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Let $(H, \langle\cdot, \cdot\rangle)$ be a Hilbert space and $P: H \to H$. In this answer, @gerw said that if $$\forall (x,y) \in H^2:\langle Px, y \rangle = \langle x, Py \rangle,$$ then $P$ is linear. Because I'm trying to prove it by myself, I've not read his/her solution.

On the other hand, From this Wikipedia link about self-adjoint operator,

In mathematics, a self-adjoint operator (or Hermitian operator) on a finite-dimensional complex vector space $V$ with inner product $\langle\cdot, \cdot\rangle$ is a linear map $A$ (from $V$ to itself) that is its own adjoint:$\langle A v, w\rangle=\langle v, A w\rangle$ for all vectors $v$ and $w$.

From the paragraph, I got that self-adjoint operator is not necessarily linear. If it was, they would not said "linear map $A$ that is its own adjoint". Could you please reconcile this difference?

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  • $\begingroup$ @IsaacRen you meant $\forall (x,y) \in H^2:\langle Px, y \rangle = \langle x, Py \rangle$ does not necessarily imply $P$ is linear? $\endgroup$ – LAD May 16 at 12:00
  • $\begingroup$ The definition of "adjoint" only applies to linear operators. $\endgroup$ – Isaac Ren May 16 at 12:00
  • $\begingroup$ No, my first comment (which I deleted) literally repeats what you said in your question, so I deleted it. $\endgroup$ – Isaac Ren May 16 at 12:01
  • $\begingroup$ @IsaacRen Let's not say about adjoint operator. Do you think $\forall (x,y) \in H^2:\langle Px, y \rangle = \langle x, Py \rangle$ implies $P$ is linear? $\endgroup$ – LAD May 16 at 12:03
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    $\begingroup$ Thank you so much @IsaacRen! OhI got it. $\forall (x,y) \in H^2:\langle Px, y \rangle = \langle x, Py \rangle$ implies $P$ is linear, but not necessarily the converse :) $\endgroup$ – LAD May 16 at 12:27
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Oh my bad. I got it :)

We have $\forall (x,y) \in H^2:\langle Px, y \rangle = \langle x, Py \rangle$ implies $P$ is linear, but the converse is not necessarily true.

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