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I am working on this problem.

Let $A \in \mathbb{R}^{n \times n}$ be an upper triangular matrix and suppose that for $i = 1, \dots, n$, $|a_{ii}| < 1$. Show that $\lim_{k \to \infty} A^k = 0$.

What I have done is the following. Let $k = mn$. Then $\lim_{k \to \infty} A^k = \lim_{m \to \infty} A^{mn} = (\lim_{m \to \infty} A^m)^n = 0$, because the limit will give me a strictly upper triangular matrix due to $|a_{ii}| < 1$, and strictly upper triangular matrices are nilpotent. Is this correct? There must be a better solution. Any help is appreciated.

user123456
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    Techincally, this only proves that if the limit exists, then it is zero: while it is true that $[A^m]{ii}\to 0$ for all $i$, you haven't proved that $[A^m]{ij}$ converges for $j>i$. – Sassatelli Giulio Jun 18 '24 at 19:22
  • I see. But then it seems I don't have enough information to conclude the limit exists. There will we entries that won't be multiplied by the diagonal entries and hence won't be bounded in any way. – user123456 Jun 18 '24 at 19:31
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    The usual idea would be to exploit the existence of a matrix $P$ such that $PAP^{-1}$ is in Jordan normal form and then calculate the powers of each Jordan block directly (namely, for $j>i$, $(J(\lambda,k)^m)_{ij}=\lambda^{m-j+i}\binom m{j-i}\stackrel{m\to\infty}\longrightarrow 0$ because $j-i<k$ and $\binom m{j-i}$ is a polynomial in $m$ of degree $< k$). – Sassatelli Giulio Jun 18 '24 at 19:52
  • Can I assume there is a Jordan form if $A$ is in $\mathbb{R}^{n \times n}$? – user123456 Jun 18 '24 at 19:58
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    @user123456 If the matrix is triangulable (which it is because it is triangular already), then yes. A matrix $X$ is tirangulable in $\Bbb F^{n\times n}$ if and only if the following equivalent conditions are met: 1) there is some $P\in GL(n,\Bbb F)$ such that $PXP^{-1}$ is upper triangular; 2) there is some $P\in GL(n,\Bbb F)$ such that $PXP^{-1}$ is lower triangular; 3) all the roots of $\det(A-tI)$ are in $\Bbb F$. – Sassatelli Giulio Jun 18 '24 at 20:06
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    There are other ways to see things. A low(er)-tech one could be to call $H_{ii}=\max\limits_i \lvert a\rvert_{ii}$, $H_{ij}=\max\limits_{h<k} \lvert a\rvert_{hk}$ if $i<j$ and $0$ if $i>j$, then prove that $\lvert [A^m]{ij}\rvert\le \lvert [H^m]{ij}\rvert$, then prove that $H^m\to 0$. That's because $H=\lambda I+N$ with $N$ nilpotent, therefore, for $m\ge n$, $H^m=\sum_{j=0}^{n-1} \lambda^{m-j}\binom mjN^j$. – Sassatelli Giulio Jun 18 '24 at 20:13
  • I will try this second thing. Thank you. – user123456 Jun 18 '24 at 20:25

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It is equivalent to prove each column $A^k e_j$ of $A^k$ goes to $0$. Set \begin{eqnarray*} c & = & \sup_{1 \leq i \leq n} | a_{i i} |\\ & < & 1. \end{eqnarray*} For $k \in \mathbb{N}$, \begin{eqnarray*} A^k e_1 & = & a_{1 1}^k e_1,\\ A^k e_2 & = & A^{k - 1} (a_{1 2} e_1 + a_{2 2} e_2)\\ & = & a_{1 2} a_{1 1}^{k - 1} e_1 + a_{2 2} A^{k - 1} e_2,\\ \| A^k e_2 \| & \leq & | a_{1 2} | c^{k - 1} + c \| A^{k - 1} e_2 \|\\ & \leq & | a_{1 2} | c^{k - 1} + | a_{1 2} | c^{k - 1} + c^2 \| A^{k - 2} e_2 \|\\ & \leq & \ldots\\ & \leq & k | a_{1 2} | c^{k - 1} + c^k . \end{eqnarray*} Hence $A^k e_1 \rightarrow 0$ and $A^k e_2 \rightarrow 0$. Similar estimates can show that $A^k e_j \rightarrow 0$ for $j = 3, \ldots, n$.

Mason
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  • Wouldn't there be a problem when you take the limit as $k \to \infty$? The right hand side of the last inequality depends directly on $k$. – user123456 Jun 19 '24 at 15:11
  • @user123456 Yeah it goes to $0$ as $k \to \infty$. For any $p > 0$, $k^p c^k \to 0$. – Mason Jun 19 '24 at 16:02