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

  • $\begingroup$ 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

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 $$ c_{\alpha}=\frac{n!}{\alpha_1!\cdot \dots \cdot \alpha_d!}=\frac{n!}{\alpha!}$$

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.