Diagonalizable unitarily Schur factorization Let $A$ be $n x n$ matrix.
What exactly is the difference between unitarily diagonalizable and diagonalizable
matrix $A$? Can that be that it is diagonalizable but not unitarily diagonalizable? 
What are the conditions for Schur factorization to exist? For a (unitarily) diagonalizable matrix is it necessary that Schur factorization exists and vice versa?
Thanks a lot!

Diagonalization means to decompose a square matrix $A$ into the form $PDP^{-1}$, where $P$ is invertible and $D$ is a diagonal matrix. If $P$ is chosen as a unitary matrix, the aforementioned decomposition is called a unitary diagonalization. It follows that every unitarily diagonalizable matrix is diagonalizable.
The converse, however, is not true in general. Note that if $A$ is unitary diagonalizable as $UDU^\ast$ ($U^\ast=U^{-1}$), then the columns of $U$ are the eigenvectors of $A$ and they are orthonormal because $U$ is unitary. Therefore, a diagonalizable matrix is unitarily diagonalizable if and only if it has an orthonormal eigenbasis. So, a matrix like
$$
A=\begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}2&0\\0&1\end{pmatrix}
\begin{pmatrix}1&1\\0&1\end{pmatrix}^{-1}
$$
is diagonalizable but not unitarily diagonalizable.
As for Schur triangulation, it is a decomposition of the form $A=UTU^\ast$, where $U$ is unitary and $T$ is triangular. Every square complex matrix has a Schur triangulation. Note that if $A=UTU^\ast$ is a Schur triangulation, then you can read off the eigenvalues of $A$ from the diagonal entries of $T$. It follows that a real matrix that has non-real eigenvalues cannot possess a Schur triangulation over the real field. For example, the eigenvalues of
$$
A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}
$$
are $\pm\sqrt{-1}$. So it is impossible to triangulate $A$ as $UTU^\ast$ using real unitary matrix (i.e. real orthogonal matrix) $U$ and real triangular matrix $T$. Yet there exist complex $U$ and complex $T$ such that $A=UTU^\ast$.
If a matrix $A$ is unitarily diagonalizable as $A=UDU^\ast$, the diagonalization is automatically a Schur triangulation because every diagonal matrix $D$ is by itself a triangular matrix. Yet the converse is not true in general. For example, every nontrivial Jordan block, such as
$$
A=\begin{pmatrix}0&1\\0&0\end{pmatrix},
$$
is a triangular matrix that is not unitarily diagonalizable.
