Let $w_{2n+2}$ is the matrix with ones on the non-principal diagonal and zeros elsewhere. Let $V$ be the $2n+2$-dimensional quadratic space with the symmetric bilinear form $\left<,\right>$ given by $w_{2n+2}$.
Let $\{e_1,e_2,\cdots,e_{n+1},e_{-n-1},\cdots,e_{-2},e_{-1}\}$ be the corresponding standard basis of $V$ such that $\left<e_{i},e_{-j} \right>=\delta_{ij}$ and $\left<e_{i},e_{j} \right>=0$.
Let $G$ be the special orthogonal group of $W_{2n+2}$, that is, $G=SO(W_{2n+2})$ and let $G'$ be the subgroup of $G$ fixing $e_{n+1}-e_{-n-1}$.
I know that the longest Weyl element of $G$ is $w_{2n+2}$. But I don't know what is the longest Weyl element $w_0'$ of $G'$.
Would you describe the shape of $w_o'$ as a $(2n+2) \times (2n+2)$- matrix?
Thank you very much!