The question stems from a problem i stumbled upon while working with eigenvalues. Asking to explain why $A^{100}$ is close to $A^\infty$

$$A= \left[ \begin{array}{cc} .6 & .2 \\ .4 & .8 \end{array} \right] $$ $$A^\infty= \left[ \begin{array}{cc} 1/3 & 1/3 \\ 2/3 & 2/3 \end{array} \right] $$ Answer being given that (skipping calculations) that $A$ has eigenvalues $\lambda_1=1$ and $\lambda_2=0.4$ with eigenvectors $x_1=(1,2)$ and $x_2=(1,-1)$, and $A^{\infty}$ has eigenvalues $\lambda_1=1$ and $\lambda_2=0$, with same eigenvectors, while $A^{100}$ has eigenvalues $\lambda_1=1$ and $\lambda_2=(0.4)^{100}$ with same eigenvectors as the others, concluding that as the eigenvectors are the same and the eigenvalues are close comparing $A^\infty$ and $A^{100}$ they must be close.

Creating the basis for my question, how can one conclude that two matrices with same eigenvectors and close/equal eigenvalues are close/equal to each other?

My initial thoughts is that two matrices with equal eigenvectors and eigenvalues founds the basis for the same transformation which is why they are equal - Am I completly off?