Let $A$ be an $n\times n$ matrix and $v_o$ be an $n$-dimensional initial starting vector. Define the sequence $(v_k)$ such that $v_k=A^nv_o$ (or alternatively $v_{k+1} = Av_k$). I'm being asked to prove that for any such sequence $(v_k)$, if the sequence converges to some vector $\lim_{k\to\infty}(v_k)=x$, then $Ax=x$.
Although I intuitively understand why this is true, I'm struggling to prove the statement because $x$ may not actually be an element of the sequence. Any help?
Hint: Matrix multiplication is continuous. In other words, whenever a sequence $x_n$ converges to $x$, we have $\lim_{n \to \infty} Ax_n = Ax$.