# The Unapologetic Mathematician

## Bilinear Forms

Now that we’ve said a lot about individual operators on vector spaces, I want to go back and consider some other sorts of structures we can put on the space itself. Foremost among these is the idea of a bilinear form. This is really nothing but a bilinear function to the base field: $B:V\times V\rightarrow\mathbb{F}$. Of course, this means that it’s equivalent to a linear function from the tensor square: $B:V\otimes V\rightarrow\mathbb{F}$.

Instead of writing this as a function, we will often use a slightly different notation. We write a bracket $B(v,w)=\langle v,w\rangle$, or sometimes $\langle v,w\rangle_B$, if we need to specify which of multiple different inner products under consideration.

Another viewpoint comes from recognizing that we’ve got a duality for vector spaces. This lets us rewrite our bilinear form $B:V\otimes V\rightarrow\mathbb{F}$ as a linear transformation $B_1:V\rightarrow V^*$. We can view this as saying that once we pick one of the vectors $x\in V$, the bilinear form reduces to a linear functional $\langle v,\underbar{\hphantom{X}}\rangle:V\rightarrow\mathbb{F}$, which is a vector in the dual space $V^*$. Or we could focus on the other slot and define $B_2(v)=\langle\underbar{\hphantom{X}},v\rangle\in V^*$.

We know that the dual space of a finite-dimensional vector space has the same dimension as the space itself, which raises the possibility that $B_1$ or $B_2$ is an isomorphism from $V$ to $V^*$. If either one is, then both are, and we say that the bilinear form $B$ is nondegenerate.

We can also note that there is a symmetry on the category of vector spaces. That is, we have a linear transformation $\tau_{V,V}:V\otimes V\rightarrow V\otimes V$ defined by $\tau_{V,V}(v\otimes w)=w\otimes v$. This makes it natural to ask what effect this has on our form. Two obvious possibilities are that $\tau_{V,V}\circ B=B$ and that $\tau_{V,V}\circ B=-B$. In the first case we’ll call the bilinear form “symmetric”, and in the second we’ll call it “antisymmetric”. In terms of the maps $B_1$ and $B_2$, we see that composing $B$ with the symmetry swaps the roles of these two functions. For symmetric bilinear forms, $B_1=B_2$, while for antisymmetric bilinear forms we have $B_1=-B_2$.

This leads us to consider nondegenerate bilinear forms a little more. If $B_2$ is an isomorphism it has an inverse $B_2^{-1}$. Then we can form the composite $B_2^{-1}\circ B_1:V\rightarrow V$. If $B$ is symmetric then this composition is the identity transformation on $V$. On the other hand, if $B$ is antisymmetric then this composition is the negative of the identity transformation. Thus, the composite transformation measures how much the bilinear transformation diverges from symmetry. Accordingly, we call it the asymmetry of the form $B$.

Finally, if we’re working over a finite-dimensional vector space we can pick a basis $\left\{e_i\right\}$ for $V$, and get a matrix for $B$. We define the matrix entry $B_{ij}=\langle e_i,e_j\rangle_B$. Then if we have vectors $v=v^ie_i$ and $w=w^je_j$ we can calculate

$\displaystyle\langle v,w\rangle=\langle v^ie_iw^je_j\rangle=v^iw^j\langle e_i,e_j\rangle=v^iw^jB_{ij}$

In terms of this basis and its dual basis $\left\{\epsilon^j\right\}$, we find the image of the linear transformation $B_1(v)=\langle v,\underbar{\hphantom{X}}\rangle=v^iB_{ij}\epsilon^j$. That is, the matrix also can be used to represent the partial maps $B_1$ and $B_2$. If $B$ is symmetric, then the matrix is symmetric $B_{ij}=B_{ji}$, while if it’s antisymmetric then $B_{ij}=-B_{ji}$.

April 14, 2009 Posted by | Algebra, Linear Algebra | 9 Comments