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I know I need to prove both directions of the biconditional, but am having some trouble in both ways.

First, looking left to right, If we assume the limit equals zero, it means that the $A$ matrix keeps getting smaller each time we apply $A$. Thus, since eigenvalue represent the effect of the matrix the maximum would have to be less than one but this doesn't seem very rigorous.

Then, I'm not sure how to proceed by assuming the spectral radius is less than 1 except for doing the backwards reasoning of what I talked about above. Any help is appreciated.

the_candyman
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buffalodude
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    Are you allowed to use the spectral radius formula? – Kavi Rama Murthy Mar 07 '21 at 23:25
  • Discussion related to @KaviRamaMurthy's answer and ways of showing the fact https://math.stackexchange.com/questions/2491626/proof-of-gelfands-formula-without-using-rhoa-1-iff-lim-an-0?rq=1 – paperskilltrees Dec 29 '21 at 16:34
  • Therein a link to another @user1551's answer, which does not involve binomial coefficients stuff: https://math.stackexchange.com/questions/4303957/schur-stability-without-the-jordan-canonical-form/4308091#4308091
    P.S. Okay, I realised that you asked for Schur specifically. I hope the links will be helpful nevertheless.     – paperskilltrees Dec 29 '21 at 16:36

2 Answers2

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$\rho(A)=\lim\sup ||A^{n}\|^{1/n}$. If $\rho(A) <1$ and $r =\frac {1+\rho (A)} 2$ then $\|A\|^{n}<r^{n}$ for $n$ sufficiently large and $r^{n} \to 0$ since $r<1$.

Suppose $\|A^{n}\| \to 0$. There exists $k$ such that $\|A^{k}\|< \frac 1 2$. If $\lambda $ is an eigen value of $A$ then $\lambda ^{k}$ is an eigen value of $A^{k}$ so $|\lambda^{k} |<1$ which implies $|\lambda|<1$. Hence, $\rho(A) <1$.

Kavi Rama Murthy
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  • I suppose, this is not an accepted answer precisely because in the OP's curriculum the proof of Gelfand's formula $\rho(A)=\lim ||A^{n}|^{1/n}$ relies on the OP's question (e.g., Meyer's book and Wikipedia's page follow this route). The statement $\rho(A)=\lim\sup ||A^{n}|^{1/n}$ is slightly different though. Could you shed light on how to establish it directly? – paperskilltrees Dec 29 '21 at 16:17
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Let $A=UTU^\ast$ be a unitary triangulation. Then $\rho(T)=\rho(A)$. Since all norms are equivalent on a finite-dimensional vector space, we may assume that the norm in question is the induced $2$-norm (i.e. the largest singular value, a.k.a. operator norm or spectral norm) of a matrix. Thus $\|A^n\|=\|T^n\|$ and the problem reduces to showing that $\lim_{n\to\infty}\|T^n\|=0$ if and only if $\rho(T)<1$ for any upper triangular matrix $T$.

Suppose $\lim_{n\to\infty}\|T^n\|=0$. Since $0\le\rho(T)^n=\rho(T^n)\le\|T^n\|$, we have $\lim_{n\to\infty}\rho(T)^n=0$ and hence $\rho(T)<1$.

Conversely, suppose $\rho(T)<1$. Let $T$ be $m\times m$ and write $T=D+F$, where $D$ and $F$ are respectively the diagonal part and strictly upper triangular part of $T$. Since the product of a diagonal matrix and a strictly upper triangular matrix is strictly upper triangular, and the product of $m$ or more $m\times m$ strictly upper triangular matrices is zero, if we expand $T^n=(D+F)^n$ as a sum of monomials in $D$ and $F$, all terms containing at least $m$ copies of $F$ are zero. It follows that when $n\ge m-1$, $$ \|T^n\| \le\sum_{k=0}^{m-1}\binom{n}{k}\|F\|^k\|D\|^{n-k} =\sum_{k=0}^{m-1}\binom{n}{k}\|F\|^k\rho(T)^{n-k}. $$ Since $\rho(T)<1$, we have $\lim_{n\to\infty}\binom{n}{k}\rho(T)^{n-k}=0$ for each $k$. Consequently, $\lim_{n\to\infty}\|T^n\|=0$.

user1551
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