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.

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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.

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From Wikipedia:

A matrix which is simultaneously triangular and normal is also diagonal. This can be seen by looking at the diagonal entries of $A^*A$ and $AA^*$, where $A$ is a normal, triangular matrix.

If $A$ is lower-triangular, then the diagonal entries of $AA^*$ are just the squared magnitudes of the diagonal entries of $A$, whereas the diagonal entries of $A^*A$ are these squared magnitudes plus those of the off-diagonal entries. For the two to be equal, all the off-diagonal entries must be zero.

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If $T$ is an operator on an inner product space, then $\ker T=\ker T^*T$. To see that $\ker T\subseteq \ker T^*T$, note that if $Tx=0$, then $T^*Tx=T^*0=0$. To see that $\ker T^*T\subseteq \ker T$, note that if $T^*Tx=0$, then $\|Tx\|^2=\langle T^*Tx,x\rangle=\langle 0,x\rangle =0$. So, if $T$ is normal, then $\ker T=\ker T^*$.

If $A$ is normal, then so is $A-\lambda I$ for each $\lambda\in\mathbb C$. Thus $\ker (A-\lambda I)=\ker (A^*-\overline{\lambda} I)$, which implies that $A$ and $A^*$ have precisely the same eigenvectors (but with eigenvalues conjugated).

This fact can be used to give an induction proof alternative to the good answer Davide Giraudo already gave. The base case $n=1$ is trivial. Suppose for some $n\geq 1$ that every lower triangular normal $n$-by-$n$ matrix is diagonal. Let $A$ be a lower triangular normal $(n+1)$-by-$(n+1)$ matrix, and write $A$ in the block form $A=\begin{pmatrix}a&0\\\ v&T\end{pmatrix}$, where $a\in\mathbb C$, $v$ is an $n$-by-$1$ matrix, and $T$ is an $n$-by-$n$ matrix. Then the first standard basis vector $e_1=(1,0,0,\ldots,0)^\mathrm{T}$ is an eigenvector for $A^*=\begin{pmatrix}\overline a&v^*\\\ 0&T^*\end{pmatrix}$. By the previous paragraph, $e_1$ is an eigenvector for $A$, so $v=0$. Since $\begin{pmatrix}|a|^2&0\\\ 0&TT^*\end{pmatrix}=\begin{pmatrix}a&0\\\ 0&T\end{pmatrix}\begin{pmatrix}\overline a&0\\\ 0&T^*\end{pmatrix}= \begin{pmatrix}\overline a&0\\\ 0&T^*\end{pmatrix}\begin{pmatrix}a&0\\\ 0&T\end{pmatrix} =\begin{pmatrix}|a|^2&0\\\ 0&T^*T\end{pmatrix}$, $T$ is normal. Since $A$ is lower triangular, so is $T$. By the induction hypothesis, $T$ is diagonal. Hence so is $A$, and this completes the proof by induction.

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