In a vector space $V$, If $v^tAw = v^tBw$ for all $v,w \in V$, does it imply $A=B$?

Question

OK. So, basically this question came up when I was trying to solve a homework question about how the matrix representation of a bi-linear form on $V$ changes if we change the basis on $V$.

Let's say that $<\cdot,\cdot>:V\times V \to {K}$ is a bi-linear form on $V$ over the field $K$.

It is easy to check that any function $T: V \times V \to K$ given by $T(v,w) = v^tAw$ for $A \in L(V,V)$ is a bilinear form. Conversely, if we fix a basis on $V$ then any bilinear form on $V$ can be represented by a matrix whose $(i,j)$ entry is $<e_i,e_j>$.

Now since a bi-linear form is linear in each argument, it gives rise to two linear transformation as the following:

1. $R_{w}(v)=<v,w>=v^t(Aw)$ where $w$ is fixed and $Aw$ is regarded as a linear transformation from $K^n$ to $K^n$.
2. $L_{v}(w)=<v,w>=(v^tA)w$ where this time $v$ is fixed and $v^tA$ is regarded as a linear transformation from $(K^n)^*$ to $(K^n)^*$.

Am I right so far?

Therefore, some questions about non-degeneracy of a bi-linear form can be studied by checking whether $R_{w}(v)$ or $L_{v}(w)$ has a trivial kernel or not.

So, back to my question, if for all $v,w \in V$ we have $v^tAw=v^tBw$ if and only if $v^t(A-B)w=0$.

Fixing $v$ and looking at $(v^t(A-B))w=0$ as a linear transformation in $w$, since it is a linear transformation that vanishes for all vectors, it must be the zero transformation. Therefore $v^t(A-B)=0$. A similar argument shows that $A-B=0$ therefore $A=B$.

Now to check that what a change of basis on $V$ does to the matrix representation of a bi-linear form we do as follows:

Suppose that $P$ is a change of coordinates matrix that receives the components of a vector in our old basis $B$ and gives its new components with respect to the new basis $C$ and $M$ is the matrix representation of our binary form in the old basis and $N$ is the new representation.

Therefore $[v]_{C}=[P]_{C}^{B}[v]_{B}$ and $[w]_{C}=[P]_{C}^{B}[w]_{B}$. Now we have:

$$<[v]_{C},[w]_{C}>=<[P]_{C}^{B}[v]_{B},[P]_{C}^{B}[w]_{B}>=<[v]_B,[w]_B>$$ $$([P]_{C}^{B})^t N [P]_{C}^{B} = M$$

Therefore our new representation is given by:

$$N = (([P]_{C}^{B})^t)^{-1} M ([P]_{C}^{B})^{-1}$$

Is it a correct proof?