. The same arguments show that for all $n_i = 0; i =0 \ldots k$ $\alpha_i$ equals $\alpha_{i+0}$. This yields that $\alpha_i=\alpha_j$ for all $i,j =0 \ldots k,$ and we obtain that $g$ is a scalar matrix.  Proof of Theorem \[TH0\] ========================  