w, v\otimes b\cdot w, \cdots, v\otimes b^{p-1}\cdot
w, b\cdot v\otimes w, b\cdot v\otimes b\cdot w, \cdots, b\cdot v\otimes b^{p-1}\cdot
w, \cdots, b^{p-1}\cdot v\otimes w, b^{p-1}\cdot v\otimes b\cdot w,
\cdots, b^{p-1}\cdot v\otimes b^{p-1}\cdot w\}$ has the form $$G_0=\left(\begin{array}{cccc}
G_{11}&G_{12}&\cdots&G_{1p}\\
0&G_{22}&\cdots&G_{2p}\\
\cdots&\cdots&\cdots&\cdots\\
0&0&\cdots&G_{pp}
\end{array}\right)$$ where each $G_{st}$ $(s\leqslant t)$ is a upper triangular $p\times p$-matrix, and $G_{ss}$ has the form $$\left(\begin{array}{ccccc}
\\\xi^{i+j}&\alpha_{12}&*&\cdots&*\\
0&\xi^{i+j}&\alpha_{23}&\cdots&*\\
0&0&\xi^{i+j}&\ldots&*\\
\cdots&\cdots&\cdots&\cdots&\cdots\\
0&0&0&\cdots&\xi^{i+j}
\end{array}\right)$$ with $\alpha_{i_1,i_1+1}\neq 0$. Hence $\xi^{i+j}$ is the unique eigenvalue of the action of $g$ on $S_i\bigotimes S_j$. Moreover, $r(\xi^{i+j}I_p-G_{ss})=p-1$. It follows that $r(\xi^{i+j}I_{p^2}-G_0)\geqslant p(p-1)$. Thus, dim$V_{\xi^{i+j}}\leqslant p$, where $V_{\xi^{i+j}}$ is the eigenspace of the action of $g$ on $S_i\otimes S_j$.

Obviously, $u_0=v\otimes w\in V_{\xi^{i+j}}$. For any $1\leqslant l\leqslant p-1$, let $u_{(l)}=b^l\cdot v\otimes w$. Then $h(u)=u(1)=l>0$. It follows from Lemma \[3.2\](1) that $g\cdot
u_{(l)}\neq\xi^{i+j}u_{(l)}$. Then by Theorem \[3.6\], there is an element $u_{(l)}'\in S_i\otimes S_j$ with $h(u_{(l)}')\leqslant l$ and $u_{(l)}'(1)<l$ such that $g\cdot u_l=\xi^{i+j}u_l$, where $u_l=u_{(l)}+u_{(l)}'$. Obviously, $u_l(1)=l$ and $h(u_l)=l$ for all $0\leqslant l\leqslant p-1$. It follows that $\{u_0, u_1, \cdots, u_{p-1}\}\subset V_{\xi^{i+j}}$ are linearly independent over $k$. Thus, $\{u_0, u_1, \cdots, u_{p-1}\}$ is a $k$-basis of $V_{\xi^{i+j}}$.

Let $v_l=g\cdot u_{(l)}-\xi^{i+j}u_{(l)}$. Then it follows from Lemma \[3.2\] that $v_l=-l\xi^{i+j}(1-\xi^i)b^{l-1}\cdot v\otimes w
+\sum\limits_{i_1<l-1}\alpha_{i_1}b^{i_1}\cdot v\otimes w$. Hence $v_l(1)=l-1$ 