# Existence of a transpose of a linear operator

Let $X,Y$ be two linear spaces and let $b:X\times Y\to\Bbb R$ be a bilinear map. For any linear operator $A:X\to X$ we can define its $b$-transpose acting on $Y$ by the system of the following equations: $$D(A^T) = \{y\in Y:\exists y'\in Y\text{ s.t. }b(x,y') = b(Ax,y) \text{ for all }x\in X\}$$ defines the domain of $A^T$ which is such that $b(x,A^Ty) = b(Ax,y)$ for all $x\in X$, $y\in D(A^T)$.

I wonder when $A^T$ is well-defined, i.e. it holds that $D(A) = Y$ and $A^Ty$ is unique for all $y\in D(A)$. For example, if we say that $y\sim y' \Leftrightarrow b(x,y-y') = 0$ for all $x\in X$ then $A^T$ is clearly defined uniquely for all $y\in Y/\sim$ whenever it is defined. However, I do not have idea how to deal with the existence.

More precisely, I wonder what are the sufficient conditions for $D(A) = Y$ e.g. if $b$ is not degenerate, i.e. if $y \sim y'$ defined above implies $y = y'$.

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