Inclusion of numerical range and translation and rotation invariance Let $T\in\mathscr{B(\mathcal{H})}$ and $X\in M_n(\mathbb{C})$. Then prove that following two are equivalent:
(i) $W(B\otimes X)\subseteq W(B\otimes T)$, for all $B\in M_n$
(ii) $W(C\otimes (aX+bI_n))\subseteq W(C\otimes (aT+bI))$, for all $C\in M_n$ and $a,b\in\mathbb{C}$.
Where $W(T):=\{\langle Tx,x\rangle: \Vert x\Vert=1\}$ is called the numerical range of $T$.
Comment: I can see this is true if we calculate numerical range in a simple tensor of $\mathbb{C}^n\otimes\mathbb{C}^n$ but in general I could not prove.
Any hint/comment is highly appreciated.
 A: My thoughts on the problem (too long for a comment):
If we decompose $x = \sum_k e_k \otimes \tilde x_k$ where $\tilde x_k \in \Bbb C^n$, then we have
$$
\begin{align*}
\langle[C \otimes X]x ,x\rangle &= \sum_{j,k = 1}^n \langle (Ce_j \otimes X\tilde x_j),(e_k \otimes \tilde x_{k})\rangle = 
\sum_{j,k=1}^n \langle Ce_j,e_k \rangle \langle X \tilde x_j,\tilde x_k \rangle
\\ & = \operatorname{trace}(C^TM)
\end{align*}
$$
where $M$ is the matrix satisfying $M_{jk} = \langle X \tilde x_j,\tilde x_k \rangle$.
Similarly, if we decompose $x = \sum_k e_k \otimes \tilde y_k$ where $\tilde y_k \in \mathcal H$, then we have
$$
\begin{align*}
\langle[C \otimes T]y ,y\rangle &= 
\sum_{j,k=1}^n \langle Ce_j,e_k \rangle \langle T \tilde y_j,\tilde y_k \rangle
 = \operatorname{trace}(C^TN)
\end{align*}
$$
where $N_{jk} = \langle T \tilde y_j,\tilde y_k \rangle$.
The difficult part is proving that (i) implies (ii). For any unit vector $x \in \Bbb C^n$, note that
$$
\langle[C \otimes (aX + bI_n)]x ,x\rangle = 
a \langle[C \otimes X]x ,x\rangle + b\langle[C \otimes I_n]x ,x\rangle
\\= a\operatorname{trace}(C^TM^{(1)}) + b\operatorname{trace}(C^TM^{(2)}).
$$
and
$$
\langle[C \otimes (aT + bI_n)]y ,y\rangle = 
a \langle[C \otimes T]y ,y\rangle + b\langle[C \otimes I_n]y ,y\rangle
\\= a\operatorname{trace}(C^TN^{(1)}) + b\operatorname{trace}(C^TN^{(2)}).
$$
Here, 
$$
M^{(1)}_{jk} = \langle X \tilde x_j,\tilde x_k \rangle, \quad M^{(2)}_{jk} = \langle \tilde x_j,\tilde x_k \rangle\\
N^{(1)}_{jk} = \langle T \tilde y_j,\tilde y_k \rangle, \quad N^{(2)}_{jk} = \langle \tilde y_j,\tilde y_k \rangle
$$
Statement (i) is equivalent to saying that for any fixed $C$: for any $x \in \Bbb C^n \otimes \Bbb C^n$, there exists a $y \in \Bbb C^n \otimes \mathcal H$ such that 
$$
\operatorname{trace}(C^T M^{(1)}(x)) = \operatorname{trace}(C^T N^{(1)}(y))
$$
Statement (ii) is equivalent to saying that for any fixed $C$: for any $x \in \Bbb C^n \otimes \Bbb C^n$, there exists a $y\in \Bbb C^n \otimes \mathcal H$ such that 
$$
\operatorname{trace}(C^T M^{(j)}(x)) = \operatorname{trace}(C^T N^{(j)}(y))
$$
for $j=1,2$.
