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

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

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

Let $n\in \mathbb{N}^*$ and ${\bf T}=(T_1,...,T_d) \in \mathcal{B}(F)^d$ such that the operators $T_k$ are not necessarily commuting, I want to prove that $$\|\mathbf{T}^n\|^2=\sum_{g\in \mathbf{G}(n,d)}\|\mathbf{T}_g\|^2\,$$ where $\mathbf{G}(n,d)$ is the set of all functions from $\{1,\cdots,n\}\to\{1,\cdots,d\}$ and $\mathbf{T}_g:=\prod_{i=1}^dT_{g(i)}$.

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  • $\begingroup$ You make a claim. Is your question asking how to prove the claim? If so, why do you believe the claim should be true? $\endgroup$
    – Yemon Choi
    Jan 19, 2019 at 12:12
  • $\begingroup$ Because I try with an example, such as d=2, n=3. $\endgroup$
    – Schüler
    Jan 19, 2019 at 12:17
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    $\begingroup$ I'm a bit confused; why isn't this true just by definition after expanding out? $\endgroup$ Jan 19, 2019 at 19:32

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I believe you meant to write $\mathbf{T}_g:=\prod_{i=1}^n T_{g(i)}$ so maybe a better symbol would be $\mathbf{T}_g^n:=\prod_{i=1}^n T_{g(i)}$.

Now, follwing @DongryulKim's suggestion, we can see why your claim holds by observing how $\mathbf{T}^n$ is formed.

Let us characterize functions $g \in G(n,d)$ by the values they take $g_{i_1, i_2, \dots , i_d} \in G(n,d)$, meaning $g_{i_1, i_2, \dots , i_d}(k) = i_k$ with $k \in {1, 2, \dots , n}$ and $i_k \in {1,2,\dots,d}$.

Clearly, when $N=1$ the components run over all the functions in $G(1,d) = \{g_1, g_2 , \dots, g_d \}$ since

$$\mathbf{T}^1 = (T_1, T_2, \dots , T_d) = \left(T_{g_1(1)}, T_{g_2(1)}, \dots, T_{g_d(1)} \right ) = \left(\mathbf{T}_{g_1}^1, \mathbf{T}_{g_2}^1, \dots, \mathbf{T}_{g_d}^1 \right ) $$

Let us assume now that this also holds for $N=n$, that is $$\mathbf{T}^n = (\mathbf{T}_{g_{1,1,\dots,1}}^n, \dots, \mathbf{T}_{g_{i_1,i_2,\dots,i_n}}^n, \dots , \mathbf{T}_{g_{d,d,\dots,d}}^n)$$

Now, going to $N=n+1$ we have

$$\mathbf{T}^{n+1} = (T_1 \mathbf{T}_{g_{1,1,\dots,1}}^n, \dots ,T_1 \mathbf{T}_{g_{d,d,\dots,d}}^n , T_2 \mathbf{T}_{g_{1,1,\dots,1}}^n, \dots, T_2 \mathbf{T}_{g_{d,d,\dots,d}}^n , \cdots , T_d \mathbf{T}_{g_{1,1,\dots,1}}^n, \dots, T_d \mathbf{T}_{g_{d,d,\dots,d}}^n )$$

It is easy now to see from this arrangement that the first block of components corresponds to terms $\mathbf{T}_g^{n+1}$ for functions $g \in G(n+1,d)$ with $g(1) = 1$ and $g(i) = g'(i-1)$ for any function $g' \in G(n,d)$ and $i>1$. Similarly, the second block corresponds to terms $\mathbf{T}_g^{n+1}$ for functions $g \in G(n+1,d)$ with $g(1) = 2$ and $g(i) = g'(i-1)$ for any function $g' \in G(n,d)$ and $i>1$. Following this all the way to the last block, where $g(1) = d$ we see that we have covered all the functions in G(n+1,d). Thus

$$\mathbf{T}^{n+1} = (\mathbf{T}_{g_{1,1,\dots,1}}^{n+1}, \dots,\mathbf{T}_{g_{i_1,i_2,\dots,i_d}}^{n+1},\dots,\mathbf{T}_{g_{d,d,\dots,d}}^{n+1} )$$

So, in general, we have

$$\mathbf{T}^n = (\mathbf{T}_{g_{1,1,\dots,1}}^n, \dots ,\mathbf{T}_{g_{i_1,i_2,\dots,i_d}}^n,\dots,\mathbf{T}_{g_{d,d,\dots,d}}^n )$$

from which we can get

$$ \|\mathbf{T}^n\|^2=\sum_{(i_1,...,i_n) \in \{1,...,d\}^n}\|\mathbf{T}_{g_{i_1,...,i_n}}^n\|^2 = \sum_{g\in \mathbf{G}(n,d)}\|\mathbf{T}_g^n\|^2 $$

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  • $\begingroup$ I don't understand why you write an element $g \in G(n,d)$ as $g_{i_1, i_2, \dots , i_d} \in G(n,d)$?Thanks. $\endgroup$
    – Schüler
    Jan 27, 2019 at 7:23
  • $\begingroup$ It is just an explanatory way to write the elements of $G(n,d)$ so that we can later range over them in a clearer way. It is just a symbolic way of saying that "$g_{i_1, i_2, \dots , i_d}$ is the function that sends $1$ to $i_1$, $2$ to $i_2$, etc." $\endgroup$
    – Sotiris
    Jan 27, 2019 at 9:21

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