# Why this equality is true?

Let $E$ be a complex Hilbert space. Let ${\bf S} = (S_1,...,S_d) \in \mathcal{L}(E)^d$. We recall that $\|{\bf S}\|$ is defined by \begin{eqnarray*} \|{\bf S}\| &:=&\sup\left\{\bigg(\displaystyle\sum_{k=1}^d\|S_kx\|^2\bigg)^{\frac{1}{2}},\;x\in E,\;\|x\|=1\;\right\}, \end{eqnarray*}

If the operators $S_k$ are commuting, why we have $$\displaystyle\sup_{\|x\|=1}\displaystyle\sum_{|\alpha|=n}\frac{n!}{\alpha!}\|{\bf S}^{\alpha}x\|^2=||{\bf S}^n||^2\;?? \;,$$ with $n\in\mathbb{N}^*,\;$ $\alpha = (\alpha_1, \alpha_2,...,\alpha_d) \in \mathbb{N}^d;\;\alpha!: =\alpha_1!...\alpha_d!,\;|\alpha|:=\displaystyle\sum_{j=1}^d\alpha_j$; ${\bf S}^\alpha:=S_1^{\alpha_1} \cdots S_d^{\alpha_d}$ and ${\bf S}^n:={\bf S}\diamond{\bf S}\diamond\cdots\diamond{\bf S}$.

Note that ${\bf S}^2 :={\bf S}\diamond{\bf S}= (S_1 S_1,\cdots,S_1 S_d,S_2S_1,\cdots,S_2S_d,S_dS_1\cdots,S_d S_d)$.

Thank you!!

• you have $n$ operators $\alpha_i$ from each how many ways can you reorder them if order matters? – Adam Nov 23 '17 at 19:01
• I think ${n \choose{\alpha_1, \alpha_2, \ldots , \alpha_d }} = \frac{n!}{\alpha_1!\ldots \alpha_d!}$. – Schüler Nov 23 '17 at 19:05
• well there is the missing factor – Adam Nov 23 '17 at 19:06
• I mean your formula $||{\bf S}^n||^2=\displaystyle\sup_{\|x\|=1}\displaystyle\sum _{k=1}^{d^n}\| S_1 ^{\alpha _1} S_2^{\alpha_2} \cdots S_d^{\alpha_d}x \|^2.$ is false as you already reordered the components (using the commutivity) without this factor – Adam Nov 23 '17 at 19:09
• Yes, but how can I get a formula which is equal to the following: $$\displaystyle\sup_{\|x\|=1}\displaystyle\sum_{|\alpha|=n}\frac{n!}{\alpha!}\|{\bf S}^{\alpha}x\|^2 ?$$ – Schüler Nov 23 '17 at 19:12

Recall that the number of ways in which you can permute $n$ objects where $\alpha_1$ are of $1$st type, $\alpha_2$ are of $2$nd type,..., and $\alpha_d$ are of $d$th type is $$\frac{n!}{\alpha_1!...\alpha_d!}=\frac{n!}{\alpha!}$$
As you already mentioned, $S^n$ is $d^n$ tuple of operators (each operator is product of $n$ operators from $\{S_1,...,S_d\}$) . Hence $$||S^n||^2=\sup_{||x||=1} \sum_{|\alpha|=n}\frac{n!}{\alpha!}||S^{\alpha}x||^2.$$
A typical operator in the $d^n$ tuple of operators will look like $S_{i_1}S_{i_2}...S_{i_n}$ where each $i_j\in \{1,2,...,d\}.$ Let $\alpha_k=$ number of $j$ such that $i_j=k .$ Then the operator $S_{i_1}S_{i_2}...S_{i_n}$ will become $S_1^{\alpha_1}...S_d^{\alpha_d}.$ Thus every operator in the $d^n$ tuple will become some $S_1^{\alpha_1}...S_d^{\alpha_d}$ for some $\alpha_1,...,\alpha_d$ with $\sum \alpha_i=n.$
Now fix any $\alpha_1,...,\alpha_d$ with $\alpha_1+...+\alpha_d=n.$ We have to count number of $S_{i_1}S_{i_2}...S_{i_n}$ in the $d^n$ tuple which will become $S_1^{\alpha_1}...S_d^{\alpha_d}$. Take $\alpha_1$ copies of $S_1, \alpha_2$ copies of $S_2$,...,$\alpha_d$ copies of $S_d.$ Let $\mathcal A$ be the set of all permutations of this operators. Note that each operator in $\mathcal A$ will appear in the $d^n$ tuple of operators and each operator (in the $d^n$ tuple of operators) which will become $S_1^{\alpha_1}...S_d^{\alpha_d}$ (after using commutative property) must be present in $\mathcal A.$ Hence number of $S_{i_1}S_{i_2}...S_{i_n}$ in the $d^n$ tuple which will become $S_1^{\alpha_1}...S_d^{\alpha_d}$ is same as the number of ways in which you can permute $n$ objects where $\alpha_1$ are of $1$st type, $\alpha_2$ are of $2$nd type,..., and $\alpha_d$ are of $d$th type. Since the later number is $\frac{n!}{\alpha!},$ we hav $$||S^n||^2=\sup_{||x||=1} \sum_{|\alpha|=n}\frac{n!}{\alpha!}||S^{\alpha}x||^2.$$