Jordan canonical form and reordering

Question

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?

Answer 1

Let $A \in \mathbb{C}^{n \times n}$ have the following two Jordan Decompositions: \begin{align*} A = X_1 J_1 X_1^{-1}, \quad A = X_2 J_2 X_2^{-1}. \end{align*} Here, $J_1, J_2 \in \mathbb{C}^{n\times n}$ is block-diagonal, and $X \in \mathbb{C}^{n\times n}$ is invertible. Then we have \begin{align*} X_1 J_1 X_1^{-1} = X_2 J_2 X_2^{-1} \implies J_1 = X_1^{-1} X_2 J_2 X_2^{-1}X_1. \end{align*} Now define $X_3 = X_1^{-1} X_2$, so that $X_3^{-1} = X_2^{-1} X_1$, then we have \begin{align*} J_1 = X_3 J_2 X_3^{-1}. \end{align*}

$J_2$ is a "reordering" of $J_1$, but since $J_1$ and $J_2$ are similar matrices, they represent the same linear operator. They will share the same properties that are characteristic of that operator, but they will not share the properties that are dependent on a choice of basis. So, for example, the geometric multiplicies of $J_1$ and $J_2$ will be the same, but the spans of the eigenvectors will probably be different. Wikipedia's discussion on matrix similarity may be enlightening.