Let $A$ and $B$ be $n\times n$ matrices over $\mathbb{C}$ . Then $A$ and $B$ are called, as I say, $\mathbb{C}^*-similar$ if there exists a non-zero scalar $k\in\mathbb{C}^*$ and an invertible matrix $P$ such that $B=kPAP^{-1}$.

Note that it is an equivalence relation. I guess that may be defined by someone before probably. But I checked it on Google but nothing was found. So I want to ask the relevant references for this. Thanks in advance.


This just means that $A$ is similar to a non-zero multiple of $B$ (or some non-zero multiple of $A$ is similar to $B$). Now the Jordan normal form of $kB$ is obtained from the Jordan normal form of $B$ by just multiplying all the eigenvalues by $k$. (To see this, observe that $kB$ is similar to $k$ times the Jordan normal form of $B$. Then it is easy to see that all the $k$'s above the main diagonal can be changed into $1$'s without changing the similiarity type) Thus the equivalence relation you propose is that the two matrices have the same Jordan normal form up to multiplication of all eigenvalues by a non-zero factor.

  • $\begingroup$ Are you sure that the Jordan form of $kB$ is $k$ times the Jordan form of $B$? The matrices $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and $\begin{bmatrix} 0 & k \\ 0 & 0 \end{bmatrix}$ both have the same Jordan form (namely the former matrix). $\endgroup$ – mechanodroid Sep 3 '18 at 8:09
  • $\begingroup$ @mechanodroid: Thanks for pointing out the mistake, I have edited the reply accordingly. $\endgroup$ – Andreas Cap Sep 3 '18 at 10:18
  • $\begingroup$ Many thanks. Is there any reference or paper about this? Because I would like to write something and want to cite this result, but I do not know which one to cite. $\endgroup$ – XSR Sep 3 '18 at 12:17
  • $\begingroup$ I don't know a reference, but what I have written there is essentially a complete argument. $\endgroup$ – Andreas Cap Sep 3 '18 at 12:52

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