From this link, I saw :
If $f: V→W$ is a linear map between vector spaces $V$ and $W$ with nondegenerate bilinear forms, we define the transpose of $f$ to be the linear map $^tf : W→V$, determined by $$B_V(v,{}^tf(w))=B_W(f(v),w) \quad \forall\ v \in V, w \in W.$$ Here, $B_V$ and $B_W$ are the bilinear forms on $V$ and $W$ respectively. The matrix of the transpose of a map is the transposed matrix only if the bases are orthonormal with respect to their bilinear forms.
If $f : V → W$ is a linear map, then the transpose (or dual) $f^* : W^* → V^*$ is defined by $$f^*(\varphi) = \varphi \circ f \, $$ for every $φ ∈ W^*$. If the linear map $f$ is represented by the matrix $A$ with respect to two bases of $V$ and $W$, then $f^*$ is represented by the transpose matrix $A^T$ with respect to the dual bases of $W^*$ and $V^*$.
I would like to get more precisions between these 2 definitions :
1.Question : Does Riesz theorem make the link between these 2 definitions above ? i.e between the definition involving the inner product and the other one involving the notion of dual linear map ?
If I take the following formula expressing the inner product $(.|.)$ between 2 vectors $X,\,Y \in V$ (for a matricial product point of view) :
$(X|Y) = X^{T}\,M\,Y$ with $$M$$ the matrix associated to the inner product, i.e $$M=\text{Mat}_{V}(e_i\cdot e_j)$$ ,
2.Question : Can I consider the product $$X^{T}\,M$$ like a dual linear form such the formula $$(X|Y)$$ represents the action of the linear form $$X^{T}\,M$$ on vector $$Y$$ ?
I don't know what's the english for "crochet de dualité" (french, maybe bracket of duality ?) that expresses simply the "dot-product" between a linear form l and a vector v : <l,v>=l(v)
Now, I would like to use the definition of the transpose application to visualize it with matrices product : the adjoint of matrix $A$ is noted $A^{*}$ and defined by :
$$\forall\ X, Y \in V : (AX|Y) = (X|A^{*}Y)$$
I consider real coefficients for matrix $A$. If basis is orthonormal, I have $$M$$ matrix equal to $$I_{d}$$ (identity matrix).
So, I have with product of matrices :
$(AX|Y) = X^{T}\,A^{T}\,I_{d}\,Y = (X|A^{*}Y) = X^{T}\,I_{d}\,A^{*}\,Y$
$=X^{T}\,A^{*}\,Y$
So I deduce $$A^{*}=A^{T}$$
But as you can see, I presupposed the rule for the transpose of a product, i.e $$(AX)^{T}=X^{T}A^{T}$$
3.Question : If someone could give me another demonstration or a way to visualize well the notion of transpose linear map, this would be nice to give it.
4.Question : a last question : in my definition of adjoint $A^{*}$ with inner product, is it always necessary to have an orthonormal basis ? because if this is not the case, I can only have the following equality :
$$(AX|Y) = X^{T}\,A^{T}\,M\,Y = (X|A^{*}Y) = X^{T}\,M\,A^{*}\,Y$$
$$\Rightarrow\quad\quad A^{T}\,M = M\,A^{*}$$
i.e not directly $A^{T} = A^{*}$.
Thanks for your help