The Unapologetic Mathematician

Mathematics for the interested outsider

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 »

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