Normal block upper triangular matrix proof

Question

Prove that if a block upper triangular matrix is normal then its off-diagonal blocks is zero and each of its diagonal blocks is normal. This question was asked before, but it got just one answer which contains a mistake(it does not take into consideration the order of multiplication of blocks). A complex square matrix $A$ is normal if it commutes with its conjugate transpose $A^*$( $A$ is normal $<=>$ $A^*A=AA^*$). So let’s say the matrix M is normal and it is block upper triangular, so it looks like this: $$M = \begin{pmatrix} A & B \\ 0 & C \end{pmatrix} $$ $$M^* = \begin{pmatrix} A^* & 0 \\ B^* & C^* \end{pmatrix} $$ We know that $A$ and $C$ are square, and we want to prove that $A$ and $C$ are normal, $B=0$. If $M$ is normal, doing the block-computations gives us four following equations 1) $AA^* + BB^* = A^*A$, 2) $A^*B=BA^*$, 3) $B^*A=CB^*$, 4) $C^*C+B^*B=CC^*$.

This is as far as I was able to get, have no idea how we can prove that $B=0$ from this. For context, I have completed a year long course of linear algebra, so I would say I know all the basic characteristics of normal matrices, such as spectral theorem and also some facts about Gram matrix of columns of matrix M(that it is a Hermitian matrix for example) and so on.

Answer 1

polynomial proof:
observe that for $k \in \mathbb N$
$M^k = \begin{pmatrix} A^k & * \\ 0 & C^k \end{pmatrix}$ and
$M=UDU^*$, $M^*=UD^*U^*$ for some unitary $U$
This means we can find [e.g. via a Vandermonde matrix or Lagrange interpolation] a polynomial $p$ such that
$p\big(D\big)=D^*\implies p\big(M\big)= Up\big(D\big)U^* =M^*$
Thus while $M^*$ is block lower triangular, we also know $p\big(M\big)=M^*$ is block upper triangular
$\implies M^*$ is block diagonal, i.e. $B=0$

more conceptual proof:
let $A$ be $r\times r$. What $M = \begin{pmatrix} A & B \\ 0 & C \end{pmatrix}$ says is that the first $r$ standard basis vectors generate $W$ an $M$-invariant subspace, i.e. $MW\subseteq W$. Since $M$ is diagonalizable $W$ has a basis given by the eigenvectors of $M$ and these are also eigenvectors of $M^*$ since $M$ is normal. Thus $M^*W\subseteq W$ i.e. $M^*=\begin{pmatrix} A^* & 0 \\ 0 & C^* \end{pmatrix}$

The key result here is that for a diagonalizable $M\in \mathbb F^{n\times n}$, then any $M$ invariant subspace $W$ has a basis given by eigenvectors of $M$. Proof:
Let $\dim W=r$ and create a basis for $W$. Then using the basis extension algorithm, append eigenvectors of $M$ until we have a basis for $\mathbb F^n$. Let those $n-r$ eigenvectors generate the image of idempotent (/projection) matrix $P$ and $\ker P =W$.
confirm that $w \in \ker P\implies PMw = MPw$ and $v \in \text{image } P \implies PM v= MPv$. Thus $PM=MP$ and both are diagonalizable so they are simultanesouly diagonalizable and in particular $W=\ker P$ has a basis given by the eigenvectors of $A$.

Answer 2

I think the following may be an argument which is roughly of the kind the OP requests: suppose that $T$ is a normal operator on $\mathbb C^n$ which has matrix $$ \left(\begin{array}{cc} A & B\\ 0 & D\end{array} \right) $$ then we claim that $B=0$ and $A$ and $D$ are normal.

Now the condition that $T^*T = TT^*$ yields the following equations in $A,B$ and $D$. $$ \begin{split} AA^*+BB^*&= A^*A, \\ BD^* &= A^*B\\ DB^* &= B^*A\\ DD^* &= B^*B+D^*D \end{split} $$ Now $\text{Mat}_n(\mathbb C)$ carries an inner product given by $\langle A,B \rangle = \text{tr}(AB^*) = \text{tr}(B^*A)$. Indeed if $A = (a_{ij})$ and $B = (b_{ij})$ then $\text{tr}(AB^*) = \sum_{i=1}^n (AB^*)_{ii} = \sum_{i,j=1}^n a_{ij}\bar{b}_{ij}$, thus the inner product is just what one obtains by identifying $\text{Mat}_n(\mathbb C)$ with $\mathbb C^{n^2}$.

But now the first of the above 4 equations implies that $\text{tr}(AA^*)+ \text{tr}(BB^*) = \text{tr}(A^*A)$. Since $\text{tr}(AA^*) = \text{tr}(A^*A)$, it follows that $\text{tr}(BB^*)=\langle B,B \rangle = 0$, and hence by definiteness, $B=0$. This immediately implies that $AA^*=A^*A$ and $DD^*= D^*D$, that is, $A$ and $D$ are normal matrices.