Please let me know if anything is unclear in my post, as it is not recieving any responses yet, I would provide additional explanation if needed. Also if you have any idea for my problem below please don't hesitate to post an answer. Any help is appreciated now.
I've been recently playing around with the linear recurrence sequences. Consider the following recurrence equation:
$$ a_n = c_1a_{n-1} + \cdots + c_ka_{n-k}, \quad \forall n > k $$
is equivalent to
$$\begin{pmatrix}a_n \\ \vdots \\ a_{n-k+1}\end{pmatrix} = \begin{pmatrix} c_1& c_2& \dots& c_{k-1}& c_k \\ 1& 0& \dots& 0& 0\\ 0& 1& \dots& 0& 0\\ \vdots& \vdots& \ddots& 0& 0\\ 0& 0& \dots& 1& 0\end{pmatrix}\begin{pmatrix} a_{n-1} \\ \vdots \\ a_{n-k} \end{pmatrix} = \dots = \begin{pmatrix} c_1& c_2& \dots& c_{k-1}& c_k \\ 1& 0& \dots& 0& 0\\ 0& 1& \dots& 0& 0\\ \vdots& \vdots& \ddots& 0& 0\\ 0& 0& \dots& 1& 0\end{pmatrix}^{n-k} \begin{pmatrix} a_k \\ \vdots \\ a_1 \end{pmatrix}, \quad \forall n > k $$
denote the R.H.S. matrix as $M$ (without the exponent).
And I want to show such $M$, regardless of the choice of $c_1, \dots , c_k$ (with $c_k \neq 0$), has only one Jordan block corresponding to each eigenvalue in its Jordan canonical form, by an alternative way.
Let $U^-1MU = \Lambda + N$ be its Jordan form with $\Lambda N = N \Lambda$. Suppose $M$ consists of $m$ different eigenvalues $\lambda_1,\dots, \lambda_m$ with algebraic multiplicity $\alpha_1,\dots,\alpha_m$; i.e. the characteristic polynomial is $$ c_M(x) = (x-\lambda_1)^{\alpha_1}\dots(x-\lambda_m)^{\alpha_m} $$ and the minimal polynomial reads $$ (x-\lambda_1)^{u_1}\dots(x-\lambda_m)^{u_m}$$ then $$\begin{pmatrix}a_n \\ \vdots \\ a_{n-k+1}\end{pmatrix} = U^{-1}\begin{pmatrix} \lambda_1^n& {n \choose 1}\lambda_1^{n-1}& \dots& {n \choose *} \lambda_1^{n-*}& 0& \dots& 0& \\ 0& \lambda_1^n& \dots& {n \choose {*-1}}\lambda_1^{n-*+1}& 0& \dots& 0\\ \vdots& \vdots& \ddots& \vdots& \vdots& \vdots& \vdots\\ 0& 0& \dots& \lambda_1^n& 0& \dots& 0\\ 0& 0& \dots& 0&\lambda_1^n& \dots& \vdots\\ 0& 0& \dots& 0& 0& \ddots& \vdots\\ 0& 0& \dots& 0& 0& \dots& \lambda_m^n\\ \end{pmatrix} U \begin{pmatrix} a_k \\ \vdots \\ a_1 \end{pmatrix}, \; \forall n > k $$ the product of three matrice R.H.S. each has all entries as a function of $n$ in the shape of $P_1(n)\lambda_1^n + ... + P_m(n)\lambda_m^n$, with each $\deg P_i \leq u_i-1 $(note that $u_i \leq \alpha_i$). Note that this equation holds as long as the first recurrence equation holds.
One may prove the following result using the fact that $M$ has only one Jordan block for each eigenvalue to derive the following fact:
When $c_1, \dots, c_k$ and $a_1. \dots, a_k$ are chosen, there must uniquely exist $Q_1, \dots, Q_m$ with $\deg Q_i \leq \alpha_i-1$ such that the equation $a_n = Q_1(n)\lambda_1^n + \dots + Q_m(n)\lambda_m^n$ always hold, and for all $\deg Q_i \leq \alpha_i-1$, the sequence given by $a_n = Q_1(n)\lambda_1^n + \dots + Q_m(n)\lambda_m^n$ should also satisfy $a_n = c_1a_{n-1} + \dots + c_ka_{n-k}$.
But I want to take the statement above for granted(one can verify the result by checking if the $k$ by $k$ matrix representation of the relation between coefficient of $P_i$s and $a_i$s is singular. And also there are other ways I know to derive this result) and prove that $M$ must only have one Jordan block for each eigenvalue(i.e. $u_i = a_i$).
Proof
For fixed $c_1, \dots, c_k$, if $u_i < \alpha_i$ for some $i$, pick $a_n = n^{\alpha_i}\lambda_i^n$. By the statement above, the recurrence relation holds, and hence (by the Jordan Form analysis) there exists $Q_1, \dots, Q_m$ such that $a_n = Q_1(n)\lambda_1^n + \dots + Q_m(n)\lambda_m^n\, \; \forall n > k$ with $\deg Q_i < u_i$, but such expression is unique, and it does not contain the term $n^{\alpha_i}\lambda_i^n$, leading to a contradiction; thus there must only be one Jordan block corresponding to each $\lambda$.
Is this proof valid? And can I write it more rigorously? I've seen other shorter proofs of it, and I would like to see if there are common concepts between this proof and the other proofs.
Short proof: Jordan basis of $A$ when $A$ is the companion matrix?
Alternate proofs: The characteristic and minimal polynomial of a companion matrix
xyz-matricestags were created by me. – Rodrigo de Azevedo Oct 29 '22 at 15:20