be the orthogonal operator part of $T$, and let the selfadjoint operator $B \in \mathbf{B}(\sK)$ be decomposed as in . Let the operators $C_0$ and $C$ be defined by and . Then $$\label{redu}
 T^*(I + i B)T = (T_{\s})^{\times}C_0^{1/2}(I+i C) C_0^{1/2}T_{\s},$$ and, consequently, $T^*(I + i B)T$ is maximal sectorial and $$\label{redu1}
 \mul T^*(I + i B)T=\mul T^{*}=\mul (T_{\s})^{\times}.$$

In order to prove the equality in , assume that $\{h,h'\}\in T^* (I + i B) T$. This means that $$\label{eka-}
\{h,\varphi\}\in T \quad \mbox{and} \quad \{(I + i B)\varphi,h'\}\in T^*$$ for some $\varphi\in\sK$. Decompose the element $\varphi$ as $$\label{eka--}
\varphi=\varphi_1+\varphi_2, \quad \varphi_{1} \in \cdom T^{*},
\,\,\varphi_2\in\mul T.$$ Since $\{0, \varphi_{2}\} \in T$, it is clear that $$\label{eka---}
\{h,\varphi\}\in T \quad \Leftrightarrow \quad  \{h,\varphi_{1}\}\in T_{\s}.$$ Using and the above decomposition of $B$, one observes that $$\{(I + i B)\varphi,h'\}
 =\left\{ \begin{pmatrix}   (I+i B_{11})\varphi_1+iB_{12}\varphi_2 \\
                                          iB_{12}^*\varphi_1+(I+i B_{22})\varphi_2
              \end{pmatrix},
   h' \right\},$$ which implies that the condition $\{(I + i B)\varphi,h'\}\in T^*$ is equivalent to $$\left\{ \begin{array}{l} \{ (I+i B_{11})\varphi_1+iB_{12}\varphi_2, h' \} \in T^*,
           \\ iB_{12}^*\varphi_1+(I+i B_{22})\varphi_2=0,
           \end{array}
           \right.$$ or, what is the same thing, $$\label{eka}
 \left\{ \begin{array}{l} \{ [I+i B_{11} + B_{12}(I+i
  B_{22})^{-1}B_{12}^*]\varphi_1, h'\} \in T^*,  \\
  \varphi_2=-i(I+i B_{22})^{-1}B_{12}^*\varphi_1.
           \end{array}
          \right.$$ Due to 