If A unitary matrix and orthogonally diagonalizable why there is a basis in whichthe linear trans. matrix is diagonal? If $A$ is a  $n\times n$ unitary matrix (above the complex field) and is orthogonally diagonalizable, why does it mean that the is an orthonormal basis $\mathbb C$ in which the matrix that represent the linear transformation of $A$ is diagonal? Thank a lot!
 A: The way you've asked the question suggests you just need to think about the definitions of orthogonal and unitary diagonalisation, so I'm not quite sure exactly what you are asking. We usually talk about orthogonal diagonalisation in $\Bbb R^n$ and unitary diagonalisation in $\Bbb C^n$.
Firstly, look at the definitions of a unitary matrix and an orthogonal matrix. 
The inverse of a unitary matrix $A$ is the conjugate transpose, $A^*$. This means it is normal, i.e. $AA^*= A^*A$. (This is important because a matrix with complex entries is unitarily diagonalisable iff it is normal.)
The inverse of an orthogonal matrix is its transpose. So it is the analogue in $\Bbb R^n$ of a unitary matrix.
(A hermitian matrix in $\Bbb C^n$ is the analogue of a symmetric matrix in $\Bbb R^n$, both of these have real eigenvalues.)
$A$ is unitary iff its rows (columns) form an orthonormal basis of $\Bbb C^n$ wrt the complex Euclidean inner product. The eigenvectors corresponding to distinct eigenvalues of a unitary matrix are orthogonal (but not necessarily orthonormal, you have to normalise them).
If $A$ is unitarily diagonalisable, then there exists a diagonal matrix $D$ and an orthogonal matrix $P$ such that $A = PDP^{-1} = PDP^*$. The diagonal entries of $D$ are the eigenvalues of $A$, and the columns of $P$ are the corresponding unit eigenvectors of $A$. 
Note that if you have a repeated eigenvalue, its eigenvectors may not be orthogonal, so you will need Gram-Schmidt to make them orthogonal.
The columns of $P$ form the orthonormal basis in which $D$ acts. So, instead of doing expensive calculations in the standard basis using $A$, we can do cheap calculations in the new basis using $D$. $P$ is the change of basis matrix. 
