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We know that the conjugacy classes of $A\in M_n(\mathbb{C})$ are determined by the characteristic polynomial of $A$ and a partition of $n$. Is there an analogous statement for upper triangular matrices? Since over an algebraically closed field a matrix $A$ is always triangularisable, I am hoping similar statements are true for upper triangular matrices.

But I am unable to find any reference for the same. If anyone knows any reference about the same, please let me know about it. Any help regarding this will be appreciated.

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  • $\begingroup$ It's not exactly what you describe. The characteristic polynomial determines a partition $n=\sum_{t\in\mathbf{C}} n_t$ and then the Jordan form determines a partition of each $n_t$. $\endgroup$
    – YCor
    Commented Jun 17, 2022 at 6:04
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    $\begingroup$ In any case, $E_{12}$ and $E_{23}$ are not conjugate by invertible upper triangular matrices. $\endgroup$
    – YCor
    Commented Jun 17, 2022 at 6:13
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    $\begingroup$ Over finite fields, the analogous problem is notoriously difficult. $\endgroup$
    – Geoff Robinson
    Commented Jun 17, 2022 at 10:27
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    $\begingroup$ I interpreted the last question differently: If the diagonal entries of $U$ and $V$ are equal to each other IN THE SAME ORDER, and are all distinct, are $U$ and $V$ conjugate? The answer is yes. Proof: We can assume that $V$ is diagonal. Let $(i,j)$ be the position with $U_{ij} \neq 0$ and $j-i$ minimal. Then, for an appropriate $t$, the matrix $(1+t E_{ij}) U (1+t E_{ij})^{-1}$ is $0$ in the $(i,j)$ position, and is still zero in all the places closer to the diagonal. (This computation uses that $U_{ii} \neq U_{jj}$. Continuing in this manner, we can make $U$ diagonal.) $\endgroup$
    – David E Speyer
    Commented Jun 17, 2022 at 14:07
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    $\begingroup$ The main issue seems to be the case of nilpotent matrices. $\endgroup$
    – YCor
    Commented Jun 19, 2022 at 7:42

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