For finite dimensions one has the theorem that a matrix is normal if and only if unitarily diagonalizable.
So now, by dint of the lemma, write your general $NΓN$ matrix as: $$π=π\tilde{π}π^β $$ with $\tilde{π}$ upper triangular and π unitary. Given $ππ^β =π^β π$, show that $\tilde{π}\tilde{π}^β =\tilde{π}^β \tilde{π}$, then $π^β $ is normal and upper triangular so it is diagonal.
To prove unitarily diagonalizable implies normal:
If linear map $A:β^N\toβ^N$ has a matrix that is unitarily diagonalizable, then by definition we have $π=ππ²π^β $ with $ππ^β =\mathrm{id}$, where $π$ is the matrix of $A$ and $π²$ is diagonal. A straightforward calculation then shows that an entity of this kind commutes with its Hermitian conjugate and is therefore normal.To prove normal implies unitarily diagonalizable:
To prove the other way, you need to make use of Schur's lemma.Every complex square matrix is unitarily similar to an upper triangular matrix.Proof: a matrix $π$ has at least one eigenvector $X$ - let's make it a unit vector $\hat{X}$ - with eigenvalue $\lambda$ and, by the Gram-Schmidt procedure, build an orthonormal matrix of the form $π_X=[\hat{X}\hat{Y}_2β―\hat{Y}_N]$ with columns $\hat{X}$,$\hat{Y}_2β―\hat{Y}_N$. Now work out $π_X^{-1}ππ_X=π_X^β ππ_X$ and you find it is a matrix of the form: $$\left(\begin{array}{c|c}\lambda & β―\\\hline O&π_1\end{array}\right)$$ where $π_1$ is an $N-1ΓN-1$ matrix. So now repeat the procedure with the matrix $π_1$ and iterate, each iteration leaving a lower right square matrix of dimension one fewer than the one from the former iteration. When we reach the lower right corner, we have an upper triangular matrix, similar to the original by the product of all the unitary transformations at each step.
So now, by dint of the lemma, write your general $NΓN$ matrix as: $$π=π\tilde{π}π^β $$ with $\tilde{π}$ upper triangular and π unitary. Given $ππ^β =π^β π$, show that $\tilde{π}\tilde{π}^β =\tilde{π}^β \tilde{π}$, then $π^β $ is normal and upper triangular so it is diagonal.
If $A$ is normal and upper triangular then it is diagonal.$\newcommand{\abs}[1]{\left|#1\right|}$ Denote $A=\begin{pmatrix}a_{11}&a_{12}&\cdots&a_{1n}\\ &a_{22}&\cdots&a_{2n}\\&&\ddots&\vdots\\&&&a_{nn}\end{pmatrix}$. Observe that the $(1,1)$-entries of $A^\ast A$ and $AA^\ast$ are $$\abs{a_{11}}^2\quad\mbox{and}\quad \sum_{i=1}^n\abs{a_{1i}}^2,$$ respectively. Since $A$ is normal, $$ \sum_{i=1}^n\abs{a_{1i}}^2=\abs{a_{11}}^2\quad\Rightarrow\quad \sum_{i=2}^n\abs{a_{1i}}^2=0\quad\Rightarrow\quad a_{12}=a_{13}=\cdots=a_{1n}=0.$$ Now, from preceding result, the $(2,2)$-entry of $A^\ast A$ is $$\abs{a_{12}}^2+\abs{a_{22}}^2=\abs{a_{22}}^2,$$ and the $(2,2)$-entry of $AA^\ast$ is $\displaystyle\sum_{i=2}^n\abs{a_{2i}}^2$. Again, since $A$ is normal, $$ \sum_{i=2}^n\abs{a_{2i}}^2=\abs{a_{22}}^2\quad\Rightarrow\quad \sum_{i=3}^n\abs{a_{2i}}^2=0 \quad\Rightarrow\quad a_{23}=a_{24}=\cdots=a_{2n}=0.$$ Continue this process, we may conclude that the upper off-diagonal entries of $A$ are all zero. Hence, $A$ is a diagonal matrix.