Show map is norm-preserving and determine Eigenvalues Can someone of you give me a solution for this?

Let $N\in \mathbb N$.
  a) We define the map $\mathfrak F:(\mathbb C^N, ||\cdot||_2)\to(\mathbb C^N, ||\cdot||_2)$ by $$(\mathfrak F(x))_k := \frac{1}{\sqrt N}\sum_{j=1}^Nx_je^{2\pi i\frac{(j-1)(k-1)}{N}} \quad \forall k\in \{1,...,N\}$$ 
  Show that $\mathfrak F$ is norm-preserving, i.e. $$||(\mathfrak F(x))_k||_2 = ||x||_2 \quad \forall k\in \{1,...,N\}$$ 
  b) For $n,m \in \{1,2,...,N\}$ we define the entries of $M\in \mathbb C^{N\times N}$ by $$M_{nm} := 
\begin{cases}
\frac{N+1}{2}, \quad \quad n=m \\
\frac{1}{e^{2\pi i \frac{m-n}{N}}-1}, \,\,n \neq m
\end{cases}$$
  Show that $1, 2, 3, ..., N$ are the Eigenvalues of M.

 A: To simplify notation, write $$\omega_j=e^{\frac{2\pi i\,j}N},\ \ \ j=0,1,\ldots,N-1,$$ the $N^{\rm th}$-roots of unity. 
For simplicity, I'll renumber the indices of $x$ and $k$ to $0,\ldots,N-1$. So, with $e_0,\ldots,e_{N-1}$ the canonical basis, 
$$
\mathfrak F(x)=\frac1{\sqrt N}\,\sum_{h=0}^{N-1}\sum_{j=0}^{N-1}\,\omega_j^kx_j\,e_k.
$$
Then
\begin{align}
\|\mathfrak F(x)\|_2^2
&=\langle \mathfrak F(x),\mathfrak F(x)\rangle
=\frac1N\,\sum_{k=0}^{N-1}\sum_{j=0}^{N-1}\sum_{s=0}^{N-1}\sum_{t=0}^{N-1}
\omega_j^kx_j\overline{\omega_s^tx_s}\langle e_k,e_t\rangle\\ \ \\
&=\frac1N\,\sum_{k=0}^{N-1}\sum_{j=0}^{N-1}\sum_{s=0}^{N-1}
\omega_j^kx_j\overline{\omega_s^kx_s}\\ \ \\
&=\frac1N\,\sum_{k=0}^{N-1}\sum_{j=0}^{N-1}\sum_{s=0}^{N-1}
\omega_j^k\omega_s^{-k}x_j\overline{x_s}\\ \ \\
&=\frac1N\,\sum_{j=0}^{N-1}\sum_{s=0}^{N-1}
x_j\overline{x_s}\,\left(\sum_{k=0}^{N-1}\omega_j^k\omega_s^{-k}\right)\\ \ \\
&=\frac1N\,\sum_{j=0}^{N-1}\sum_{s=0}^{N-1}
x_j\overline{x_s}\,\left(\sum_{k=0}^{N-1}\omega_{j-s}^k\right)\\ \ \\
&=\sum_{j=0}^{N-1}|x_j|^2=\|x\|_2^2
\end{align}
(note that the sum in brackets is zero when $j\ne s$ and $N$ when $j=s$). 
(No more time now, I'll look at the second part tomorrow)
