# Notion of a normal operator

I understand that a normal operator is an operator such that

$$AA^\dagger = A^\dagger A$$ where $\dagger$ is the conjugate transpose.

However, what is the most intuitive way to "characterise" this? For example,

1. $SO(3)$ is the group of rotations in $\mathbb R^3$
2. A unitary matrix is one that represents an isometry
3. A hermitian matrix is a generalization of symmetric, and is the "nicest" (diagonalizable, real eigenvalues, etc) of all matrices over $\mathbb C$

I was hoping for some sort of intuitive explanation of why I would care about normal matrices over $\mathbb C$ (maybe other reasons than the spectral theorem? Something more fundamental / geometric perhaps)

• Normal operators are the ones which can be diagonalized via an unitary. In this case they behave like finite sequences of numbers (namely, the diagonal), and they have very well-behaved analytical properties. Aug 10, 2016 at 16:02
• @LuizCordeiro - so the reason I would choose to care is because there's a unitary change of basis that makes them diagonalizable? Aug 10, 2016 at 16:07
• Yes. Moreover, since an unitary change of basis preserves the inner product, the geometry of the vector space is also unchanged, which can be quite useful. Aug 10, 2016 at 16:10

The term "normal" here refers to orthogonality. Geometrically, a normal operator on $\mathbb C^n$ represents scaling by possibly different (complex) factors along different axes of an orthogonal coordinate frame. This, I think, is more intuitive than a Hermitian operator (which, of course, is also normal).
A square complex matrix $A$ is normal iff it has an orthonormal basis of eigenvectors. That's why you would care about it. Unitary and selfadjoint matrices are special cases. Normal is the most general case.