Let $V, W$ be $n$-dimensional $K$-vector spaces with a non-degenerate bilinear form $(\cdot, \cdot) : V \times W \to K$. We call a basis $(\beta_1, \dots, \beta_n)$ for $W$ dual to a basis $(\alpha_1, \dots, \alpha_n)$ for $V$ with respect to $(\cdot, \cdot)$ if for all $i, j$ we have $(\alpha_i, \beta_j) = \delta_{i j}$, where $\delta_{i j}$ denotes the Kronecker delta.
The above I've found is fairly standard terminology. I've seen assertions that such a dual basis always exists and it is unique, but I haven't really seen a proof.
Can anyone point me in the right direction for this proof? Is it constructive?
Best Answer
Thanks to Daniel Fischer, making everything slightly more verbose:
Existence: The bilinear form gives $\Phi : W \to V^*$, where $\Phi(w) = \lambda_w \in V^*$, for $\lambda_w(v) = (v, w)$. This is a linear map, by linearity of $(\cdot, \cdot)$. It is injective by non-degeneracy of $(\cdot, \cdot)$, and since $\dim W = \dim V^*$, it must also be surjective. A bijective linear map is an isomorphism, so we have a linear inverse $\Phi^{-1}$. Note for any $w \in W$ we have $$(v, \Phi^{-1}(\lambda_w)) = (v, w) = \lambda_w(v)$$ and since $\Phi$ is surjective this works for any $\lambda \in V^*$.
We can construct the dual basis of $(\alpha_1, \dots, \alpha_n)$ in $V^*$ by setting $\lambda_i(\alpha_j) = \delta_{i j}$ and taking $(\lambda_1, \dots, \lambda_n)$ (see for example Prop 1.0.2 of these notes).
Now, $(\beta_1, \dots, \beta_n)$, where $\beta_i = \Phi^{-1}(\lambda_i)$, gives a basis for $W$ that satisfies the required duality: $$ (\alpha_i, \beta_j) = (\alpha_i, \Phi^{-1}(\lambda_j)) = \lambda_j(\alpha_i) = \delta_{i j} $$ It is a basis because we're applying an isomorphism to elements that also constitute a basis.
Uniqueness: Suppose $(\beta_1', \dots, \beta_n')$ is another basis of $W$ satisfying the duality condition. Since the $\beta_1, \dots, \beta_n$ span $W$ we can write $\beta_j' = \sum_k c_{j k} \beta_k$ for all indices $j$. Applying $(\alpha_i, \cdot)$ to that equation we get: $$\delta_{j i} = \delta_{i j} = (\alpha_i, \beta_j') = \sum_k c_{j k} (\alpha_i, \beta_k) = c_{j i}$$ hence $\beta_j' = \beta_j$ for all indices $j$.