Suppose that $X$ is a finite dimensional Hilbert space. Let $A\_{1},\dots,A\_{r}:X\rightarrow X$ be linear operators. Then define the multi-spectral radius of $(A\_{1},\dots,A\_{r})$ to be
$$\limsup\_{n\rightarrow\infty}(\sum\_{a\_{1},\dots,a\_{n}\in\{1,\dots,r\}}\|A\_{a\_{1}}\dots A\_{a\_{n}}\|)^{1/n}.$$


If $r=1$, then the multi-spectral radius of $(A\_{1},\dots,A\_{r})$ is simply the spectral radius of $A\_{1}$ which is equal to $\max(|\lambda\_{1}|,\dots,|\lambda\_{s}|)$ where $\lambda\_{1},\dots,\lambda\_{s}$ is an enumeration of the eigenvalues of $A\_{1}$. Is there a similar characterization of the multi-spectral radius in terms of eigenvalues or something similar?


Since this question can be formulated in terms of Banach algebras or other spaces, feel free to answer this question in a more general context.


If an answer in the general case is hard to obtain, then I would still be interested in the case where $A\_{1},\dots,A\_{r}$ are satisfy some condition (normal operators etc.), but I want the operators $A\_{1},\dots,A\_{r}$ to not commute with each other.


As we all expect, this question is motivated by some questions I have about large cardinals.



Spectral radius for multiple linear operators
Yes. There is a characterization of the multi-spectral radius in terms of the ordinary spectral radius.


Suppose that $A$ is a unital Banach algebra. If $a\_{1},\dots,a\_{r}\in A$, then define the multi-spectral radius of $a\_{1},\dots,a\_{r}$ to be
$\limsup\_{n\rightarrow\infty}(\sum\_{i\_{1},\dots,i\_{n}\in\{1,\dots,r\}}\|a\_{i\_{1}}\dots a\_{i\_{n}}\|)^{1/n}$.


I claim that the multi-spectral radius of $a\_{1},\dots,a\_{r}$ is the maximum value of a spectral radius $\rho(x\_{1}\iota(a\_{1})+\dots+x\_{n}\iota(a\_{r}))$ where $\iota:A\rightarrow B$ is an isometric embedding of unital Banach algebras and where $x\_{i}\iota(a\_{j})=\iota(a\_{j})x\_{i}$ for all $i$ and $j$ and where $\|x\_{i}\|\leq 1$ for all $i$.


Observe that if $\iota:A\rightarrow B$ is an isometric embedding of unital Banach algebras and where $x\_{i}\iota(a\_{j})=\iota(a\_{j})x\_{i}$ for all $i$ and $j$ and where $\|x\_{i}\|\leq 1$ for all $i$, then


$$\rho(x\_{1}\iota(a\_{1})+\dots+x\_{r}\iota(a\_{r}))$$
$$=\lim\_{n\rightarrow\infty}\|(x\_{1}\iota(a\_{1})+\dots+x\_{r}\iota(a\_{r}))^{n}\|^{1/n}
=\lim\_{n\rightarrow\infty}\|\sum\_{i\_{1},\dots,i\_{n}\in\{1,\dots,r\}}x\_{i\_{1}}\iota(a\_{i\_{1}})\dots x\_{i\_{n}}\iota(a\_{i\_{n}})\|^{1/n}$$
$$\leq\limsup\_{n\rightarrow\infty}[\sum\_{i\_{1},\dots,i\_{r}\in\{1,\dots,n\}}\|x\_{i\_{1}}\dots x\_{i\_{n}}\iota(a\_{i\_{1}})\dots\iota(a\_{i\_{n}})\|]^{1/n}$$
$$\leq\limsup\_{n\rightarrow\infty}[\sum\_{i\_{1},\dots,i\_{r}\in\{1,\dots,n\}}\|x\_{i\_{1}}\|\dots \|x\_{i\_{n}}\|\cdot\|\iota(a\_{i\_{1}})\dots\iota(a\_{i\_{n}})\|]^{1/n}$$
$$\leq\limsup\_{n\rightarrow\infty}[\sum\_{i\_{1},\dots,i\_{r}\in\{1,\dots,n\}}\|\iota(a\_{i\_{1}})\dots\iota(a\_{i\_{n}})\|]^{1/n}
\leq\limsup\_{n\rightarrow\infty}[\sum\_{i\_{1},\dots,i\_{r}\in\{1,\dots,n\}}\|a\_{i\_{1}}\dots a\_{i\_{n}}\|]^{1/n}.$$


From the following proposition, one can choose $x\_{1},\dots,x\_{r},B,\iota$ such that $\rho(x\_{1}\iota(a\_{1})+\dots+x\_{r}\iota(a\_{r}))$ is the multi-spectral radius of $A$.


Proposition: For each unital Banach algebra $A$ and cardinal $\lambda$, there is a Banach algebra $B$ and an isometric embedding $\iota:A\rightarrow B$ along with a subset $X\subseteq B$ where $|X|\geq\lambda$ such that $\|x\|\leq 1$ for each $x\in X$ and such that $$\|(\iota(a\_{1})x\_{1}+\dots+\iota(a\_{r})x\_{r})^{n}\|=\sum\_{i\_{1},\dots,i\_{n}\in\{1,\dots,r\}}\|a\_{i\_{1}}\dots a\_{i\_{n}}\|$$ whenever
$a\_{1},\dots,a\_{r}\in A$ and $x\_{1},\dots,x\_{r}$ are distinct elements in $X$.


Proof: Let $X$ be a set disjoint from $A$ with $|X|\geq\lambda$. Let $B$ be the collection of all sums of the form $\sum\_{w\in X^{\*}}w\cdot a\_{w}$ where $a\_{w}\in A$ for all $w\in X^{\*}$ and where $\sum\_{w\in X^{\*}}\|a\_{w}\|<\infty$. Define a norm on $B$ where we set $\|\sum\_{w\in X^{\*}}w\cdot a\_{w}\|=\sum\_{w\in X^{\*}}\|a\_{w}\|$.


$B$ is clearly a vector space. Define multiplication by letting
$$(\sum\_{u\in X^{\*}}u\cdot a\_{u})\cdot(\sum\_{v\in X^{\*}}v\cdot a\_{v})
=\sum\_{u,v\in X^{\*}}uv\cdot a\_{u}a\_{v}.$$
Observe that multiplication on $B$ satisfies the inequality $\|\mathbf{b}\mathbf{c}\|\leq\|\mathbf{b}\|\cdot\|\mathbf{c}\|$. One can now check that $B$ is a Banach space, but if you don't want to check that $B$ is a Banach space, then the proof still goes through when you replace $B$ with its completion.


The mapping $\iota:A\rightarrow B$ where $\iota(a)=\epsilon\cdot a$ and $\epsilon$ denoted the empty word is a Banach algebra embedding. All the conditions of the theorem are satisfied.


Q.E.D.


From the above proposition, we know that


$$\rho(x\_{1}\iota(a\_{1})+\dots+x\_{r}\iota(a\_{r}))=\lim\_{n\rightarrow\infty}\|(x\_{1}\iota(a\_{1})+\dots+x\_{r}\iota(a\_{r}))^{n}\|^{1/n}$$


$$=\lim\_{n\rightarrow\infty}(\sum\_{i\_{1},\dots,i\_{n}\in\{1,\dots,r\}}\|a\_{i\_{1}}\dots a\_{i\_{n}}\|)^{1/n},$$


so the limit superior in the definition of the multi-spectral radius is actually just a limit.


Here is a conjecture that I did not have too much time to think about.


Conjecture: Whenever $A\_{1},\dots,A\_{r}$ are complex matrices, then the multi-spectral radius of $A\_{1},\dots,A\_{r}$ is the supremum of the spectral radii of matrices of the form $A\_{1}\otimes U\_{1}+\dots+A\_{r}\otimes U\_{r}$ where there exists an $n$ where $U\_{1},\dots,U\_{r}$ are $n\times n$-unitary matrices .


