# Is self-adjoint operator necessarily linear?

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.

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?

• @IsaacRen you meant $\forall (x,y) \in H^2:\langle Px, y \rangle = \langle x, Py \rangle$ does not necessarily imply $P$ is linear? – LAD May 16 at 12:00
• The definition of "adjoint" only applies to linear operators. – Isaac Ren May 16 at 12:00
• No, my first comment (which I deleted) literally repeats what you said in your question, so I deleted it. – Isaac Ren May 16 at 12:01
• @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? – LAD May 16 at 12:03
• 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 :) – LAD May 16 at 12:27

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.