 Solution

The original PDF uses the conjugate transpose $N^{†}$, rather than the transpose found in the transcription. The printed diagonalisation criterion is correct when diagonalisation means unitary diagonalization of a normal matrix. Since $N^{†}=A-iB$,Consequently $N{N}^{†}=N^{†}N$ if and only if $AB=BA$. By part (ii), this gives a common unitary matrix $U$ andwhich is diagonal. Conversely, if $N=UD{U}^{†}$ for a diagonal $D$ and a unitary matrix $U$, thenThus the precise conclusion isIf arbitrary invertible changes of basis are allowed, the printed assertion is false. For example,satisfy $N=A+iB$. Both Hermitian matrices have distinct eigenvalues: $(3\pm \sqrt{2})/2$ for $A$ and $\pm 1/2$ for $B$. The distinct eigenvalues $1,2$ make $N$ diagonalizable by similarity, but $N{N}^{†}=\left(\begin{array}{cc}2& 2\\ 2& 4\end{array}\right)$ and $N^{†}N=\left(\begin{array}{cc}1& 1\\ 1& 5\end{array}\right)$ differ. This distinguishes ordinary diagonalization from the required unitary version.