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Suppose $A(t)$ is a time-varying square matrix which is always invertible, and we have $$ \lim_{t\rightarrow \infty}(X(t)A(t) - I) = 0. $$ Can we prove that $$ \lim_{t\rightarrow \infty}(X(t) - A^{-1}(t)) = 0 $$?

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BTW: kudos for the avatar :-) – leonbloy Dec 5 '12 at 13:51
@leonbloy I'm a big fan of Miyazaki and his cartoons! lol – Chen Dec 5 '12 at 14:19
up vote 3 down vote accepted

$$\lim_{t\rightarrow \infty}(X_t - A^{-1}_t) = \lim_{t\rightarrow \infty}((X_t A_t-I) \, A_t^{-1}) = \lim_{t\rightarrow \infty}(B_t A^{-1}_t)$$ where $B_t=X_t A_t-I$.

Hence we want to know if $\lim_{t\rightarrow \infty} B_t=0$, implies $\lim_{t\rightarrow \infty} B_t C_t=0$ for arbitrary $C_t$. Of course, that's not true in general, not even in the scalar case. Not unless we can bound $C_t$.

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Thanks very much!! – Chen Dec 5 '12 at 14:06

Did you try in dimension 1 ? For instance, you can take $X(t) = 1+t$ and $A(t) = \frac{1}{t}$.

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