5
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ A brief remark concering the scaling in your formula: if $A_1 = \dots = A_r = \operatorname{id}$, then the multi-spectral radius of $(A_1,\dots,A_r)$ becomes $r$ rather than $1$. Is this consistent with your intentions when defining the multi-spectral radius? $\endgroup$ Feb 17, 2019 at 16:35
  • $\begingroup$ @JochenGlueck. Yes. I intend for the multi-spectral radius to be a sort of "sum of spectral radiuses." $\endgroup$ Feb 17, 2019 at 16:45
  • 1
    $\begingroup$ If we call your quantity $R$, then, trivially, $R\ge \max r(A_j)$, and I think that's all you can say in general. For example, take $A_1, A_2$ as (multiples of) projections onto orthogonal vectors. $\endgroup$ Feb 17, 2019 at 17:58
  • 1
    $\begingroup$ On the other hand, if we take a $2\times 2$ Jordan block with eigenvalue $0$ as one matrix and its adjoint as the second, then $R=1$ even though the individual matrices have spectral radius $0$. So the individual spectra don't really tell you that much about $R$. $\endgroup$ Feb 18, 2019 at 16:28
  • $\begingroup$ If $X$ is a Banach algebra, $U$ is an open subset of $\mathbb{C}$ and $f:U\rightarrow\mathbb{C}$ is holomorphic, then the mapping $z\mapsto\log(\rho(f(z)))$ is subharmonic. Therefore, by an $r$-fold application of the maximum principle, if $A_{1},...,A_{r}\in X$, then $$\max\{\rho(e_{1}A_{1}+\dots+e_{r}A_{r}):|e_{1}|\leq 1,\dots,|e_{r}|\leq 1\}$$ $$=\max\{\rho(e_{1}A_{1}+\dots+e_{r}A_{r}):|e_{1}|=1,\dots,|e_{r}|=1\}\leq R.$$ $\endgroup$ Feb 18, 2019 at 19:45

1 Answer 1

0
$\begingroup$

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 .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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