equation (\[J\]) we have $s=1$ and $\bold{k}_1=\bold{k}=
(\vec{k}_1,\dots,\vec{k}_r)$, $\vec{k_i}\in\mathbb N^{d_i}$, $1\le
i\le r$. In this situation we denote by $F^{st}_{ij}\in
Mat(k^i_s,k^j_t)$, $1\le s,t\le r$ the matrix with the entries $F^{st}_{ij;pq}=F_{p'q'}$, $p'=\sum_{l=1}^{s-1}|\vec{k_l}|+
\sum_{l=1}^{i-1} k_s^l+p$, $q'=
\sum_{l=1}^{t-1}|\vec{k_l}|+\sum_{l=1}^{j-1} k_s^l+ q$. In these notations the following lemma holds

\[kerSJ\] Let $J$ be the matrix given by (\[J\]) with $s=1$ and $\bold{k}_1=\bold{k}=(\vec{k}_1,\dots,\vec{k}_r)$. Then $F\in
\ker S_J$ if and only if $$\begin{gathered}
F^{st}_{ij}=0, \mbox{ if } t-s\ne 1,\\ F^{s,s+1}_{ij}=
(\sum_{l=0}^{k_s^i-1}
c^{s}_{ij;l}J_{k_s^i}^l(0))D^{k_s^i,k_{s+1}^j}_\tau \mbox{ if }
k_s^i\le k_{s+1}^j\label{ker1},\\ F^{s,s+1}_{ij}=
D^{k_s^i,k_{s+1}^j}_\tau(\sum_{l=0}^{k_{s+1}^j-1}
c^{s}_{ij;l}J_{k_{s+1}^j}^l(0)) \mbox{ if } k_s^i>
k_{s+1}^j\label{ker2},\end{gathered}$$ where $c^{s}_{ij;l}\in \mathbb C$, $J^l_{k_s^i}(0)$ ( and $J^l_{k_{s+1}^j}(0)$) is the $l$-th power of the Jordan block matrix, and $D^{k_s^i,k_{s+1}^j}\in Mat(k_s^i,k_{s+1}^j)$ is given by formula $$\begin{gathered}
D^{k_s^i,k_{s+1}^j}_{\tau;pq}=\delta_{p+k_{s+1}^j,q+k_s^i}\tau^{2-2p}
\mbox{ if } k_s^i\le k_{s+1}^j\\
D^{k_s^i,k_{s+1}^j}_{\tau,pq}=\delta_{p,q}\tau^{2-2p} \mbox{ if }
k_s^i> k_{s+1}^j.\end{gathered}$$

The system of linear equations $S_J(F)=0$ is equivalent to the collection of linear systems: $$\tau J_{k_s^i}(\lambda\tau^{2s-2})
F^{st}_{ij}-\tau^{-1}F^{st}_{ij}
J_{k_t^j}(\lambda\tau^{2t-2})=0,\quad 1\le s,t \le r,$$ because $J$ has a block structure. The equations for the entries of $F^{st}_{ij}$ are of the simple form: $$\label{blok}
F^{st}_{ij;pq}\lambda(\tau^{2s-1}-\tau^{2t-3})=
\tau(\delta_{p,k^s_i}-1)F^{st}_{ij;p+1,q}-
\tau^{-1}(\delta_{q,1}-1)F^{st}_{ij;p,q-1}.$$

First consider the case $t-s\ne 1$. Then $\tau^{2s-1}-\tau^{2t-3}\ne 0$ and equations (\[blok\]) express the entries of the $i$-th diagonal through the entries of the $(i-1)$-th diagonal. It easy to see that 