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 $B_{ij}=-B_{ji}$.

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Posted by John Armstrong | Algebra, Linear Algebra

9 Comments	»

1. [...] Inner Products Now that we’ve got bilinear forms, let’s focus in on when the base field is . We’ll also add the requirement that our [...]

Pingback by | April 15, 2009 | Reply

2. [...] Inner Products Now consider a complex vector space. We can define bilinear forms, and even ask that they be symmetric and nondegenerate. But there’s no way for such a form to [...]

Pingback by | April 22, 2009 | Reply

3. [...] start with either a bilinear or a sesquilinear form on the vector space . Let’s also pick an arbitrary basis of . I want [...]

Pingback by | June 24, 2009 | Reply

4. [...] of matrices and forms and try to tie both views of our matrix together. We’re considering a bilinear form \$ on a vector space over the real or complex numbers, which we can also think of as a linear [...]

Pingback by | June 25, 2009 | Reply

5. [...] back in on a real, finite-dimensional vector space and give it an inner product. As a symmetric bilinear form, the inner product provides us with an isomorphism . Now we can use functoriality to see what this [...]

Pingback by | October 29, 2009 | Reply

6. [...] I said above, this is a bilinear form. Further, Clairaut’s theorem tells us that it’s a symmetric form. Then the spectral [...]

Pingback by | November 24, 2009 | Reply

7. [...] this as a bilinear function which takes in two vectors and spits out a number . That is, is a bilinear form on the space of tangent vectors at [...]

Pingback by | September 20, 2011 | Reply

8. [...] the next three families of linear Lie algebras we equip our vector space with a bilinear form . We’re going to consider the endomorphisms such [...]

Pingback by | August 9, 2012 | Reply

9. [...] can now define a symmetric bilinear form on our Lie algebra by the [...]

Pingback by | September 3, 2012 | Reply

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