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I've come across the following question and am not sure why the answer makes sense.

Let $f,g \in End(\mathbb{C}^4)$ be neither nilpotent nor invertible with their characteristic and minimal polynomials being equal. Do they have the same Jordan form?

The answer to this question in my lecture notes is yes which implies that they are similar (correct me if I'm wrong). My theory is that because they have the same characteristic polynomial that means that they have the same eigenvalues with the same algebraic multiplicity. Furthermore, having the same minimal polynomial means that the size of the largest Jordan block is the same for both Jordan-forms. Could someone please expand on this/correct me?

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Your argument isn't complete because you haven't used the fact that the vector space is $4$-dimensional. In higher dimensions the statement can be wrong, as the following pair of matrices illustrates: $$ \pmatrix{0&1\\ 0&0\\ &&0&1\\ &&0&0\\ &&&&1}, \ \pmatrix{0&1\\ 0&0\\ &&0&0\\ &&0&0\\ &&&&1}. $$ On $\mathbb C^4$, suppose one of the two endomorphisms --- say $f$ --- has exactly two non-trivial Jordan blocks in its Jordan form. Since $\mathbb C^4$ is $4$-dimensional, the Jordan form of $f$ must be $J_2(\lambda_1)\oplus J_2(\lambda_2)$ for some $\lambda_1,\lambda_2\in\mathbb C$, where the expression $J_k(\lambda)$ denotes a $k\times k$ Jordan block for an eigenvalue $\lambda$. Note that $\lambda_1\ne \lambda_2$, otherwise $f$ is either nilpotent when $\lambda_1=0$, or invertible when $\lambda_1\ne0$. Hence the common minimal polynomial of $f$ and $g$ is $(x-\lambda_1)^2(x-\lambda_2)^2$ and the Jordan form of $g$ is $J_2(\lambda_1)\oplus J_2(\lambda_2)$ too.

The remaining case, where each of $f$ and $g$ has at most one non-trivial Jordan block, should be easy.

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  • $\begingroup$ Thanks for replying! What do you mean by non-trivial Jordan block? $\endgroup$ – RakoonBerry Jul 4 at 15:58
  • $\begingroup$ @RakoonBerry Any Jordan block $J_k(\lambda)$ of size $k>1$, i.e. any Jordan block whose characteristic/minimal polynomial is $(x-\lambda)^k$ for some $\lambda\in\mathbb C$ and some $k>1$. $\endgroup$ – user1551 Jul 4 at 16:31

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