# Norm of a tuple of operators

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 $$\|{\bf A}\|^2=\sum_{k=1}^d\|A_k\|^2.$$

For $${\bf T}=(T_1,...,T_d) \in \mathcal{B}(F)^d$$ and $${\bf S}=(S_1,\cdots,S_m)\in \mathcal{B}(F)^m$$, we 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 $${\bf T}=(T_1,...,T_d) \in \mathcal{B}(F)^d$$ such that the operators $$T_k$$ are not necessarily commuting, I claim that $$\|\mathbf{T}^n\|^2=\sum_{g\in \mathbf{G}(n,d)}\|\mathbf{T}_g\|^2\,$$ where $$g\in \mathbf{G}(n,d)$$ and $$\mathbf{T}_g:=\prod_{i=1}^dT_{g(i)}$$.

Note that $$\mathbf{G}(n,d)$$ denotes the set of all functions from $$\{1,\cdots,n\}$$ into $$\{1,\cdots,d\}$$.

How to prove that the claim is true?

Note that if the operators $$T_k$$ are commuting, it can be seen that $$\|\mathbf{T}^n\|^2=\sum_{|\alpha|=n}\frac{n!}{\alpha!}\|\mathbf{T}^{\alpha}\|^2.$$ 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 $$\mathbf{T}^\alpha:=T_1^{\alpha_1} \cdots T_d^{\alpha_d}$$.

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

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

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

• 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. – Schüler Jan 27 at 7:23
• 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." – Sotiris Jan 27 at 9:21