Writing $T:V \rightarrow \mathbb{F}$ as an inner product. Let $V$ be a finite dimensional vector inner product space over $\mathbb{F}$, and let $g:V \rightarrow \mathbb{F}$ be a linear transformation. Then there exists a unique vector $y \in V$ such that $g(x)=\langle x,y\rangle$.
This vector is found to be: $$y=\sum_{i=1}^{n}\overline{g(v_i)}v_i$$ Where $\{v_1,v_2,...v_n\}$ is an orthonormal basis for $V$. I think the intuition is basically:
\begin{align}
g(x)&=\langle x,y\rangle\\
g(x)&=\langle x, \sum{a_iv_i} \rangle \iff \\
g(v_j)&= \langle v_j, \sum{a_iv_i} \rangle \\
g(v_j)&=\overline{a_j}
\end{align}
Where I put the if and only if to indicate that the statement is true if and only if the basis vectors of the two linear transformations agree.
So this $y$ is determined by an orthonormal basis and the transformation and more over there is only one such $y$ for any linear transformation over an inner product space (From $V$ to $\mathbb{F}$). 
My question is: Given A finite dimensional vector space $V$, and a linear transformation $T$, will $y$ be guaranteed different for different inner products over the same space? For example if we are considering $P^2(\mathbb{R})$ (the polynomials of up to degree two with real coefficients) with the inner product $\langle f,g \rangle= \int_0^1 f(t)g(t) \ dt$ and $g(f)=f(0)+f'(0),$ then we find that $g(f)=\langle f,210x^2-204x+33 \rangle.$ If I change this inner product (but remain in the same vector space), will $y$ be guaranteed to be different? 
Or maybe I should ask:
let $T=x\cdot y$,   $T=x+y'$ (sorry bad notation)  be the representation of $T$ in the inner product space $(V, \cdot)$ and $(V,+)$, ($T:V \rightarrow \mathbb{F}$). Then, is it true that $y = y'$ if and only if $\cdot$ and $+$ are the same inner product?
 A: Given an n-dimensional real vector-space and a basis $\{e_1,\ldots,e_n\}$, you can define an inner product by
$$
\langle x,y\rangle_e=\sum x_i y_i,
$$
if $x=\sum x_i e_i, y=\sum y_i e_i$. If we have another basis $\{e_1,f_2,\ldots,f_n\}$, it is clear that the linear functional $g(x)=\langle x,e_1\rangle_e=\langle x,e_1\rangle_f$ is represented by the same vector (namely $e_1$). So having the information for just one linear functional is not sufficient to ensure that two inner products agree.
[Edit] For the argument above to work, you actually need $\{e_1,f_2,\ldots,f_n\}$ to be an orthonormal basis with respect to $\langle .,.\rangle_e$. In dimension strictly larger than 1, such a choice can always be made. In dimension two, consider the two bases $\{e_1,e_2\},\{e_1,-e_2\}$, where $e_1$ and $e_2$ denotes the standard basis on $\mathbb R^2$ and $\langle.,.\rangle_e$ is the usual standard inner product (dot product).
On the other hand, if you have such information for all linear functionals (as opposed to just one), then it follows that $\langle x,y\rangle_e=\langle x,y\rangle_f$ for all $x,y\in V$ (since $x\mapsto \langle x,y\rangle_e$ is a linear functional which is represented by $y$ with respect to $\langle .,.\rangle_e$), which clearly implies that the two inner products agree.
