# Can an upper triangular matrix be put in jordan form using upper triangular matrices and permutations only.

Give an upper triangular matrix A does there always exist $P, P^{-1}$ and $U, U^{-1}$ such that $PUAU^{-1}P^{-1}$ is in Jordan canonical form and U is upper triangular and P is a permutation matrix.

I believe the answer is yes as when we have distinct eigenvalues we can place them together and make diagonal block and that block can be diagonalized by an upper triangular matrix so we need only concern ourselves with the repeated eigenvalues case. when we have repeated eigenvalues we can use permutation matrices in a block say $B_j$ where $j=1,...,q$ where q is the number of distinct eigenvalues. if we put with the eigenvalues down the diagonals (which we can) then we consider $(B_j-\lambda )^i e_i$ for $i=1,2,...$ we have a collection of i generalized eigenvectors for each distinct eigenvalue. this will allow us to find n generalized eigenvectors (the distinct case is obvious) where in the $v_i$th vector will have 0's in the all components after the ith entry. I believe this is equivalent to the desired result.

Field is obviously $\Bbb C$ arbitrary $n \times n$ matrix.

Notice that it is not true for A not upper triangular.

Edit!: I guess we could technically do this over an arbitrary field given that it is upper triangular at least over $\Bbb R$ would be ok as all the eigenvalues would already be real if it's in upper triangular form.

We bring a matrix to JNF by changing to a basis of (generalized) eigenvectors, so we want to show that such a change of basis can be implemented by an upper triangular matrix. Equivalently, we want to show that there is a basis $v_1, \ldots , v_n$ of generalized eigenvectors, such that the components of $v_j$ are zero beyond the $j$th entry.
The eigenvalues of $A$ are its diagonal entries, and if these never repeat, then this is trivial. So the only challenge are multiple eigenvalues, and let's say we have a block that looks like $$B = \begin{pmatrix} a & * & * \\ 0 & a & * \\ 0 & 0 & a \end{pmatrix} ,$$ and $$A = \begin{pmatrix} A_0 & * & * \\ 0 & B & * \\ 0 & 0 & A_1 \end{pmatrix} ,$$ with the $A_j$'s upper triangular also. Of course, in the actual matrix, these entries $a$ don't need to follow each other, but for this, we can make use of the permutation matrix.
Again, it's clear that $(B-a)^j e_j = 0$ for $j=1,2,3$, and by giving these vectors suitable earlier components, we obtain generalized eigenvectors of $A$ also.