I don’t understand intuition behind this.

From my script.

Adjoint operator. Let A∈L(V,W)A\in L(V,W). Now, for every linear functional g∈W′=L(W,K)g\in W’=L(W,K) we define linear functional by v↦g(Av),∀v∈Vv \mapsto g(Av),\quad \forall v\in V

We can easily prove that we defined linear functional on VV. We will call this functional as f=A′gf=A’g

There is more, but I don’t understand what does this have to do with adjoint operators? Can somebody explain me how this works?

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I didn’t post all of it, just introduction, A′∈L(W′,V′)A’ \in L(W’,V’) from above is adjoint operator in final chapter.
– Rush ThaMan
2 days ago

Also, f(v)=(A′g)(v)=g(Av),∀v∈Vf(v)=(A’g)(v)=g(Av), \forall v \in V
– Rush ThaMan
2 days ago

Thanks, that is exactly what I have in my book. I understand that this is mathematics manipulation, and it turns out, it is adjoint operator with all wanted properties. But how did someone think about defining it this way
– Rush ThaMan
2 days ago

Yea, and I think the key here is that we can only write f(v)=(A′g)(v)=g(Av),∀v∈Vf(v)=(A’g)(v)=g(Av), \forall v\in V and we can’t rewrite this as f(v)=A′(g(v))…f(v)=A'(g(v))… because A′A’ maps W′W’ to V′V’, I think usually when we have composition of linear operators we can switch brackets
– Rush ThaMan
2 days ago

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