Operator norm of a family of matrices Let $c$ be a complex number. Consider the family of $n\times n$ matrices $M_n$ which have $c$'s on one off-diagonal, $\bar{c}$'s on the other off-diagonal, and zero everywhere else. So $M_4$ looks like this:$$\left(\begin{array}{cccc}0&c&0&0\\\bar{c}&0&c&0\\0&\bar{c}&0&c\\0&0&\bar{c}&0\end{array}\right)$$How does one find a general formula for the operator norm of $M_n$?
 A: Write $c=\alpha|c|$, with $|\alpha|=1$. Then 
$$
M_4=|c|\,\begin{bmatrix}
1&0&0&0\\
0&\alpha&0&0\\
0&0&\alpha^2&0\\
0&0&0&\alpha^3\end{bmatrix}
\begin{bmatrix}
0&1&0&0\\
1&0&1&0\\
0&1&0&1\\
0&0&1&0
\end{bmatrix}
\begin{bmatrix}
1&0&0&0\\
0&\alpha&0&0\\
0&0&\alpha^2&0\\
0&0&0&\alpha^3\end{bmatrix}^*,
$$
and so $M_4(c)$ is unitarily equivalent to $|c|\,M_4(1)$. Thus we only need to determine the spectrum of $M_4(1)$. The same argument applies for $M_n$ for any $n$. 
The eigenvalues of $M_n(1)$ are $\lambda_k=2\cos\frac{k\pi}{n+1}$, where an eigenvector for $\lambda_k$ is $\sum_{s=1}^n\sin\frac{ks\pi}{n+1}e_s$, and $e_1,\ldots,e_n$ is the canonical basis. 
As $M_n(1)$ is selfadjoint, its norm is the greatest eigenvalue in absolute value, so 
$$
\|M_n(c)\|=|c|\,\|M_n(1)\|=2|c|\,\cos\frac{\pi}{n+1}. 
$$

$\newcommand\abajo{\\ \ \\}$
Edit: The Eigenvalues of $M_n(1)$.
Let $e_1,\ldots,e_n$ be the canonical basis of $\mathbb C^n$. Let $\eta_k=\sum_{\ell=1}^n\sin\frac{k\ell\pi}{n+1}\,e_\ell$, $k=1,\ldots,n$. Then
\begin{align*}
M_n(1)\eta_k
&=\sum_{s=1}^{n-1}\sin\frac{sk\pi}{n+1}\,e_{s+1}+\sum_{s=2}^{n}\sin\frac{sk\pi}{n+1}\,e_{s-1} \abajo
&=\sin\frac{2k\pi}{n+1}\,e_1+\sum_{s=2}^{n-1}\left(\sin\frac{(s-1)k\pi}{n+1}+\sin\frac{(s+1)k\pi}{n+1}\right)\,e_{s}
+\sin\frac{(n-1)k\pi}{n+1}\,e_{n} \abajo
&=2\cos\frac{k\pi}{n+1}\sin\frac{k\pi}{n+1}\,e_1+\sum_{s=2}^{n-1}2\cos\frac{k\pi}{n+1}\sin\frac{sk\pi}{n+1}\,e_s+2\cos\frac{k\pi}{n+1}\sin\frac{nk\pi}{n+1}\abajo
&=2\cos\frac{k\pi}{n+1}\,\eta_k.
\end{align*}
All the above requires is the sine of a sum formula. 
