class? One can also take the case when the spectral decomposition of $T$ equals to the resolution of the identity.
Not every normal operator admits an orthonormal basis of eigenvectors; in fact, normal (even selfadjoint and positive) operators exist that have no eigenvalues whatsoever, the easiest example being the multiplication operator $M_x\in B(L^2[0,1])$ given by $(M_xf)(t)=tf(t)$. A linear combination of pairwise orthogonal projections is normal, and a limit of normals is normal. Which means that any operator of the form $\sum_j\lambda_jP_j$ is normal. If all $\lambda_j$ are equal then you get that $T$ is a scalar multiple of the identity, or a scalar multiple of a projection is $0$ is an eigenvalue.