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

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 ?

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 $$$$A = \begin{pmatrix} 2 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix} .$$$$ Its Jordan normal form is $$$$\begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix} .$$$$ Thus, in order to bring it into its Jordan normal form, the two $$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.
• @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. – darij grinberg Jan 10 at 13:09