Some proprieties of a peculiar scalar product In the old notes of a friend of mine I've found this definitions without any other explanations:

Let $V$ with $g_1$ and $W$ with $g_2$ be two euclidean vector spaces.
   Let $B_1=\{v_1,...,v_n\}$ be an ortonormal basis of $V$ and
  $B_2=\{w_1,...,w_n\}$. 
Consider the bilinear form $h:
Homomorphisms(V,W) \times Homorphisms(V,W) \to \mathbb{R}$ such that
  $$\forall f_1,f_2 \in Homomorphism(V,W), \quad \text{we have:} \quad
h(f_1,f_2)= \sum_{j=1}^n g_2(f_1(v_j),f_2(v_j)).$$
In addition we note that for $i,j=1,...n$, it is true that $f_{(i,j)}(v_s)=\delta_{i,s}w_j$ is a basis of $Homomorphism(V,W)$, where $\delta$ is "Kroneker's delta".


I would like to prove that (1) this definition is well-posed, that is that it doesn't depend on the choice of $B$; that (2) $h$ is symmetric and positive defined, but I'm unsure about how I should go about these proofs. Can you help me? 
Basides, I don't know why we have that $f_{(i,j)}(v_s)$ (for $i,j=1,...n$) is a basis of $Homomorphism(V,W)$. Could you show me why this statement is true?
 A: Here is a specific example of the last statement in your question which you may find helpful.
Let $V = W = \mathbb{R}^2$ be real vector spaces, and let $B_1$ and $B_2$ both be the standard basis, i.e. $v_1 = w_1 = (1,0)$ and $v_2 = w_2 = (0,1)$. 
We want to show that the $f_{i,j}$ span $Hom(V,W)$ and are linearly independent. Suppose that $g \in Hom(\mathbb{R}^2,\mathbb{R}^2).$ This means that $g$ is a linear transformation from $V$ to $W$, so that $g$ is determined by the values it takes on the basis of $V$. In other words, the two values
$$g((1,0)) = (a, b)$$
$$g((0,1)) = (c, d)$$
determine $g$ as a function.
Now the $f_{i,j}$ are members of $Hom(\mathbb{R}^2,\mathbb{R}^2),$ so we can describe them by the values they take on the basis.
$$f_{1,1}((1,0)) = (1, 0), f_{1,1}((0,1)) = (0, 0),$$
$$f_{1,2}((1,0)) = (0, 1), f_{1,2}((0,1)) = (0, 0),$$
$$f_{2,1}((1,0)) = (0, 0), f_{2,1}((0,1)) = (1, 0),$$
$$f_{2,2}((1,0)) = (0, 0), f_{2,2}((0,1)) = (0, 1)$$.
Therefore we have that $g = af_{1,1} + bf_{1,2} + cf_{2,1} + df_{2,2}$, so the $f_{i,j}$ are indeed a spanning set. If we look back at the expression of $g$ from earlier, this shows that $g$ is the zero function iff $g((1,0)) = (a,b) = (0,0)$ and $g((0,1)) = (c,d) = (0,0)$ iff $a=b=c=d=0$, so the $f_{i,j}$ are also linearly independent.
A: 2) In fact the $f_{ij}$'s for $i,j\in [1,n]$ satisfy :
$$h(f_{ij},f_{kl})=\sum_r g_2(f_{ij}(v_r),f_{kl}(v_r))=\sum_r g_2(\delta_{ir}w_j,\delta_{kr}w_l)=\sum_r \delta_{ir}\delta_{kr} \delta_{jl}=\delta_{ik}\delta_{j,l}.$$
Therfore the basis is ortonormal and $h$ is positive defined.
1) This is to much computations (not a theoretic idea check the comment at the end). 
Note that : $$f_{ij}(x)=g_1(v_i,x)w_j.\qquad (A)$$ Let $v_1',\cdots v_n'$ be another basis of $V$, computing using $(A)$ with this new basis gives :
$$h(f_{ij},f_{kl})=\sum_r g_2(f_{ij}(v'_r),f_{kl}(v'_r))=\sum_r g_2(g_1(v_i,v'_r)w_j,g_1(v_k,v'r)w_l)\\=\delta_{jl}\sum_r g_1(v_i,v_r')g_1(v_k,v_r')\\=\delta_{ij}g_1(v_i,v_k)=\delta_{ij}\delta_{k,l}.$$
To go from the second line to the third line you can use a change of basis argument :
$$v_i=\sum_r g_1(v_i,v_r')v'r,$$
i.e insert it in $g_1(v_i,v_k)$ and do the same for $v_k$.
In fact this show that $h(f_1,f_2)$ is right $O(g_1)$ (orthogonal group) invariant i.e :
$$h(f_1\circ S,f_2 \circ S)=h(f_1,f_2), \: for \: S\in O(g_1).$$
