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It seems that a Jordan basis, for, say, an operator $φ:V→V$ with $V$ a vector space can have several Jordan bases. However I don't see how this is true and I didn't achieve to prove it. Indeed I tried assuming the uniquness of a Jordan basis of $\varphi$ to achieve a contradiction but I did not find any.. I also looked up on mathstack but the only thing I found was this:

Jordan basis unique?

And it doesn't realy help because I'm more looking for the general case, which is unclear to me. Could anyone explain why this is true and help me proving it ?

Rhaena
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    What do you mean by "you're looking for the general case"? Your question is whether it's true that the Jordan basis is unique for every operator $\varphi$, and the question that you found gives a counterexample to this statement. – Ben Grossmann Apr 18 '21 at 10:05
  • Well there are several good explanations in that question of why a Jordan basis can never be unique. – ancient mathematician Apr 18 '21 at 10:05
  • You are saying "I tried assuming the uniqueness of a Jordan basis of $\varphi$ to achieve a contradiction but I did not find any" while at the same time giving a link to a question that explicitly lists multiple different Jordan bases for the same endomorphism, which of course contradicts uniqueness. Please clarify how your question isn't already answered by what you linked to. I'm voting to close this as a duplicate. – Christoph Apr 18 '21 at 10:15
  • Indeed @BenGrossmann I didn't see it that way ! And we can see it clearly since it's not true for the identity in fact, I think I had just confused myself so thanks. – Rhaena Apr 18 '21 at 10:17
  • @Christoph I agree with you I just confused myself while working on antoher question, it is indeed a duplicate. – Rhaena Apr 18 '21 at 10:20

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A simple proof: for any operator $\varphi$, any basis $\mathcal B_1 = \{v_1,\dots,v_n\}$, and any non-zero constant $k$, the basis $$ \mathcal B_2 = \{kv_1,\dots,kv_n\} $$ is such that the matrix of $\varphi$ with respect to $\mathcal B_1$ is equal to the matrix of $\varphi$ with respect to $\mathcal B_2$. Thus, if $\mathcal B_1$ is a Jordan basis for $\varphi$ (over any algebraically closed and therefore infinite field), we can now generate infinitely many other Jordan bases.

Ben Grossmann
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