A matrix A is said to be normal if $AA^H=A^HA$.
Theorem of normal matrix is: A matrix A is normal if and only if A possesses a complete orthonormal set of eigenvectors.
I don't know how to understand "A possesses a complete orthonormal set of eigenvectors". Is this the same as "A can be diagonalizable"? Because A can be diagonalizable if and only if we can find enough eigenvectors. Here enough means complete. i.e., as long as we exclude those matrices which have the shortage of eigenvectors when geometric multiplicity is below algebraic multiplicity, then A must possesses a complete orthonormal set of eigenvectors? Do I think correct? I don't know.
No; it's stronger than asserting that $A$ is diagonalizable. Asserting that $A$ is diagonalizable simply means that it has a complete set of eigenvectors. For instance, the matrix $A=\left[\begin{smallmatrix}1&1\\0&2\end{smallmatrix}\right]$ is diagonalizable: $\{(1,0),(1,1)\}$ is a complete set of eigenvectors. But it is not an orthogonal set and, in fact, $A$ has no complete orthogonal set of eigenvectors.