# Orthogonal projection to graph of Hilbert space

Let H be a Hilbert space and A be a bounded linear operator on H. Let G be the graph of A , i.e a subset of the direct sum of H with itself of the form (h,Ah). Then we know G is a closed linear subspace (This is one direction of closed graph theorem), so we can define the orthogonal projection P from the direct sum of H with itself onto G. My question is that do we have an explicit form of P? I have no idea how to express it explicitly. Any help would be appreciated.

The orthogonal projection of $$(y,z)\in\mathcal{H}\times\mathcal{H}$$ onto the graph $$\mathcal{G}(A)=\{ (x,Ax)\in\mathcal{H}\times\mathcal{H} : x\in\mathcal{H} \}$$ is the unique $$(x,Ax)\in\mathcal{H}\times\mathcal{H}$$ such that the following orthogonality conditions hold in $$\mathcal{H}\times\mathcal{H}$$: $$((y,z)-(x,Ax)) \perp (x',Ax'), \;\;\; x'\in\mathcal{H}.$$ That is, the following hold for all $$x'\in\mathcal{H}$$: $$(y-x,z-Ax)\perp(x',Ax'),\;\;\; x'\in\mathcal{H}, \\ \langle y-x,x'\rangle=\langle Ax-z,Ax'\rangle,\;\;\; x'\in\mathcal{H}, \\ \langle y-x,x'\rangle=\langle A^*(Ax-z),x'\rangle,\;\;\; x'\in\mathcal{H}, \\ y-x=A^{*}(Ax-z) \\ y+A^*z =(A^*A+I)x \\ x=(A^*A+I)^{-1}(y+A^*z).$$ Therefore, if $$P$$ denotes the orthogonal projection of $$(y,z)$$ onto the graph $$\mathcal{G}(A)$$, then $$P(y,z) = (x,Ax) \\ = ((A^*A+I)^{-1}(y+A^*z),A(A^*A+I)^{-1}(y+A^*z)) \in\mathcal{G}(A)\subset\mathcal{H}\times\mathcal{H}.$$ It's ugly and pretty all at the same time.
• @biden : That would be the set of all $(y,z)\in\mathcal{H}\times\mathcal{H}$ such that $y+A^*z=0$ or $y=-A^*z$. In other words, it s a transpose of the graph of $-A^*$. Commented Oct 5, 2020 at 3:35