space of linear operators to the space of bilinear functions Let V be a finite-dimensional Euclidean space with an orthonormal basis ${e_1,..., e_n}$, equipped with an inner product (positive definite symmetric bilinear function) which is denoted $(\ ,\ )$.
My book A Course in Algebra by Vinberg says:

To each vector $a \in V$, there corresponds the linear function
  $\varphi_a(x)=(x,a)$ ..(some argument).. the map $a \mapsto \varphi_a$ is a canonical isomorphism between the spaces $V$ and $V^*$.

I have understood the above statement, what confused me is the following statements:

Similarly, to each linear operator $\mathcal{A}$ on the space V, there corresponds a bilinear function $\varphi_{\mathcal{A}}(x,y)=(x,\mathcal{A}y)$ ...(some argument)... It follows that the map $\mathcal{A} \mapsto \varphi_{\mathcal{A}}$ is a isomorphism from the space of linear operators to the space of bilinear functions on V. This isomorphism does not depend on the choice of a basis.

I have two questions:


*

*What does the word "Similarly" means? Is there an intrinsic relation between the two statements?

*Why $\mathcal{A} \mapsto \varphi_{\mathcal{A}}$ is also a canonical isomorphism?


Indeed, I wonder if there is a "big picture"(relation) in these two statements?
Found a probably related question but I could not reason further
symmetric bilinear function and linear operator
 A: 1) The first excerpt describes an isomorphism $\varphi: V \overset{\sim}{\to} V^*$ sending $a \mapsto \varphi_a$. The second describes an isomorphism $\DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\Bil}{Bil} \Phi: \Hom(V,V) \overset{\sim}{\to} \Bil(V,F)$ sending $A \mapsto \Phi_A = (\rule{7pt}{1pt} \, , A \rule{7pt}{1pt})$, where $F$ is the base field and $\Bil(V,F)$ is the set of bilinear forms on $V$. As shown in the post you linked, $\Bil(V,F) \cong \Hom(V, V^*)$, so we have $\Hom(V,V) \cong \Hom(V, V^*)$. Thus we've really just applied $\Hom(V, \, \rule{7pt}{1pt})$ to the isomorphism given in the first excerpt.
But the connection is even a bit tighter than that. Given a linear map $\psi: U \to W$, applying $\Hom(V, \, \rule{7pt}{1pt})$ gives us a map $\psi_*: \Hom(V, U) \to \Hom(V, W)$ given by $f \mapsto \psi \circ f$. (This is what it means to be a functor.) Applying $\Hom(V, \, \rule{7pt}{1pt})$ to the isomorphism $\varphi$, we obtain the map $\varphi_*: \Hom(V,V) \to \Hom(V, V^*)$, $A \mapsto \varphi \circ A$. Given $y \in V$, we have
$$
\varphi_*(A)(y) = (\varphi \circ A)(y) = \varphi_{A(y)} = (\rule{7pt}{1pt} \, , A(y))
$$
so $\varphi_*(A) = \Phi_A$ for all $A \in \Hom(V,V)$. Thus we see that, under the identification $\Hom(V,V^*) \cong \Bil(V,F)$, $\varphi_*$ is exactly $\Phi$! I think this qualifies $\varphi$ and $\Phi$ as being quite similar.
2) I'm not sure which part of the statement is the problem. I assume the (some argument) you've omitted proves the isomorphism, so maybe it's the "canonical" part. To be canonical roughly means that it is independent of the choice of basis, as remarked in the quoted passage. (There is also a more precise notion of a map being natural.)
