In a paper it is claimed that the matrix algebra generated by a matrix $$J=\oplus_i J_{\lambda_i,m_i}$$ in Jordan normal form "is the algebra of block diagonal matrices with the same block partition, where each block contains an arbitrary upper triangular Toeplitz matrix".
I take issue with the arbitrary Toeplitz blocks - an explicit calculation for any polynomial $$p(J)=\sum_k p_k J^k=p\left(\bigoplus_i J_{\lambda_i,m_i}\right)=\bigoplus_i p\left(J_{\lambda_i,m_i}\right) \ ,$$ and thus any element of the generated matrix algebra yields $$\sum\limits_k\begin{bmatrix}p_k\lambda^k&p_kk\lambda^{k-1}&p_k(k^2-k)\lambda^{k-2}&\ddots&\ddots &\ddots&\ddots\\0&p_k\lambda^k&p_kk\lambda^{k-1}&\ddots&\ddots& p_kk!\lambda^{k-l}/(l!(k-l)!)&\ddots\\0&0&p_k\lambda^k&\ddots&\ddots&\ddots&\ddots\\0&0&0&\ddots&\ddots&\ddots&\ddots\\0&0&0&0&p_k\lambda^k&p_kk\lambda^{k-1}&p_k(k^2-k)\lambda^{k-2}\\0&0&0&0&0&p_k\lambda^k&p_kk\lambda^{k-1}\\0&0&0&0&0&0&p_k\lambda^k\end{bmatrix}$$
for the Toeplitz blocks. But the same polynomial has to be applied to all blocks and so how can the full matrix be composed of arbitrary upper triangular matrix blocks?
For example, I fail to see how one would obtain a single diagonal block with all other blocks being trivial zero Toeplitz submatrices.
What am I missing here?
If you refer to the paper Markova. O. V. & Novochadov, D. Yu., Generating Systems of the Full Matrix Algebra That Contain Nonderogatory Matrices, J. Math. Sci. 262, 99–107 (2022), available at https://doi.org/10.1007/s10958-022-05802-2, where this statement appears at the bottom of page 101, I note that in their context the Jordan matrix subject to this statement is assumed to be "nonderogatory," i.e., that different blocks correspond to different eigenvalues. I will use this hypothesis below.
Suppose then the matrix $J$ is made up of blocks $B_1, \dots, B_m$ with the block $B_j$ corresponding to the eigenvalue $\lambda_j$ and having size $k_j \times k_j$ (so that $B_j = \lambda_j I_{k_j} + J_{k_j}$ where $I_{k_j}$ denotes the $k_j \times k_j$ identity matrix and $J_{k_j}$ denotes the $k_j \times k_j$ matrix with $1$s immediately above its main diagonal and zeros everywhere else), and that the eigenvalues $\lambda_1, \dots, \lambda_m$ are distinct.
Fix $1 \leq i \leq m$. Letting $q(x) = \prod_{j \neq i} (x - \lambda_j)^{k_j}$, we have that $q(B_j) = 0$ for all $j \neq i$, and our assumption that none of the $\lambda_j$ for $j \neq i$ are eigenvalues for $B_i$ implies that the matrix $q(B_i)$ is a product of invertible matrices hence invertible. Thus $q(J)$ is the block matrix whose $i$th block is the invertible matrix $q(B_i)$ and whose other blocks are zero. There is a polynomial $p$ such that $p(q(B_i)) = q(B_i)^{-1}$ (see e.g. this MSE post for why the inverse of a matrix is a polynomial in that matrix), and it follows that $A := p(q(J)) \cdot q(J)$ is the block matrix whose $i$th block is $I_{k_j}$ and whose other blocks are zero (i.e., an example of "a single diagonal block with all other blocks being trivial zero Toeplitz submatrices").
It follows further that $p(q(J)) \cdot q(J) \cdot J$ is the block matrix whose $i$th block is $B_i$ and whose other blocks are zero, and that $B := p(q(J)) \cdot q(J) \cdot J - \lambda_i A$ is the block matrix whose $i$th block is what we called $J_{k_i}$ and whose other blocks are zero. The block matrix whose $i$th block is upper triangular and Toeplitz with arbitrary first row $a_0, a_1, \dots, a_{k_i - 1}$ and whose other blocks are zero, also known as $a_0 A + \sum_{n=1}^{k_i - 1} a_n B^n$, is thus in the algebra generated by $J$, because we have shown that $A$ and $B$ are.
Having done this for fixed $i$ in the $i$th block, we can just as easily do it every block and add the results together - arriving at what the authors refer to as the set of all "block diagonal matrices with the same block partition, where each block contains an arbitrary upper triangular Toeplitz matrix."
ADDED IN RESPONSE TO OP'S COMMENT:
I don't see anything that "goes wrong" with your reasoning, but I also don't see how your reasoning leads to the conclusion that one couldn't get arbitrary upper triangular Toeplitz matrices as blocks in the nonderogatory case. (The main thing I get out of that formula is that the general expression for $p(J)$ in terms of the coefficients of $p$ is not very easy to analyze.)
To be clear, the hypothesis that $J$ be nonderogatory is necessary: should $J$ have two blocks that correspond to the same eigenvalue, your formula shows that (at least) the diagonals of these two blocks will be equal for every element of the algebra (so in particular, those blocks cannot be taken to be an arbitrary pair of upper triangular Toeplitz matrices). More generally, you appear to observe (correctly) that at least in some circumstances the blocks of an element of the algebra might (by virtue of the fact that they have to arise out of the same polynomial in $J$) be "constrained," such that different blocks cannot be assigned arbitrarily. The key is that those circumstances are exactly what is ruled out by the hypothesis that $J$ be nonderogatory.
Consider an analogy to polynomial interpolation, specifically Lagrange interpolation (the special case of your problem where the Jordan blocks all have size $1$). If the complex numbers $b_1, \dots, b_m$ are distinct and arbitrary complex numbers $t_1, \dots, t_m$ are given, there is a polynomial $f$ satisfying $f(b_i) = t_i$ for all $1 \leq i \leq m$. One way to construct such an $f$ is first to construct polynomials $f_i$, $1 \leq i \leq m$, such that $f_i(b_j) = 1$ whenever $i=j$ and $f_i(b_j) = 0$ otherwise, and one way to construct such $f_i$ is to form $\prod_{j \neq i} (b_i - b_j)^{-1} \prod_{j \neq i} (x - b_j)$, where the idea for $f_i$ comes from noting that the polynomial $\prod_{j \neq i} (x - b_i)$ certainly vanishes at all of the $b_j$, $j \neq i$, and that if we scale it by the inverse of its value at $b_i$ (which exists by our distinctness hypothesis on the $b_1, \dots, b_m$), the resulting polynomial will still take the value $0$ at all of the $b_j$ with $j \neq i$, and now also take the value $1$ at $b_i$. It is then easy to check that $f := \sum_{i=1}^n t_i f_i$ solves the interpolation problem.
Our construction above does something similar, where the input data is Jordan blocks $B_1, \dots, B_m$ instead of scalars $b_1, \dots, b_m$ and the hypothesis that $J$ is nonderogatory takes the place of the distinctness of the scalars. For $1 \leq i \leq m$, the polynomial $q_i(x) = \prod_{j \neq i} (x - \lambda_j)^{k_j}$ has the property that it vanishes at each of the $B_j$, $j \neq i$, and its value at $B_i$ is invertible (as $J$ is nonderogatory). If we scale $q_i(x)$ by $p_i(q_i(x))$ (where $p_i$ is as in the above argument: its value at $q_i(B_i)$ is the inverse of $q_i(B_i)$), the resulting polynomial $f_i(x) := p_i(q_i(x)) q_i(x)$ will still take the value $0$ at all of the $B_j$ with $j \neq i$, and now also take the value $I_{k_i}$ at $B_i$. If arbitrary upper triangular Toeplitz blocks $T_1, \dots, T_m$ of the appropriate sizes are given, it is easy to find for each $1 \leq i \leq m$ a polynomial $t_i$ such that the $i$th block of $t_i(J)$ is $T_i$. Indeed, as we implicitly used in the above argument (or as your formula shows), if the $i$th block has size $k_i$ and corresponds to the eigenvalue $\lambda_i$, and the desired first row of $T_i$ is $a^i_0, \dots, a^i_{k_i - 1}$, then the polynomial $t_i(x) := \sum_{n=0}^{k_i - 1} a^i_n (x - \lambda_i)^n$ will do the job. We can then use the polynomials $f_i$ constructed from the $p_i$ and $q_i$ above to construct a single polynomial $f$ such that the $i$th block of $f(J)$ is $T_i$ for all $1 \leq i \leq m$ at the same time: the polynomial $f(x) := \sum_{i=1}^m t_i(x) f_i(x)$ fairly clearly does the job.
To fulfill the request for explicit polynomials, suppose that $J$ is a sum of $m$ Jordan blocks $B_i$ corresponding to the distinct eigenvalues $\lambda_i$ and having sizes $k_i$, and suppose that for each $1 \leq i \leq m$ the (arbitrary) upper triangular Toeplitz matrix $T_i$ with first row $a^i_0, a^i_1, \dots, a^i_{k_i - 1}$ is given. Then, for $1 \leq i \leq m$ letting $p_i$ and $q_i$ be the polynomials constructed in the above argument, the single polynomial $$ f(x) = \sum_{i=1}^m \left(\sum_{n=0}^{k_i - 1} a^i_n (x - \lambda_i)^n \right) p_i(q_i(x)) q_i(x) $$ has the property that the $i$th block of $f(J)$ is $T_i$ for all $1 \leq i \leq m$.
The following table, which records properties of the polynomials constructed in the argument, might be helpful. \begin{array}{c|c|c} \text{Polynomial $h(x)$} & h(B_i) & h(B_j), j \neq i \\\hline q_i(x) & \text{invertible} & 0 \\ p_i(q_i(x)) q_i(x) & I_{k_i} & 0 \\ \sum_{n=0}^{k_i - 1} a^i_n (x - \lambda_i)^n & T_i & * \\ \left(\sum_{n=0}^{k_i - 1} a^i_n (x - \lambda_i)^n \right) p_i(q_i(x)) q_i(x) & T_i & 0 \\ \end{array} where $*$ denotes some matrix that does not appear with nonzero coefficient in the formula for $f(x)$.
Note that while the above formula makes the action of $f(x)$ on $J$ relatively easy to "see," it is not easy to "see" what the coefficients of $f(x)$ actually are. As I noted at the beginning of this comment, I am not sure that it helps to think in terms of the coefficients of $f$ as opposed to its mapping properties. Insisting on this perspective may make the problem more difficult (as it seems to do even in the case of Lagrange interpolation).