Operator Norm Is Unchanged by Spectator Registers

Write the specified orthonormal bases as $f_0,\dots ,f_{a-1}$ for $S$ and $e_0,\dots ,e_{d-1}$ for $R$; $a,d\ge 1$. By [F1], $f_k\otimes e_j$ form an orthonormal basis of $S\otimes R$. Expand $v={\sum }_{k<a,j<d}{c}_{kj}{f}_k\otimes e_j$ and set $v_j={\sum }_{k<a}{c}_{kj}{f}_k$. Grouping this finite sum gives $v={\sum }_{j<d}{v}_j\otimes e_j$, while orthonormality gives $\parallel v{\parallel }^2={\sum }_{j<d}\parallel v_j{\parallel }^2$. By [F2], $(T\otimes I_R)v={\sum }_{j<d}T{v}_j\otimes e_j$. Hence [F3] gives $\parallel (T\otimes I_R)v{\parallel }^2={\sum }_{j<d}\parallel T{v}_j{\parallel }^2\le \parallel T{\parallel }^2{\sum }_{j<d}\parallel v_j{\parallel }^2=\parallel T{\parallel }^2\parallel v{\parallel }^2$, so $\parallel T\otimes I_R\parallel \le \parallel T\parallel$.

Choose a unit vector $s\in S$ attaining the maximum in [F3], and choose a unit vector $e\in R$. The product inner product gives $\parallel s\otimes e\parallel =1$, and [F2] gives $\parallel (T\otimes I_R)(s\otimes e)\parallel =\parallel Ts\parallel =\parallel T\parallel$. Thus $\parallel T\otimes I_R\parallel \ge \parallel T\parallel$, proving the first equality.

If the tensor register is zero-dimensional, both $A$ and $PA{P}^{-1}$ have operator norm $0$ by [F3], proving the conjugation equality. Otherwise [F5] shows that $P$ and $P^{-1}$ preserve norms, so each has operator norm $1$ by [F3]. Submultiplicativity [F4] gives $\parallel PA{P}^{-1}\parallel \le \parallel A\parallel$. Applying the same inequality to $A=P^{-1}(PA{P}^{-1})P$ gives the reverse inequality. Therefore $\parallel PA{P}^{-1}\parallel =\parallel A\parallel$ in every case.