# 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 $B:V\times W\to F$ 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 $w\in W$.

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 $V,W$ 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 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 $A^{\prime}=(\beta_{ij})$ 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 $A^{\prime}=A^{T}$, 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. ∎

Title duality with respect to a non-degenerate bilinear form DualityWithRespectToANondegenerateBilinearForm 2013-03-22 16:23:02 2013-03-22 16:23:02 alozano (2414) alozano (2414) 8 alozano (2414) Definition msc 15A99 BilinearForm PolaritiesAndForms