Prove that $w(R_n)\le w(T\otimes S_n)$ where $S_n$ denote cyclic forward shift on $\Bbb{C}^n$, $R_n\in M_n(B(H))$ with subdiagonal entris $T$ This is an exercise of Vern Paulsen's book Completely Bounded Maps and Operator Algebras (Exercise 3.13 (iii))
Let's first define for $T$ a bounded operator on hilbert space $H$, the numerical radius of $T$
$$w(T):=\text{sup}\{|\langle Tx,x\rangle|:\ \lVert x\rVert=1\}$$
Let $S_n$ be the cyclic forward shift defined as $S_n(e_j)=e_{j+1\text{(mod }n\text{)}}$ for $j=0,1,\ldots,n-1$ where $\{e_0,\ldots,e_{n-1}\}$ is standard basis of $\Bbb{C}^n$. Matrix of $S$ with respect to standard ordered basis is the matrix whose subdiagonal entries are $1$ and zero elsewhere.
Let $T$ be a bounded operator on hilbert space $H$, then $T\otimes S_n$ is bounded operator on $H\otimes \Bbb{C}^n$. Let $R_n$ be the $n\times n$ operator matrix whose subdiagonal entries are $T$ and that is $0$ elsewhere. Then $R_n\in M_n(\mathcal{B}(H))=\mathcal{B}(H^{(n)})$
We have to prove $w(R_n)\le w(T\otimes S_n)$.
Let $\tilde{x}=(x_0,x_1,\ldots,x_{n-1})\in H^{(n)}$. Then I have computed that $$\left|\langle R_n\tilde{x},\tilde{x}\rangle_{H^{(n)}}\right|=\left|\sum\limits_{i=1}^{n-1}\langle T x_i, x_{i+1}\rangle\right|$$
I have to show this is $\le w(T\otimes S_n)$.
There is a hint provided which says consider $x_\lambda=\lambda x_1\oplus\cdots\oplus\lambda^n x_n\in H^{(n)}$ with $|\lambda|=1$. I cannot find any way-out to use this hint.
Can anyone help me with any idea to solve this? Thanks for your help in advance.
 A: The matrix of $S$ is not the matrix whose subdiagonal entries are 1 and zero elsewhere. You are missing a 1 in the 0,n-1 entry. Without that, you cannot have possibly done parts (i) and (ii).
The computation you did should say
$$\tag1
\left|\langle R_n\tilde{x},\tilde{x}\rangle_{H^{(n)}}\right|=\left|\sum\limits_{i=0}^{n-2}\langle T x_i, x_{i+1}\rangle\right|
$$
On the other hand,
\begin{align}
\Big\langle (T\otimes S_n)\sum_{k=0}^{n-1}x_k\otimes e_k,\sum_{k=0}^{n-1}x_k\otimes e_k\Big\rangle
&=\langle Tx_0,x_{n-1}\rangle+\sum_{k=1}^{n-2}\langle Tx_k,x_{k+1}\rangle
\end{align}
Now let $\lambda\in\mathbb T$ with $$\,\sum\limits_{i=0}^{n-2}\langle T x_i, x_{i+1}\rangle=\lambda\,\Bigg|\sum\limits_{i=0}^{n-2}\langle T x_i, x_{i+1}\rangle\Bigg|.$$
Then, since $$\langle T(\lambda ^kx_k),\lambda^{k+1}x_{k+1}\rangle=\lambda^{-1}\langle Tx_k,x_{k+1}\rangle,$$
we have, with $y=(0,\lambda x_1,\lambda^2x_2,\ldots,\lambda^{n-1}x_{n-1})$
\begin{align}
|\langle R_n\tilde x,\tilde x\rangle|&=\Bigg|\sum_{k=0}^{n-2}\langle Tx_k,x_{k+1}\rangle\Bigg|
=\lambda^{-1}\,\sum\limits_{i=0}^{n-2}\langle T x_i, x_{i+1}\rangle\\[0.3cm]
&=\sum_{k=0}^{n-2}\langle Ty_k,y_{k+1}\rangle
=\langle Ty_0,y_1\rangle+\sum_{k=0}^{n-2}\langle Ty_k,y_{k+1}\rangle\\[0.3cm]
&=\langle (T\otimes S_n)y,y\rangle\leq\omega(T\otimes S_n). 
\end{align}
