=
        \begin{pmatrix}
        \frac{1}{(k-1)!}\frac{d^{k-1}}{d {\lambda}^{k-1}} \sum_{i=0}^\infty a_i {\lambda}^i \\
        \frac{1}{(k-2)!}\frac{d^{k-2}}{d {\lambda}^{k-2}} \sum_{i=0}^\infty a_i {\lambda}^i \\
        \vdots \\
        \frac{d}{d {\lambda}} \sum_{i=0}^\infty a_i {\lambda}^i \\
        \sum_{i=0}^\infty a_i {\lambda}^i
        \end{pmatrix}.$$

It suffices to show that $$\label{eq:jordan}
T_{a_0}\dots T_{a_n}\begin{pmatrix} 0\\ \vdots \\ 0\end{pmatrix}=
\begin{pmatrix}
\sum_{i=0}^n \binom{i}{k-1} a_i{\lambda}^{i-k+1} \\
\sum_{i=0}^n \binom{i}{k-2} a_i{\lambda}^{i-k+2}\\
\vdots \\
\sum_{i=0}^n ia_i{\lambda}^{i-1}\\
\sum_{i=0}^n a_i {\lambda}^i
\end{pmatrix},$$ with the usual convention that $\binom ij=0$ if $i<j$. We prove this by induction: for $n=0$ we have $$T_{a_0}
\begin{pmatrix} 0\\ \vdots \\ 0 \\ 0\end{pmatrix} = \begin{pmatrix} 0\\ \vdots \\ 0 \\ a_0
\end{pmatrix},$$ which is what we need. Assume (\[eq:jordan\]) holds for $n-1$; then, given that $T_{a_0}(v)=Mv+a_0(0,0,\dots,0,1)^T$, $$\begin{aligned}
T_{a_0}\begin{pmatrix}
\sum_{i=0}^{n-1} \binom{i}{k-1} a_{i+1}{\lambda}^{i-k+1} \\
\sum_{i=0}^{n-1} \binom{i}{k-2} a_{i+1}{\lambda}^{i-k+2}\\
\vdots \\
\sum_{i=0}^{n-1} ia_{i+1}{\lambda}^{i-1}\\
\sum_{i=0}^{n-1} a_{i+1} {\lambda}^i
\end{pmatrix}
&=
\begin{pmatrix}
\sum_{i=0}^{n-1} \left(\binom{i}{k-1}+\binom{i}{k-2}\right) a_{i+1}{\lambda}^{i-k+2}  \\
\sum_{i=0}^{n-1} \left(\binom{i}{k-2}+\binom{i}{k-3}\right) a_{i+1}{\lambda}^{i-k+3}\\
\vdots \\
\sum_{i=0}^{n-1} (i+1)a_{i+1}{\lambda}^i\\
\sum_{i=0}^{n-1} a_{i+1} {\lambda}^{i+1}+a_0
\end{pmatrix}
\\
&=\begin{pmatrix}
\sum_{i=0}^n \binom{i}{k-1} a_i{\lambda}^{i-k+1} \\
\sum_{i=0}^n \binom{i}{k-2} a_i{\lambda}^{i-k+2}\\
\vdots \\
\sum_{i=0}^n ia_i{\lambda}^{i-1}\\
\sum_{i=0}^n a_i {\lambda}^i
\end{pmatrix},\end{aligned}$$ as required.

It is easy to see how this would generalize to multiple Jordan blocks.

Return to the general case of $M$ given by (\[eq:M\]). The following theorem is along the lines of [@HS Theorem 3.1] and is based on the ideas from [@Gunturk] (originally) and 