<s>[INST] Normal and Lower triangular matrix implies diagonal matrix A lower triangular complex matrix $A$ satisfies $AA^*=A^*A$.
I would like to show that $A$ is diagonal. I know there exists a unitary matrix $P$ such that $PAP^*$ is diagonal. But I don't know how to show $A$ itself is diagonal.
 [/INST] We can show it by induction on the dimension. For $n=2$, let $A=\begin{pmatrix}a&0\\\ b&c\end{pmatrix}$ such a matrix. Then 
\begin{align*}A^* A-AA^* &=\begin{pmatrix}\bar a&\bar b\\ 0&\bar c\end{pmatrix}\begin{pmatrix}a&0\\ b&c\end{pmatrix}-\begin{pmatrix}a&0\\ b&c\end{pmatrix}\begin{pmatrix}\bar a&\bar b\\ 0&\bar c\end{pmatrix}\\ 
&=\begin{pmatrix}|a|^2+|b|^2-|a|^2&\bar bc-a\bar b\\ \bar cb-b\bar a&|c|^2-(|b|^2+|c|^2)\end{pmatrix}\\ 
&=\begin{pmatrix}|b|^2&\bar bc-a\bar b\\ \bar cb-b\bar a&-|b|^2\end{pmatrix},
\end{align*}
so $b=0$ and $A$ is diagonal. If the result is true for $n\geq 2$, let $A=\begin{pmatrix}T&0\\\ v&a\end{pmatrix}$, where $T$ is a $n\times n$ triangular matrix, $v$ a $1\times n$ matrix and $a$ a complex number. Since $A^*A=AA^*$, we have 
\begin{align*}
\begin{pmatrix}0&0\\ 0&0\end{pmatrix}&=\begin{pmatrix}T^* &v^* \\ 0&\bar a\end{pmatrix}\begin{pmatrix}T&0\\ v&a\end{pmatrix}-\begin{pmatrix}T&0\\ v&a\end{pmatrix}\begin{pmatrix}T^* &v^* \\ 0&\bar a\end{pmatrix}\\
&=\begin{pmatrix}T^* T+v^* v &v^* a \\ \bar a v&|a|^2\end{pmatrix}-
\begin{pmatrix}TT^* &Tv^* \\ vT^* &|v|^2+|a|^2\end{pmatrix}\\
&=\begin{pmatrix}T^* T-TT^* +v^* v &v^* a-Tv^*  \\ \bar av-v T^* &-|v|^2,\end{pmatrix}
\end{align*}
hence $v=0$ and $T$ is normal. Since $T$ is lower triangular, $T$ is normal and by induction hypothesis $T$ is diagonal. We conclude that $A$ is diagonal.
</s>