Given a basis $a_i$ to a space, does knowing $\langle f, a_i \rangle$ for each $i$ determine $f$? Let $A=\text{span}(a_1, ..., a_n)$ where $a_i$ is a basis (not orthonormal) wrt inner product $\langle ,\rangle$. Suppose $f \in A$. 
If I know $\langle f, a_i \rangle$ for each $i$, does that uniquely determine $f$? I think so because we get $n$ equations for $n$ coefficients, but how to prove that those $n$ equations are all "different" to each other?
 A: Suppose you know that $\langle f,a_i \rangle=b_i$ for all $1 \leq i\leq n$. Let $f,g$ be two vectors satisfying these conditions. Hence
$$\langle f,a_i \rangle=\langle g,a_i \rangle $$
for all $1 \leq i\leq n$. Subtracting and using linearity of the inner product gives
$$\langle f-g,a_i \rangle =0$$ for all $1 \leq i \leq n$.
But there is only one vector which is perpendicular to all $a_i$...
EDIT:
In order to show the existence express $f$ as a linear combination
$$f=f_1 a_1+\dots f_n a_n .$$
Your equations become $$f_1 \langle a_1,a_i \rangle+ \dots+f_i \langle a_i,a_i \rangle + \dots +f_n \langle a_n,a_i \rangle=b_i $$
for $1 \leq i \leq n$. These can be written in matrix form as  $$G \mathbf{x}=\mathbf{b} $$
where $G$ is the Gram matrix of the basis vectors. Since it is invertible there is a unique solution for the coefficients $\{f_i \}$.
A: No.  Simply knowing $\left<f,a_i\right>$ is not quite enough to write down the vector $f$.  One also needs to know $\left<a_i, a_j\right>$ for $0\leq i,j,\leq n$.  This is exactly why orthonormal bases are nice.  If $\{a_1,...,a_n\}$ is an orthonormal basis, then $\left<a_i,a_j\right> = \delta_{i,j}$, where $\delta_{i,j} =1$ if $i=j$ and $0$ otherwise.  In this case, we have
\begin{equation*}
f = \left<f,a_1\right>a_1 + ...+\left<f,a_n\right>a_n = \sum_{i=1}^n \left< f,a_i\right> a_i.\end{equation*}
It is true, however, as noted in the answer provided by user1337, that $f$ is in fact uniquely determined, but this is still not enough information to write down $f$ in terms of the basis $\{a_1,...,a_n\}$.
A: Let $M$ be the $n\times n$ matrix with the $a_i$ as columns.  The values $\langle f,a_i\rangle$ are the entries of $M^Tf$.  This determines $f$ because $M$ (and hence $M^T$) is invertible; explicitly,
$$f = (M^T)^{-1} \left[\begin{matrix} \langle f,a_1\rangle \\ \vdots \\ \langle f,a_n\rangle \end{matrix}\right]$$
