24 - Dual Basis Matrix Transpose Theorem

Dual Basis Matrix Transpose Theorem

Let $V$ and $W$ be finite dimensional
Let ${\mathcal{B}}_W$ and ${\mathcal{B}}_V$ be bases for $W$ and $V$ then

For any linear map $T:V\to W$

$$({}_{{\mathcal{B}}_W}[T{]}_{{\mathcal{B}}_V}{)}^T={}_{{\mathcal{B}}_V^{\prime }}[T{]}_{{\mathcal{B}}_W^{\prime }}$$

where ${\mathcal{B}}_W^{\prime }$ and ${\mathcal{B}}_V^{\prime }$ are the dual bases

Proof

Let ${\mathcal{B}}_V=\{e_1,\cdots ,e_n\}$ and ${\mathcal{B}}_W=\{x_1,\cdots ,x_m\}$ and

$${}_{{\mathcal{B}}_W}[T{]}_{{\mathcal{B}}_V}=A=(a_{ij})$$

then

$$T(e_j)=\sum _{i=1}^m{a}_{ij}{x}_i\text{ and hence }{x}_i^{\prime }(T(e_j))=a_{ij}$$

Let

$${}_{{\mathcal{B}}_V^{\prime }}[T^{\prime }{]}_{{\mathcal{B}}_W^{\prime }}=B=(b_{ij})$$

then

$$T^{\prime }(x_i^{\prime })=\sum _{j=1}^n{b}_{ji}{e}_j^{\prime }\text{ and hence }(T^{\prime }(x_i^{\prime }))(e_j)=b_{ji}$$

As by definition then

$$(T^{\prime }(x_i^{\prime }))(e_j)=x_i^{\prime }(T(e_j))$$

Hence

$$a_{ij}=b_{ji}⟹A^T=B$$