Binding eigenvalue properties with normal operators Let $N$ be a normal transformation over $\Bbb C$.
A) True/False:$$N = N^* \Leftarrow\Rightarrow \lambda \in \Bbb R$$
(where $\lambda$ is any eigenvalue of $N$)
B) True/False:$$N^* = N^{-1} \Leftarrow\Rightarrow |\lambda|=1$$
(where $\lambda$ is any eigenvalue of $N$)

In both cases, I am able to prove the direction $\Rightarrow$ quite easily.
but in the other direction i am having a struggle (perhaps its false?).
for ex. statement B: I thought of showing that since $N$ is normal, then it is diagonalizable, therefore $N^*$ is also diagonalizable with the conjugate eigenvalues on its diagonal, therefore for any $\lambda = (a+bi)$ then $NN^*$ has $(a+bi)(a-bi)$ on its diagonal, which is the identity matrix (since $|\lambda|=1$) and therefore $N$ is unitary, but i'm not sure my transitions are valid.
Anyway, any help much appreciated on both statements!
 A: Since $N$ is a normal $n \times n$ - matrix, there is a unitary matrix $U$ such that $N=UDU^{*}$ , where $D=diag (\lambda_1,..., \lambda_n)$ and the $ \lambda_k$ are the eigenvalues of $N$.
A) if all $ \lambda_k$ are real, then $D^{*}=D$, hence $N^{*}=UD^{*}U^{*}=UDU^{*}=N$.
B) if for all $ \lambda_k$ we have $|\lambda_k|=1$, then $\overline{\lambda_k}= \frac{1}{\lambda_k}$, thus $N^{*}=UD^{*}U^{*}=UD^{-1}U^{*}=N^{-1}$.
A: A): Since $\mathcal N$ is normal, there is a eigenbasis $(e_j)$ of the vector space. Since $(\mathcal N - \mathcal N^*)e_j = (\lambda - \lambda) e_j = 0$ for all $j = 1, \ldots, n$ [assuming the vector space $V$ has dimension $n < \infty$]. Hence 
$$
\mathcal Nx = \mathcal N^*x
$$
for all $x \in V$, i.e. $\mathcal N = \mathcal N^*$.
B) If $|\lambda| = 1$, then applied the operator $\mathcal N\mathcal N^*$ to $e_j$ gives 
$$
\mathcal N \mathcal N^* e_j = \mathcal N (\overline \lambda e_j) = \lambda \overline \lambda e_j = e_j\quad [j = 1, \ldots, n].
$$
Thus $\mathcal N \mathcal N^* = \mathcal I$ [the identical mapping].
P.S. the symbol $\iff$ is available as "\iff" .
