= Solution
The original PDF uses the <conjugate transpose> $N^\dagger$, rather than the transpose found in the transcription. The printed diagonalisation criterion is correct when diagonalisation means <unitary diagonalization of a normal matrix>. Since $N^\dagger=A-iB$,
$$
NN^\dagger=A^2+B^2-i(AB-BA),\qquad N^\dagger N=A^2+B^2+i(AB-BA).
$$
Consequently $NN^\dagger=N^\dagger N$ if and only if $AB=BA$. By part (ii), this gives a common <unitary matrix> $U$ and
$$
U^\dagger NU=D_A+iD_B,
$$
which is diagonal. Conversely, if $N=UDU^\dagger$ for a diagonal $D$ and a <unitary matrix> $U$, then
$$
NN^\dagger=UDD^\dagger U^\dagger=UD^\dagger DU^\dagger=N^\dagger N.
$$
Thus the precise conclusion is
$$
\boxed{N\text{ is unitarily diagonalizable}\quad\Longleftrightarrow\quad N\text{ is normal}\quad\Longleftrightarrow\quad AB=BA.}
$$
If arbitrary invertible changes of basis are allowed, the printed assertion is false. For example,
$$
N=\begin{pmatrix}1&1\\0&2\end{pmatrix},\quad A=\begin{pmatrix}1&1/2\\1/2&2\end{pmatrix},\quad B=\begin{pmatrix}0&-i/2\\i/2&0\end{pmatrix}
$$
satisfy $N=A+iB$. Both <Hermitian matrices> have distinct <eigenvalues>: $(3\pm\sqrt2)/2$ for $A$ and $\pm1/2$ for $B$. The distinct <eigenvalues> $1,2$ make $N$ diagonalizable by similarity, but $NN^\dagger=\begin{pmatrix}2&2\\2&4\end{pmatrix}$ and $N^\dagger N=\begin{pmatrix}1&1\\1&5\end{pmatrix}$ differ. This distinguishes ordinary <diagonalization> from the required unitary version.
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