Is the transformation matrix of an upper triangular matrix to its Jordan normal form always triangular?

Question

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 ?

Answer 1

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.