17 - Dual Basis Theorem

Dual Basis

Let $V$ be finite dimensional with basis $\mathcal{B}=\{e_1,\cdots ,e_n\}$

Define the Dual of $e_i^{\prime }$ of $e_i$ (relative to $\mathcal{B}$) by

$$e_i^{\prime }(e_j)={\delta }_{ij}=\{\begin{array}{ll}1& \text{ if }i=j\\ 0& \text{ if }i\ne j\end{array}$$

So

$${\mathcal{B}}^{\prime }:=\{e_1^{\prime },\cdots ,e_n^{\prime }\}\text{ is a basis for }{V}^{\prime }$$

where ${\mathcal{B}}^{\prime }$ is the dual basis

The assignment $e_i\mapsto e_i^{\prime }$ defines an isomorphism of vector spaces
Hence

$$\dim V=\dim V^{\prime }$$

Proof

Assume for some $a_i\in \mathbf{F}$ such that $\sum a_i{e}_i^{\prime }=0$ then for all $j$ then

$$0=0(e_j)=\left(\sum a_i{e}_i^{\prime }\right)(e_j)=\sum a_i{e}_i^{\prime }(e_j)=a_j$$

Hence ${\mathcal{B}}^{\prime }$ is linearly independent

Let $f\in V^{\prime }$ and define $a_i:=f(e_i)$
So

$$f=\sum a_j{e}_j^{\prime }$$

As $f$ evaluates to $a_i$ on $e_i$ and a linear map is determined by values on any basis then

$${\mathcal{B}}^{\prime }\text{ is spanning}$$

So ${\mathcal{B}}^{\prime }$ is a basis for $V^{\prime }$