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Assume that $A$ is an upper triangular matrix. In the case where $A$ is 2x2, I've checked that a transformation matrix $P$ such that $J = P^{-1}AP$, with $J$ Jordan normal form of $A$, is always upper triangular. I think the same is true for every matrix of dimension $n$ (probably a proof by induction), but I've not seen that anywhere. Any source ?

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Pedantic note: You don't want to say that every matrix $P$ satisfying $J = P^{-1} A P$ will be upper-triangular (this is false already for $n = 2$). You want to say that there exists some invertible upper-triangular matrix $P$ satisfying $J = P^{-1} A P$.

Anyway, this is false for $n = 3$. Indeed, take \begin{equation} A = \begin{pmatrix} 2 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix} . \end{equation} Its Jordan normal form is \begin{equation} \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix} . \end{equation} Thus, in order to bring it into its Jordan normal form, the two $2$'s on its diagonal need to be brought together. But conjugation by an upper-triangular matrix cannot do this: Indeed, it is easy to check that if you conjugate an upper-triangular matrix $B$ by an upper-triangular matrix, then the diagonal entries of $B$ remain unchanged.

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  • $\begingroup$ It would be interesting to find the additional assumptions that make this claim work, maybe, for example, "all the diagonal coefficients are equal". $\endgroup$ – MikeTeX Jan 10 at 13:04
  • $\begingroup$ @MikeTeX: Equality of diagonal entries is not enough. $A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ is a counterexample. This time I don't have a slick proof; I just computed the result of conjugating $A$ by a general upper-triangular matrix: $\begin{pmatrix} a & x & y\\ 0 & b & z\\ 0 & 0 & c \end{pmatrix} ^{-1}\begin{pmatrix} 1 & 0 & 1\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} a & x & y\\ 0 & b & z\\ 0 & 0 & c \end{pmatrix} = \begin{pmatrix} 1 & 0 & \dfrac{c}{a}\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} $, which is not a Jordan normal form. $\endgroup$ – darij grinberg Jan 10 at 13:09

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