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Are there two $ n\times n $ matrices $A$ and $B$ such that $ \lim_{m\to\infty} A^m$ and $ \lim_{m\to\infty} B^m $ both exists but $ \lim_{m\to\infty} (A \cdot B)^m $ doesn't ?

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2 Answers 2

up vote 3 down vote accepted

It's not true for $n=2$, take $A=\pmatrix{0&1\\ 0&0}$ and $B=\pmatrix{0&0\\\ -1&0}$. Then for $m\geq 2$ $A^m=B^m=0$ but $AB=\pmatrix{-1&0\\\ 0&0}$ so $\lim_{m}(AB)^m$ doesn't exists.

For a general $n\geq 2$, just take $A$ and $B$ with other entries $0$.

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Consider $$ A = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 0\end{pmatrix} \qquad B = A^T $$ Then $\forall m \geqslant 3$, $A^m = B^m = 0$, but since $$ A \cdot B = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 2\end{pmatrix} $$

$$ \lim_{m \to \infty} (A \cdot B)^m = \lim_{m \to \infty} \begin{pmatrix} 0 & 0 & 0 \\ 0 & \phi ^{2-2 m}+\phi ^{2 m-2} & -\phi ^{1-2 m}+\phi ^{2 m-1} \\ 0 & -\phi ^{1-2 m}+\phi ^{2 m-1} & \phi ^{-2 m}+\phi ^{2 m} \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & \infty & \infty \\ 0 & \infty & \infty\end{pmatrix} $$

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