TITLE: Dual basis existence and uniqueness.
QUESTION [5 upvotes]: In Wikipedia, on Dual Basis they say:

"Algebraically, a dual set always exists, and gives an injection from $V$ into $V^*$. However, a dual basis exists if and only if a vector space is finite dimensional, and each basis has a unique dual basis..."

The following afterwards I can't understand. Could you please give me some direct proof for that interesting result of existence and uniqueness of dual basis in finite dimensional vector space?
Thanks.

REPLY [8 votes]: Say you have a basis $b_1,\ldots b_n$ for a $n$-dimensional vector space $V$. Let $b_1^*,\ldots,b_n^* \,:\, V \to \mathbb{R}$ be the unique mappings for which $$
  x = \sum_{k=1}^n b_k^*(x)b_k \text{,}
$$
i.e. the mappings which take a a vector $x$ to the coefficient of $b_k$ in the unique representation of $x$ as a sum of scaled basis vectors. That those mappings exists and are unique follows immediatly from that fact that $\{b_k\}$ is a basis. It also follows that $$
  b_j^*(b_i) = \delta_{i,j} \text{.}
$$
The $b_k^*$ are obviously linear, and thus members of $V^*$. The also form a basis of $V^*$, since for every $f \in V^*$ you have $$
  f(x) = f\left(\sum_{k=1}^n b_k^*(x)b_k\right) \overset{\text{linearity of $f$}}{=} \sum_{k=1}^n f(b_k)b_k^*(x) \text{,}
$$ where the right-hand side is a representation of $f$ as a sum of scaled $b_k^*$.