a role in . In applications involving Theorem \[s-repr0o\], it is therefore useful to recall the following corollary.

\[newnew\] Let $T'$ be a closed linear operator from the Hilbert space $\sH$ to a Hilbert space $\sK'$ and let the operator $B' \in \mathbf{B}(\sK')$ be selfadjoint. Assume that the form $\st$ in Theorem \[s-repr0o\] is also given by $$ \st[h,k]=((I+iB')T' h ,T' k),
 \quad h,k \in \dom \st=\dom T'.$$Then there is a unitary mapping $U$ from $\cran T$ onto $\cran T'$, such that $$T'=UT, \quad B_{bb}'=UB_{bb}U^*,$$ where $B_{bb}$ and $B_{bb}'$ stand for the compressions of $B$ and $B'$ to $\cran T$ and $\cran T'$, respectively.

By assumption $((I+iB')T' h ,T' k)= ((I+iB)T h ,T k)$ for all $h,k \in \dom \st$. This leads to $$(T' h ,T' k)= (T h ,T k)  \quad \mbox{and} \quad (B' T' h ,T' k)= (B T h ,T k)$$ for all $h,k \in \dom \st$. Hence the mapping $Th \mapsto T'h$ is unitary, and denote it by $U$. Then $T'=UT$ and it follows that $(B' T' h ,T' k)= (B U^*T' h , U^*T' k)$.

A matrix decomposition for $T^*(I+iB)T$ {#sect3}
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Let $T$ be a linear relation from $\sH$ to $\sK$ which is closed; 