# prove that $\lim_{t\rightarrow \infty}(X(t) - A^{-1}(t)) = 0$

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

$$\lim_{t\rightarrow \infty}(X_t - A^{-1}_t) = \lim_{t\rightarrow \infty}((X_t A_t-I) \, A_t^{-1}) = \lim_{t\rightarrow \infty}(B_t A^{-1}_t)$$ 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 $C_t$.
Did you try in dimension 1 ? For instance, you can take $X(t) = 1+t$ and $A(t) = \frac{1}{t}$.