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I am trying to understand the adjoint of a linear operator geometrically. Since the graph of the adjoint can be constructed as the orthogonal complement of a "rotated" copy of the graph of the operator itself (details follow) I am wondering how orthogonality of the graphs translates to algebraic or topological relations between two operators.

What I mean by the above introductory remarks is that using the "rotation" $$U\colon H\oplus H\to H\oplus H, \qquad (x,y)\mapsto (-y,x)$$ we can get the relation $$\mathrm{Graph}(A^*) = U(\mathrm{Graph}(A))^\perp.$$

This question is similar to this one but since I am also interested in the infinte dimensional case I would like to avoid the singular value decomposition which is used in the answers there.

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Suppose $A$ and $B$ have orthogonal graphs. It means that $(x, Ax)$ and $(y, By)$ are always orthogonal in $H\oplus H$, hence

$$0 = \langle x, y\rangle + \langle A x, B y\rangle =\langle x, y\rangle + \langle B^*A x, y\rangle$$ It follows that $$-B^*A = Id$$ So that $A$ is invertible and $B = - (A^{-1})^*$.

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  • $\begingroup$ Okay, this answers the question but does not really help me with my original problem: There $U(\mathrm{Graph}(A))$ is the graph of an operator iff $A$ is invertible then it is the graph of $-A^{-1}$ so your answer provides that $A^*=-((-A^{-1})^{-1})^*$ which does not provide additional information on $A^*$. $\endgroup$ – Christian Sep 22 '17 at 7:02
  • $\begingroup$ Then may be I don't completely understand the question. Which exact property do you assume about $A$'s graph? $\endgroup$ – Gribouillis Sep 22 '17 at 19:07

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