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Let $E$ be a complex Hilbert space and $\mathcal{L}(E)$ be the algebra of all bounded linear operators on $E$.

For ${\bf A} = (A_1,...,A_d) \in \mathcal{L}(E)^d$, the norm of ${\bf A}$ is given by $$\|{\bf A}\|^2=\sum_{k=1}^d\|A_k\|^2.$$

For ${\bf T}=(T_1,...,T_d) \in \mathcal{L}(E)^d$ and ${\bf S}=(S_1,\cdots,S_m)\in \mathcal{L}(E)^m$, we set $$\mathbf{T}\mathbf{S}:=(T_1S_1,\cdots,T_1S_m,T_2S_1,\cdots,T_2S_m,\cdots,T_dS_1,\cdots,T_dS_m).$$ Let $\mathbf{T}^2=\mathbf{T}\mathbf{T}$ and we define by induction $\mathbf{T}^{n+1}=\mathbf{T}\mathbf{T}^n$ for $n\in \mathbb{N}^*$.

Let $n\in \mathbb{N}^*$ and ${\bf T}=(T_1,...,T_d) \in \mathcal{L}(E)^d$ be such that $T_iT_j=T_iT_j$ for all $i,j\in 1,\cdots,d$. I want to prove that $$\|\mathbf{T}^n\|^2=\sum_{|\alpha|=n}\frac{n!}{\alpha!}\|\mathbf{T}^{\alpha}\|^2.$$ Note that for $\alpha = (\alpha_1,\cdots,\alpha_d) \in \mathbb{N}^d$, we write $\alpha!: =\alpha_1!\cdots\alpha_d!,\;|\alpha|:=\displaystyle\sum_{j=1}^d|\alpha_j|$ and $\mathbf{T}^\alpha:=T_1^{\alpha_1} \cdots T_d^{\alpha_d}$.

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  • $\begingroup$ have you tried induction in $n$ or $d$? $\endgroup$ – supinf Jan 17 at 11:58
  • $\begingroup$ @supinf Yes I try induction on $n$ but from $n$ to $n+1$ it is not evident. $\endgroup$ – Schüler Jan 17 at 12:03
  • $\begingroup$ Also, this question is the same as your old question math.stackexchange.com/questions/3050154/… , except that you are now squaring all monomials (but that is tantamount to squaring all variables, i.e., setting $X_i = T_i^2$). $\endgroup$ – darij grinberg Jan 17 at 14:54
  • $\begingroup$ @Schüler: Yes, it is true then, more or less by the definition. $\endgroup$ – darij grinberg Jan 17 at 22:19
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The reason why there is a factor $\frac{n!}{\alpha!}$ is that with the multi-index notation, the order with which the operator components are composed is undistinguished. For instance, if $d=2$, then since $T_iT_j=T_jT_i$ for all $i,j$, we have $$T_1T_2=T^{(1,1)}=T_2T_1 $$ Therefore, you have $$\|T^n\|^2=\sum_{|\alpha|=n}c_{\alpha}\|T^{\alpha}\|^2 $$ where $c_{\alpha}$ is the number of possible permutations of a composition of $n$ components of $T=(T_1,\dots,T_d)$ which correspond to the multi-index $\alpha$.

\begin{align*}c_{\alpha}&=\#\left\{(j_1,\dots,j_n):T_{j_1}\dots T_{j_n}=T^{\alpha},\,j_k\in \left\{0,\dots,d\right\}\right\} = \\ &=\#\left\{(j_1,\dots,j_n):(\#\left\{j_k=1\right\}=\alpha_1 \land \dots \land \#\left\{j_k=d\right\}=\alpha_d)\right\}\\ \end{align*} To compute the cardinality of the above set in an intuitive way, we may write out the multi-index $\alpha$ in the following way $$\underbrace{1,\dots, 1}_{\alpha_1\text{ times }},\underbrace{2,\dots, 2}_{\alpha_2\text{ times }},\dots, \underbrace{d,\dots, d}_{\alpha_d\text{ times }} $$ Notice that the total amount of numbers written in the above line is $\alpha_1+\dots+\alpha_d=n$. Then $c_{\alpha}$ is simply the number of possible permutations of this list, where copies of the same number are undistinguished. And this is just $$ c_{\alpha}=\frac{n!}{\alpha_1!\cdot \dots \cdot \alpha_d!}=\frac{n!}{\alpha!}$$

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  • $\begingroup$ For every $\alpha = (\alpha_1,\cdots,\alpha_d) \in \mathbb{N}^d$, I think that $\mathbf{T}^\alpha$ is always defined to be $T_1^{\alpha_1} \cdots T_d^{\alpha_d}$. I don't understand why you write ''$T^{\alpha}$ makes sense precisely because of the assumption that $T_iT_j=T_jT_i$ for all $i,j$.''. $\endgroup$ – Schüler Jan 17 at 14:15
  • $\begingroup$ Also I don't understand why $$\|T^n\|^2=\sum_{|\alpha|=n}c_{\alpha}\|T^{\alpha}\|^2 ?$$ Thank you. $\endgroup$ – Schüler Jan 17 at 14:15
  • $\begingroup$ 1. because otherwise how could you express e.g. the component $T_2T_1$ (if $d=2$)? In the notation $T_1^{\alpha_1}\dots T_d^{\alpha_d}$ the $T_1$ component always comes first.\\ 2. This follows from the definition of the norm of an operator with multiple components, as all the components of $T_n$ are in the form $T^{\alpha}$ with $|\alpha|=n$ (again assuming that $T_iT_j=T_jT_i$). $\endgroup$ – Lorenzo Quarisa Jan 17 at 14:22
  • $\begingroup$ You mean ''$T^{\alpha}$ makes sense precisely because of the assumption that $T_iT_j=T_jT_i$ for all $i,j$.'' when $\alpha = (\alpha_1,\cdots,\alpha_d) \in \mathbb{N}^d$ which is defined as $$\alpha_k=\#\{j\in\{1,\cdots,n\}\,;\;i_j=k\},$$ do you agree with me? $\endgroup$ – Schüler Jan 17 at 14:41
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    $\begingroup$ I think here '' which correspond to the multi-index $\alpha$, '' $\alpha$ is not arbitrary but it is given as $$\alpha_k=\#\{j\in\{1,\cdots,n\}\,;\;i_j=k\},$$ for all $k=1,\cdots,d$. $\endgroup$ – Schüler Jan 17 at 14:57

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