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?

  • $\begingroup$ If $J_1$ and $J_2$ are two Jordan Normal Forms of a matrix $A$, then $J_1$ and $J_2$ must be similar and have the same spectrum. Basically, one can get from $J_1$ to $J_2$ by similarity transformations. $\endgroup$ – amarney Oct 30 '17 at 21:15
  • $\begingroup$ Also, I just wanted to add, everything you said is basically correct. The eigenspaces do get scrambled as you say (similarity transformations can be seen as a change of basis, so the eigenvector basis gets mapped to a new basis). The advantage of using the Jordan Canonical form is that it allows one to study $J$ instead of $A$, which may in general be very complicated looking. $J$ tells you everything you could want to know about the eigenvalues of a linear operator, allowing one to prove many things. $\endgroup$ – amarney Oct 30 '17 at 21:40

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.


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