Conceptual question on Schur Theorem According to my textbook, Schur Theorem is stated as follows:
Let T be a linear operator on a finite dimensional inner product space V. Suppose that the characteristic polynomial of T splits. Then there exists an orthonormal basis $\beta$ for V such that the matrix $[T]_\beta$ is upper triangular.
Then it goes on to say that if such an orthonormal basis $\beta$ exists, then $[T]_\beta$ is a diagonal matrix.
My question is if that is the case why not the theorem just says that the matrix $[T]_\beta$ is diagonal? Why does it say upper triangular instead? What am I misunderstanding here? 

You are confusing Schur's theorem with diagonalization. Schur's theorem says if there are $n$ eigenvalues (a result of c.h.p. being split), then it is possible to find an orthonormal basis $\beta$ for $V$ s.t. $[T]_\beta$ is upper triangular. See for example here-Thm 3.7 for a neat proof. If $[T]_\beta$ is diagonal, then $\beta$ are necessarily eigenvectors of $T$ (check it yourself). Schur's theorem never says $\beta$ is the eigenvectors of $T$ (but it could be). So $[T]_\beta$ is not necessarily diagonal. 
Certain additional conditions can ensure $\beta$ to be the eigenvectors of $T$, in which case $[T]]_\beta$ is diagonal. For example, if:


*

*$T$ is a symmetric (hermitian) matrix

*$T$ is a skew-symmetric (skew-hermitian) matrix

*$T$ is a normal matrix.
(You may want to check Peter Lax's Linear Algebra)
It is possible that $T$ is diagonal in the general case, for example all eigenvalues are distinct (leaving no possibility to have Jordan blocks). However, the diagonalization should be done by a similarity transformation $M^{-1}TM$ instead of a congruence transformation $M^*TM$ in your case.
