Definition 1   Let $V$ and $W$ be finite dimensional vector spaces over a field $F$ and let $B:V\times W \to F$ be a non-degenerate bilinear form. Then we say that $V$ and $W$ are dual with respect to $B$ .

Example 1   Let $V$ be a finite dimensional vector space and let $W=V^\ast$ be the dual space of $V$ , i.e. $W$ is the vector space formed by all linear transformations $V\to F$ . Let $B:V\times V^\ast \to F$ be defined by $B(v,f)=f(v)$ for all $v\in V$ and all $f:V\to F$ in $V^\ast$ . Then $B$ is a non-degenerate bilinear form and $V$ and $V^\ast$ are dual with respect to $B$ .

Definition 2   Let $f:V\to V$ and $g:W\to W$ be linear transformations. We say that $f$ and $g$ are transposes of each other with respect to $B$ if $$B(f(v),w)=B(v,g(w))$$ for all $v\in V$ and $w\in W$ .

Theorem 1   Let $V,W$ be finite dimensional vector spaces over $F$ which are dual with respect to a non-degenerate bilinear form $B:V\times W \to F$ . Then there exist canonical isomorphisms $V \cong W^\ast$ and $W\cong V^\ast$ given by $$W\to V^\ast,\ w\mapsto (v\mapsto B(v,w));\quad V\to W^\ast,\ v\mapsto (w\mapsto B(v,w)).$$

Theorem 2   Let $V,W$ be finite dimensional vector spaces over $F$ which are dual with respect to a non-degenerate bilinear form $B:V\times W \to F$ . Moreover, suppose $f:V\to V$ and $g:W\to W$ are transposes of each other with respect to $B$ . Let $\mathcal{B}=\{v_1,\ldots,v_n\}$ be a basis of $V$ and let $\mathcal{C}=\{w_1,\ldots,w_n\}$ be the basis of $W$ which maps to the dual basis of $\mathcal{B}$ via the isomporphism $W\cong V^\ast$ defined in the previous theorem. If $A$ is the matrix of $f$ in the basis $\mathcal{B}$ then the matrix of $g$ in the basis $\mathcal{C}$ is $A^T$ , the transpose matrix of $A$ .

Proof. [Proof of Theorem 2.] Let $V$ and $W$ be dual with respect to a non-degenerate bilinear form $B$ and let $f$ and $g$ be transposes of each other, also with respect to $B$ so that: $$B(f(v),w)=B(v,g(w))$$ for all $v\in V$ and $w \in W$ . By Theorem 1, we have $W\cong V^\ast$ . Let $\mathcal{B}=\{v_1,\ldots,v_n\}$ be a basis for $V$ and let $\mathcal{C}=\{w_1,\ldots,w_n\}$ be a basis for $W$ which corresponds to the dual basis of $V^\ast$ via the isomorphism $W\cong V^\ast$ . Then $B(v_i,w_j)=1$ for $i=j$ and equal to $0$ otherwise. Let $A=(\alpha_{ij})$ be the matrix of $f$ with respect to $\mathcal{B}$ . Then $$f(v_j)=\sum_{i=1}^n \alpha_{ij}v_i.$$ Let $A'=(\beta_{ij})$ be the matrix of $g$ with respect to $\mathcal{C}$ so that $g(w_j)=\sum_i \beta_{ij}w_i$ . We will show that $A'=A^T$ , the transpose of $A$ . Indeed: $$B(f(v_j),w_k)=B(\sum_i \alpha_{ij}v_i,w_k)=\alpha_{kj}$$ and also $$B(f(v_j),w_k)=B(v_j,g(w_k))=B(v_j,\sum_i \beta_{ik}w_i)=\beta_{jk}.$$ Therefore $\beta_{jk}=\alpha_{kj}$ for all $k$ and $j$ , as desired.