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:
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 ?
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.