TITLE: If two matrices have the same eigenvalues and eigenvectors are they equal?
QUESTION [33 upvotes]: 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?

REPLY [15 votes]: Besides needing a complete base of eigenvectors (as others pointed out), order is important if your "equal" is componentwise-equal. The following two matrices have the same set of eigenvalues and the same set of eigenvectors:
$$A  = \begin{pmatrix}
2&0\\
0&1\\
\end{pmatrix}\;\;\text{and}\;\;
B  = \begin{pmatrix}
1&0\\
0&2\\
\end{pmatrix}\,.
$$
(Conversely, having different eigenvectors does not necessarily mean the matrices are different. Any base of each eigenspace works as eigenvectors. (Think scaling or repeated eigenvalues.))