Lower triangular block matrix is unitarily diagonalizable, then blocks on diagonal is unitarily diagonalizable Suppose $M=\bigl( \begin{smallmatrix} A & 0 \\ C & B \end{smallmatrix} \bigr)\in\mathbb{C}^{n\times n}$ is unitarily diagonalizable, that means there exists an unitary matrix $U$ s.t. $U^*MU$ is a diagonal matrix. Show $A,B$ are unitarily diagonalizable too.
My method is as following:
Since a matrix $T$ is unitarily diagonalizable iff $T$ is normal, that means $T^*T=TT^*$. I try to prove $AA^*=A^*A$ and $BB^*=B^*B$. Since we know $M^*=\bigl( \begin{smallmatrix} A^* & C^* \\ 0 & B^* \end{smallmatrix} \bigr)$. Then $M^*M=MM^*$ implies $A^*A+C^*C=AA^*,C*B=AC^*,B^*C=CA^*,B^*B=CC^*+BB^*.$ It seems we cannot deduce $AA^*=A^*A$ and $BB^*=B^*B$ in this way.
 A: Use trace to prove $C=0$ and you are close. Since $A^*A+C^*C=AA^*$
$$
Tr(A^*A+C^*C)=Tr(A^*A)+Tr(C^*C)=Tr(AA^*)
$$
Since $Tr(A^*A)=Tr(AA^*)$
$$
Tr(C^*C)=0\quad\text{ and thus }\quad C=0
$$
So $A^*A=AA^*$ and $A$ is normal. Likewise $B$ is normal too. 
A: You can do this without using the characterisation of unitarily diagonalisable matrices as those that commute with their conjugate transpose, using a more geometric argument. The question splits into two parts: first showing that $A$ and $B$ are diagonalisable, and then that if so, there is an orthonormal basis of eigenvectors (because in that case the change of basis matrix will be unitary). The latter will be possible if and only if the eigenspaces for distinct eigenvalues are orthogonal to each other (in which case it suffices to choose orthonormal bases in each eigenspace). Since it is given that the eigenspaces of $M$ are orthogonal to each other, this will follow automatically for the eigenspaces of $A$ and $B$, so the latter part is easy, and one can focus on the former part. (To be honest, this "follows automatically" is easier to see for $B$, whose eigenspaces are isometrically embedded into the eigenspaces of $M$, than it is for $A$ where one needs to consider dual eigenspaces. But it is easy to see that "unitarily diagonalisable" is invariant under (conjugate) transposition, so if it works for $B$, it should work for $A$ as well.)
What is left is to show that if a block triangular matrix$~M$ is diagonalisable, then so are its diagonal blocks. This can be proved by forming the polynomial $P=\prod_\lambda(X-\lambda)$ where $\lambda$ runs over all distinct eigenvalues of$~M$. By construction $P$ is split with simple roots, and it annihilates$~M$. But then it also annihilates the diagonal blocks of $M$, and any square matrix annihilated by a polynomial that is split with simple roots is diagonalisable. This part is related to this question, among whose answers you will find alternative arguments.
