Consider the Jordan canonical form below \begin{equation*} J = \begin{pmatrix} J_2(\lambda_1)&0& 0& 0\\ 0&J_1(\lambda_2) &0& 0\\ 0& 0& J_3(\lambda_1)& 0\\ 0 &0 &0 &J_2(\lambda_2) \end{pmatrix} \end{equation*} \begin{equation*} = \begin{pmatrix} \lambda_1& 1& 0& 0& 0& 0& 0& 0\\ 0 &\lambda_1 &0& 0& 0& 0& 0& 0\\ 0 &0& \lambda_2 &0& 0& 0& 0& 0\\ 0& 0& 0& \lambda_1& 1& 0& 0 &0\\ 0 &0& 0& 0& \lambda_1 &1& 0& 0\\ 0 &0& 0& 0& 0& \lambda_1 &0& 0\\ 0 &0& 0& 0& 0& 0 &\lambda_2 &1\\ 0& 0& 0& 0& 0& 0& 0& \lambda_2 \end{pmatrix}_{8\times 8} \in M_8(F) \end{equation*} where $\lambda_1 \neq \lambda_2$ in a field $F$.
Let \begin{equation*} B := (e_1 = \begin{pmatrix}1\\0\\\vdots\\0\end{pmatrix},\cdots e_8 = \begin{pmatrix}0\\0\\\vdots\\1\end{pmatrix} ) \end{equation*} be the standard basis of $F_c^8$ . Then we can calculate: \begin{eqnarray*} p_J(x) &=& (x - \lambda_1)^5(x - \lambda_2)^3,\\ m_J(x) &=& (x - \lambda_1)^3(x - \lambda_2)^2,\\ V_{\lambda_1} (J)& =& \text{Span}\{e_1, e_4\},\\ V_{\lambda_2} (J) &= &\text{Span}\{e_3, e_7\}. \end{eqnarray*}
Then my query is after reordering of $J$, we can also have basis of eigenspace of $\lambda_1$ to be $span\{e_1,e_3\}$ and basis of eigenspace of $\lambda_2$ to be $span\{e_6,e_7\}$ and everytime you rearrange it will be different. Will that cause any problem? The book said the Jordan canonical form matrix $J$ are still similar after reordering, but how come their eigenspace basis is so different for each reordering? Is it normal?