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Let $V$ be a normed vector space, and let $T : V \to V$ be a bounded linear operator. Then the spectral radius of $T$, call it $r(T)$ is defined to be $\lim_{n \geq 1} \|T^n\|^\frac{1}{n}$, where $\|\cdot\|$ is the canonical operator norm. I would like to know what does this definition tell us intuitively.

For finite-dimensional linear operators, we can treat them as matrices and it is simply the largest absolute value of the eigenvalues (as a result of Gelfand's formula). I see it intuitively as the largest extent in which $T$ "expands" the vectors in $V$. However, in the infinite-dimensional case, there may not be any eigenvalues, so I'm not sure how to tweak my intuition for this case.

If possible, I would also like to have an explanation of the intuition behind Gelfand's formula. That is, why is the max of $|\lambda_i|$, the set of eigenvalues, precisely $\lim_{n \geq 1}\|T^n\|^\frac{1}{n}$?

Any help is appreciated.

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To answer your first question: Bounded operators over a Banach space fall in the category of Banach algebras, i.e. they have a "multiplication", which is simply the composition of operators. In any algebra that has a unit (this can be generalized to algebras that do not admit a unit, but that's partly irrelevant), one can define the spectrum of an element as $$\sigma(a)=\{\lambda\in\mathbb{C}: \lambda1_A-a\text{ is not invertible in } A\} $$ This can be done in any algebra. Why is this set interesting? Note for example that matrices fall in the category of Banach algebras as well, and, elementary linear algebra (or compact operator theory) yields that the spectrum of a matrix is equal to the set of its eigenvalues. Also, note that the space $C(X)$ of continuous functions over a compact Hausdorff space is also a Banach algebra, and here the spectrum of a function is its image. So the spectrum seems to unify important characteristics of elements of algebras in one notion.

Note that noone assures us that in an arbitrary algebra the spectrum is non-empty. A very important result by Gelfand is that in Banach algebras, the spectrum is always non-empty. So it always makes sense and is indeed interesting to know the quantity $\max|\lambda|$, which is precisely the spectral radius. Also the spectrum is compact (this is relatively easy) and it is contained in the closed disk $D(0,\|a\|)\subset\mathbb{C}$.

Now the question of interest: how do we estimate the spectral radius of an element? As said, a first estimate is $r(a)\leq\|a\|$.

If $c_0+c_1z+\dots+c_nz^n=p(z)\in\mathbb{C}[z]$ is a polynomial and $a\in A$ is an element of a unital Banach algebra, set $p(a):=c_01_A+c_1a+\dots+c_na^n$. Using the Fundamental theorem of algebra and the fact that two commuting elements are invertible iff their product is invertible, one gets the interesting equation $\sigma(p(a))=p(\sigma(a))$, that is the image of $\sigma(a)$ through $p(z)$. Observe now that if $\lambda\in\sigma(a)$ and $n\in\mathbb{N}$ we have $\lambda^n\in\sigma(a^n)$, therefore $|\lambda^n|\leq r(a^n)\leq\|a^n\|$. Thus $|\lambda|\leq \|a^n\|^{1/n}$. Taking supremum as $\lambda$ ranges over $\sigma(a)$ yields $r(a)\leq\liminf_{n\to\infty}\|a^n\|^{1/n}$. I believe this is enough to show why one would think that this limit exists and why it should be equal to $r(a)$: people probably couldn't find any example to counter this guess (that is reasonable due to this estimate), until Gelfand and Beurling proved this formula:

$$r(a)=\lim_{n\to\infty}\|a^n\|^{1/n}.$$ Hope this helps.

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In infinite dimensions the set of eigen values gets replaced by the so-called spectrum. Note that in finite dimensions $A-\lambda I$ has an inverse if $\lambda $ is not an eigen value of $A$. In infinite dimensions non-eigen values are replaced by $\lambda $ such that $T-\lambda I$ has a bounded inverse. The remaining complex numbers form the spectrum of $T$. The spectral radius is nothing but $\sup \{|\lambda|: \lambda \text { belongs to the spectrum of}\, T\}$.

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  • $\begingroup$ Okay, so intuitively why is $\sup\{|\lambda| : \lambda \in \sigma(T)\} = \lim_{n \geq 1} \|T^n\|^\frac{1}{n}$? $\endgroup$ Mar 21, 2020 at 6:15

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