TITLE: proofreading for Proof normal matrix is diagonalizable or is there any shorter proof?
QUESTION [3 upvotes]: $$\text{Prove normal matrix is diagonalizable over }\mathbb{C}$$
$$\textbf{If }\textbf{A}^\ast\textbf{A}=\textbf{A}\textbf{A}^\ast$$
$$\textbf{Then }\textbf{A}=\textbf{P}^\ast\Lambda\textbf{P}$$
$$\text{Proof}$$
$$\textbf{A}\text{ is normal matrix}$$
$$\Rightarrow\textbf{A} = \textbf{P}^\ast\Delta\textbf{P}\tag{1}$$
$$\Delta\text{ is upper triangle matrix }\textbf{(Schur decomposition)}$$ 
$$\textbf{P}\text{ is unitary matrix}$$
$$\text{First we need to show if }\textbf{A}\text{ is }\textbf{normal}\text{ then }\Delta\text{ is }\textbf{normal}$$
$$\textbf{A} = \textbf{P}^\ast\Delta\textbf{P}\tag{2}$$
$$\Rightarrow\textbf{A}^\ast = (\textbf{P}^\ast\Delta\textbf{P})^\ast$$
$$\Rightarrow\textbf{A}^\ast = \textbf{P}^\ast(\textbf{P}^\ast\Delta)^\ast$$
$$\Rightarrow\textbf{A}^\ast = \textbf{P}^\ast(\Delta^\ast\textbf{P})\tag{3}$$
$$\text{from (2) and (3)}$$
$$\Rightarrow\textbf{A}\textbf{A}^\ast = (\textbf{P}^\ast\Delta\textbf{P})(\textbf{P}^\ast\Delta^\ast\textbf{P})$$
$$\Rightarrow\textbf{A}\textbf{A}^\ast = \textbf{P}^\ast\Delta(\textbf{P}\textbf{P}^\ast)\Delta^\ast\textbf{P}$$
$$\textbf{P} \text{ is unitary matrix}$$
$$\Rightarrow\textbf{P}^\ast = \textbf{P}^{-1}$$
$$\Rightarrow\textbf{A}\textbf{A}^\ast = \textbf{P}^\ast\Delta\Delta^\ast\textbf{P}\tag{4}$$
$$\Rightarrow\textbf{A}^\ast\textbf{A} = \textbf{P}^\ast\Delta^\ast\textbf{P}\textbf{P}^\ast\Delta\textbf{P}$$
$$\Rightarrow\textbf{A}^\ast\textbf{A} = \textbf{P}^\ast\Delta^\ast\Delta\textbf{P}\tag{5}$$
$$\textbf{From (4) and (5)}$$
$$\Rightarrow\textbf{A}\textbf{A}^\ast - \textbf{A}^\ast\textbf{A} = \textbf{P}^\ast\Delta\Delta^\ast\textbf{P}-\textbf{P}^\ast\Delta^\ast\Delta\textbf{P}$$
$$\Rightarrow\textbf{A}\textbf{A}^\ast - \textbf{A}\textbf{A}^\ast = \textbf{P}^\ast(\Delta\Delta^\ast-\Delta^\ast\Delta)\textbf{P} = \textbf{0}$$
$$\Rightarrow \Delta\Delta^\ast-\Delta^\ast\Delta = \textbf{0}$$
$$\Rightarrow\Delta\Delta^\ast = \Delta^\ast\Delta\tag{6}$$
$$\Rightarrow\Delta \textbf{ is normal matrix}$$
$$\textbf{From (6)}$$
$$\Rightarrow\vec{e_i}^\ast\Delta\Delta^\ast\vec{e_i} = \vec{e_i}^\ast\Delta^\ast\Delta\vec{e_i}$$
$$\Rightarrow\langle\Delta^\ast\vec{e_i},\Delta^\ast\vec{e_i}\rangle = \langle\Delta\vec{e_i}, \Delta\vec{e_i}\rangle$$
$$\Rightarrow\lVert\Delta^\ast\vec{e_i}\rVert^{2} = \lVert\Delta\vec{e_i}\lVert^{2}$$
$$\Rightarrow\lVert\Delta^\ast\vec{e_i}\rVert = \rVert\Delta\vec{e_i}\lVert$$
$$\Rightarrow\textbf{The length of ith column and ith row in }\Delta\textbf{ are same} \tag{7}$$
$$\Delta\textbf{ is upper triangle matrix}$$
$$\textbf{Let i to be the first row with nonzero off-diagonal element}$$
$$
\Delta_{n,n} = 
\begin{pmatrix}
a_{1,1}     &     0     & \cdots & 0      & \cdots & 0      \\
   0        & a_{2, 2}  & \cdots & 0      & \cdots & 0      \\
   0        &     0     & a_{i,i}& \ast   & \cdots & \ast   \\
\vdots      &   \vdots  & \vdots & \vdots & \cdots & \vdots \\
   \cdot    &    \cdot  & \cdot  & \vdots & \vdots & a_{n,n}
\end{pmatrix}
$$
$$\textbf{If }\Delta\textbf{ is not diagonal matrix, then the ith column is }\lvert a_{i,i} \rvert \neq \lvert a_{i,i} \rvert + \lvert \ast \rvert + \cdots + \rvert \ast \lvert$$
$$\textbf{This contracts our previous (7), therefore }\Delta\textbf{ must be diagonal matrix}$$
$$\textbf{Therefore }\textbf{A}\textbf{A}^\ast = \textbf{A}^\ast\textbf{A}\Rightarrow\textbf{A}=\textbf{P}^\ast\Lambda\textbf{P}\quad\square$$

REPLY [1 votes]: Presumably the ground field is $\mathbb{C}$, otherwise the problem statement is not true. Your proof looks fine to me then.
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