Eigenvectors of a normal matrix According to the spectral theorem every normal matrix can be represented as the product of a unitary matrix $\mathbb{U}$ and a diagonal matrix $\mathbb{D}$
 $$\mathbb{A} = \mathbb{U}^H\mathbb{D}\mathbb{U}$$
meaning that every normal matrix is diagonalizable.
Does it necessarily mean that the unitary matrix has to be composed from the eigenvectors of $\mathbb{A}$ ? 
I presume that not, because then the eigenvectors of every normal matrix would form an orthonormal set (rows and columns of a unitary matrix are orthonormal in $\mathbb{C}^n$). So am I right that only the set of eigenvectors of a hermitian (or symmetric while in $\mathbb{R}^n$) matrix is orthonormal?
 A: An $n\times n$ matrix $A$ over the field $\mathbf{F}$ is diagonalizable if and only there is a basis of $\mathbf{F}^n$ of eigenvectors of $A$. This occurs if and only if there exists an invertible $n\times n$ matrix $Q$ such that $Q^{-1}AQ$ is a diagonal matrix; the column of $Q$ form the basis made up of eigenvectors, and conversely, if you take a basis made up of eigenvectors and arrange them as columns of a matrix, then the matrix is invertible and conjugating $A$ by that matrix will yield a diagonal matrix.
An $n\times n$ matrix $A$ with coefficients in $\mathbb{R}$ or $\mathbb{C}$ is orthogonally diagonalizable if and only if there is an orthonormal basis of eigenvectors of $A$ for $\mathbb{R}^n$ (or $\mathbb{C}^n$, respectively).
An $n\times n$ matrix with coefficients in $\mathbb{C}$ is orthogonally diagonalizable over $\mathbb{C}$ if and only if it is normal; a square matrix is orthogonally diagonalizable over $\mathbb{R}$ if and only if it is Hermitian. "Unitary" is usually reserved for complex matrices, with "orthogonal" being the corresponding term for real matrices. 
A: Yes, the columns of $U^\dagger$ are always eigenvectors of $A$ whenever $AA^\dagger = A^\dagger A$ and $A=U^\dagger D U$ with a diagonal $D$. It's not hard to see why. Take $v_n$ to be the $n$-th column of $U^\dagger$. Then 
$$ U v_n = (UU^\dagger)_{n{\rm -th\,\,column}} = 1_{n{\rm -th\,\,column}}$$
so 
$$ A v_n = U^\dagger D U^\dagger v_n = U^\dagger D\cdot 1_ {n{\rm -th\,\,column}} = D_{nn} U^\dagger \cdot 1_{n{\rm -th\,\,column}} = D_{nn} U_{n{\rm -th\,\,column}} = D_{nn} v_n $$
Looking at the first and last expression of this long equation, we see that $v_n$ is an eigenstate of $A$ with eigenvalue $D_{nn}$ - a diagonal element of the matrix $D$. I have only used that $U$ is unitary; indeed, this is equivalent to the basis' being orthonormal. 
More precisely, one may always find an orthogonal basis of eigenvectors of any normal matrix $A$ (the only extra work is that we must make orthogonal every higher-dimensional space corresponding to the same eigenvalue) and by normalizing the vectors, we get an orthonormal basis of eigenvectors of $A$. The basis vectors  are the same thing as columns of $U^\dagger$.
A: You can always find $n$ orthonormal eigenvectors for a normal $n\times n$ matrix $A$. Indeed, since you know $A=U^HDU$, if $e_1,\dots,e_n$ is the standard orthonormal basis of $\mathbb{C}^n$ then these vectors are eigenvectors of the diagonal marix $D$, hence $U^He_1,\dots,U_He_n$ (the columns of $U^H$) are orthonormal eigenvectors of $A$.
A: Here is what I think is correct:
Normal matrices are matrices that have orthogonal eigenvectors.
Hermitian matrices are normal matrices that have real eigenvalues.
So this answers your first question in positive: Yes, the unitary matrix in your decomposition has the same eigenvectors as your original matrix.
