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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 $\mathbf{T}^\alpha:=T_1^{\alpha_1} \cdots T_d^{\alpha_d}$.

Schüler
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  • have you tried induction in $n$ or $d$? – supinf Jan 17 '19 at 11:58
  • @supinf Yes I try induction on $n$ but from $n$ to $n+1$ it is not evident. – Schüler Jan 17 '19 at 12:03
  • Also, this question is the same as your old question https://math.stackexchange.com/questions/3050154/proof-explanation-related-to-multinomial-coefficients , except that you are now squaring all monomials (but that is tantamount to squaring all variables, i.e., setting $X_i = T_i^2$). – darij grinberg Jan 17 '19 at 14:54
  • @Schüler: Yes, it is true then, more or less by the definition. – darij grinberg Jan 17 '19 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 $$ c_{\alpha}=\frac{n!}{\alpha_1!\cdot \dots \cdot \alpha_d!}=\frac{n!}{\alpha!}$$

Lorenzo Q
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  • 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 $i,j$.''. – Schüler Jan 17 '19 at 14:15
  • Also I don't understand why $$|T^n|^2=\sum_{|\alpha|=n}c_{\alpha}|T^{\alpha}|^2 ?$$ Thank you. – Schüler Jan 17 '19 at 14:15

  • 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.\

  • 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 $T_iT_j=T_jT_i$).

    • – Lorenzo Q Jan 17 '19 at 14:22
  • 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? – Schüler Jan 17 '19 at 14:41
  • I am not saying that $T^{\alpha}$ is ill-defined without that assumption. It is always well defined. What I mean is you cannot describe all the components of the vector-valued operator $T^n$ using just elements of the form $T^{\alpha}$ without that assumption. But I agree that the way I put it was a bit misleading, so I have edited that part. – Lorenzo Q Jan 17 '19 at 14:44
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    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 $k=1,\cdots,d$. – Schüler Jan 17 '19 at 14:57
  • Yes, this follows directly from the definition of $T^{\alpha}$. So? I don't understand your point. – Lorenzo Q Jan 17 '19 at 15:00
  • Perhaps when someone read the proof, he believe that $\alpha$ is arbitrary. – Schüler Jan 17 '19 at 15:01
  • The equation you've written above is not a definition for $\alpha_k$; it is merely a property which follows from the definition of $T^{\alpha}$ (which is a different, separate object). Therefore one does not need to cite it before defining $\alpha$. At all effects $\alpha$ is an arbitrary multi-index with $|\alpha|=n$. – Lorenzo Q Jan 17 '19 at 15:04
  • I think that $\mathbf{T}^\alpha$ is always defined to be $T_1^{\alpha_1} \cdots T_d^{\alpha_d}$ even if the operators $T_k$ are not commuting. For example if $d=3$ and $\alpha=(2,1,3)$, then $\mathbf{T}^\alpha=T_1^2T_2T_3^3$. Also, frankly I don't understand this sentences:'' $c_{\alpha}$ 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 $\alpha$.'' Thank you for your help. – Schüler Jan 17 '19 at 15:38
  • For your first question I have already answered above, so please read my previous comments again. For your second question, the meaning of the sentence you quoted is written explicitly in mathematical terms in the line below. – Lorenzo Q Jan 17 '19 at 15:42
  • Please see my edit. Could you please help me to show that the claim is true?Thanks. – Schüler Jan 18 '19 at 20:14
  • This is a different question from the original request, so why don't you just open a new question? – Lorenzo Q Jan 18 '19 at 20:59