From the definition of the normal matrix using the property $A^\dagger A=AA^\dagger$, where $A^\dagger$ is the conjugate transpose of matrix A, it is obvious that every hermitian and skew-hermitian matrix is also normal, since (skew-)hermitian matrices satisfy $A^\dagger=(-)A$. What I'm wondering is: is there a theorem which explicitly states why normal matrices are not necessarily hermitian, i.e. why doesn't it go the other way around? I'm aware of some counterexamples which prove that some normal matrices are not hermitian, and I don't question that they aren't the same, I just want to know why that's so in a general case.
Also, if for some regular matrices $A,B \in \mathcal{M}_{n\times n}$ we have $(AB)^\dagger=B^\dagger A^\dagger$, does that mean that for a normal matrix $A$ the product $C=A^\dagger A$ is neccesarily hermitian, since $C^\dagger=(A^\dagger A)^\dagger=A^\dagger A^{\dagger \dagger}=A^\dagger A=C$? Is there a way to relate that to the above question?
Best Answer
The basic example of a normal matrix is any diagonal matrix $A$; then $A^\dagger$ is the diagonal matrix with complex conjugate entries, which commutes with $A$ since any two diagonal matrices commute. Such an $A$ is Hermitian iff all the diagonal entries are real and skew-Hermitian iff all the diagonal entries are purely imaginary. This example demonstrates how Hermitian matrices are very unusual and special among the normal matrices, much as real numbers are very unusual and special among the complex numbers.
(In fact, the spectral theorem says that this is essentially the only example: a matrix is normal iff it is conjugate by a unitary to a diagonal matrix. Similarly, a matrix is Hermitian iff it is conjugate by a unitary to a diagonal matrix with real entries.)