I have this proposition from Karlheinz Spindler's Abstract algebra with applications vol. 1, page 303. It is proved there, but I tried it by myself based on that prove. I would like to know if my attempt is entirely correct. Any suggestion is welcome.
Let $A:V\rightarrow W$ be a linear mapping between finite-dimensional inner product spaces over $\mathbb K = \mathbb R$ or $\mathbb C$. Then there is a unique linear mapping $A^*:W\rightarrow V$ such that $$\langle Av,w\rangle =\langle v,A^*w\rangle$$ for all $v\in V$ and all $w\in W$, called the adjoint of A.
My prove:
Let $\{w_1,\dots,w_n\}$ be a basis of $W$. Let $[n]:=\{1,\dots,n\}$. By the Riesz' Representation Theorem $$(\forall i\in [n])(\exists! f_i\in W^*)(f_i=\langle\cdot,w_i\rangle).$$ Then $$(\forall i\in[n])(f_iA\in V^*)$$ $$\Rightarrow (\forall v\in V) (\forall i\in [n])((f_iA)v=\langle Av,w_i\rangle)........(1)$$ On the other hand, also by the Riesz' Representation Theorem we have $$(\forall i\in [n])(\exists! a_i\in V)(f_iA=\langle\cdot,a_i\rangle)$$ $$\Rightarrow(\forall v\in V)(\forall i\in [n])(\exists! a_i\in V)((f_iA)v=\langle v,a_i\rangle)........(2)$$ Then by $(1)$ and $(2)$ $$(\forall v\in V)(\forall i\in [n])(\exists! a_i\in V)(\langle Av,w_i\rangle =\langle v,a_i\rangle)........(3)$$ Now, since $\{w_i\}_{i=1}^n$ is a basis of W and $\{a_i\}_{i=1}^n$ is a set of vectors in V, there is a unique linear mapping $A^*:W\rightarrow V$ such that $$(\forall i\in [n])(A^*w_i=a_i).$$ Then in $(3)$ $$(\forall v\in V)(\forall i\in [n])(\langle Av,w_i\rangle =\langle v,A^*w_i\rangle).$$ Now let $w\in W$ be an arbitrary vector in $W$, then $w=\sum _{i=1}^nb_iw_i$ for some $b_i\in \mathbb K$ and $$\langle Av,w\rangle = \langle Av,\sum _{i=1}^nb_iw_i\rangle = \sum _{i=1}^n \bar{b_i}\langle Av,w_i\rangle = \sum _{i=1}^n \bar{b_i}\langle v,A^*w_i\rangle = \langle v,A^*(\sum _{i=1}^nb_iw_i)\rangle = \langle v,A^*w\rangle$$ for all $v\in V.\square$