in this case (\[blok\]) implies $F^{st}_{ij;k_s^i,1}=0$, that is, the first diagonal is zero. Moving from the left to the right we get that all the diagonals of $F^{st}_{ij}$ are zero.

If $s+1=t$ then equation (\[blok\]) is a linear relation between the neighboring entries on the diagonal. It is easy to derive equations (\[ker1\]), (\[ker2\]) from this fact.

Indeed, let us consider the case $k_s^i\le k_{s+1}^j$. Then equation (\[blok\]) for $p=k_i^s$, $1< q\le k_i^s$ says $F^{s,s+1}_{ij;k_i^s,q-1}=0$. Moving along the diagonal from the bottom to the top and using equation (\[blok\]) we get that the first $k_i^s-1$ diagonals of the matrix $F^{s,s+1}_{ij}$ are zero. For the rest of the diagonals equation (\[blok\]) implies $F^{s,s+1}_{ij;p+1,q+p}=F^{s,s+1}_{ij;1,q}\tau^{-2p}$. Putting $c^s_{ij;l}=F^{s,s+1}_{ij;1,l+k_{s+1}^j-k_s^i+1}$ we get equation (\[ker1\]).

Obviously $Z\in Im S_J$ if and only if $tr(ZF)=0$ for all $F\in
ker \bar{S}_J$, $\bar{S}_J(F)=\tau^{-1} JF-\tau F J$. The space $ker \bar{S}_J$ has a description similar to the one of $ker S_J$ (to get $ker\bar{S}_J$ from $ker S_J$ it is enough to change the order of the Jordan blocks in $J$) and one can easily derive

$Z\in Im S_J$ if and only if following equations hold $$\sum^{u-1}_{l=0}Z_{ij;k_s^i-l,u-l}^{s,s+1}\tau^{2l}=0, \quad
u=1,\dots,min\{k^i_s,k^j_{s+1}\},$$ where $s=1,\dots,r-1$.

The lowest nonzero diagonal of a rank one matrix contains only 