duality with respect to a non-degenerate bilinear form

Definition 1.

Let $V$ and $W$ be finite dimensional vector spaces over a field $F$ and let $B\mathrm{:}V\mathrm{\times }W\mathrm{\to }F$ 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 $f\mathrm{:}V\mathrm{\to }V$ and $g\mathrm{:}W\mathrm{\to }W$ 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 $v\mathrm{\in }V$ and $w\mathrm{\in }W$.

The reasons why the terms “dual” and “transpose” are used are explained in the following theorems (here $V^{\ast }$ denotes the dual vector space of $V$). Notice that for a fixed element $w\in W$ one can define a linear form $V\to F$ which sends $v$ to $B(v,w)$.

Theorem 1.

Let $V\mathrm{,}W$ be finite dimensional vector spaces over $F$ which are dual with respect to a non-degenerate bilinear form $B\mathrm{:}V\mathrm{\times }W\mathrm{\to }F$. Then there exist canonical isomorphisms $V\mathrm{\cong }{W}^{\mathrm{\ast }}$ and $W\mathrm{\cong }{V}^{\mathrm{\ast }}$ given by

$$W\to V^{\ast },w\mapsto (v\mapsto B(v,w));V\to W^{\ast },v\mapsto (w\mapsto B(v,w)).$$

Theorem 2.

Let $V\mathrm{,}W$ be finite dimensional vector spaces over $F$ which are dual with respect to a non-degenerate bilinear form $B\mathrm{:}V\mathrm{\times }W\mathrm{\to }F$. Moreover, suppose $f\mathrm{:}V\mathrm{\to }V$ and $g\mathrm{:}W\mathrm{\to }W$ are transposes of each other with respect to $B$. Let $\mathrm{B}\mathrm{=}\mathrm{\{}{v}_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{v}_n\mathrm{\}}$ be a basis of $V$ and let $\mathrm{C}\mathrm{=}\mathrm{\{}{w}_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{w}_n\mathrm{\}}$ be the basis of $W$ which maps to the dual basis of $\mathrm{B}$ via the isomporphism $W\mathrm{\cong }{V}^{\mathrm{\ast }}$ defined in the previous theorem. If $A$ is the matrix of $f$ in the basis $\mathrm{B}$ then the matrix of $g$ in the basis $\mathrm{C}$ is $A^T$, the transpose matrix of $A$.

Proof of Theorem 2..

Let $V$ and $W$ be dual with respect to a non-degenerate bilinearform $B$ and let $f$ and $g$ be transposes of each other, alsowith 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$ℬ=\{v_1,\mathrm{\dots },v_n\}$ be a basis for $V$ and let$𝒞=\{w_1,\mathrm{\dots },w_n\}$ be a basis for $W$ whichcorresponds 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 thematrix of $f$ with respect to $ℬ$. Then

$$f(v_j)=\sum _{i=1}^n{\alpha }_{ij}{v}_i.$$

Let $A^{\prime }=({\beta }_{ij})$ be the matrix of $g$ with respect to$𝒞$ so that $g(w_j)={\sum }_i{\beta }_{ij}{w}_i$. We willshow that $A^{\prime }=A^T$, 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.∎