# Abstract Nonsense

## Bilinear Forms

Point of post: In this post I give a quick overview of the very basic concepts in the topic of “bilinear forms”. The main things covered being what a bilinear form on the product of two vector spaces $\mathscr{U}\times\mathscr{V}$ is, and finding a basis for it. Incidental little facts may pop up.

Remark: If $\mathscr{U,V}$ are vector spaces, we follow Roman’s (in the author’s opinion) better notation $\mathscr{U}\boxplus\mathscr{V}$ for the external direct sum and reserve $\mathscr{U}\oplus\mathscr{V}$ for the internal direct sum. Also, we shall use the notation $A\leqslant B$ to denote “$A$ is a linear subspace of $B$“. Lastly, for $k\in\mathbb{N}$ we define $[k]=\left\{1,\cdots,k\right\}$

Motivation

We know that given a vector space $\mathcal{V}$ over a field $F$ it is fruitful to study the dual space of $\mathscr{V}$, denoted as always $\text{Hom}\left(\mathscr{V},F\right)$ or $\mathscr{V}^*$, which is the set of all linear mappings $\varphi:\mathscr{V}\to F$.  We have also seen that it is interesting to study $\mathscr{U}\boxplus\mathscr{V}$ given vector spaces $\mathscr{U,V}$ over a field $F$. It makes sense then to ask questions about maps $B:\mathscr{U}\boxplus\mathscr{V}\to F$. This kind of mapping has arisen before in our studies, for example the map

$f:\mathscr{V}\boxplus\mathscr{V}^{*}\to F:(x,\varphi)\mapsto \varphi(x)$

Even more common, is the dot product on usual real coordinate space

$\displaystyle \langle\cdot,\cdot\rangle:\mathbb{R}^n\boxplus\mathbb{R}^n\to\mathbb{R}:\left((x_1,\cdots,x_n),(y_1,\cdots,y_n)\right)\mapsto \sum_{j=1}^{n}x_jy_j$

Maybe less obvious, but another very important example is the determinant of $2\times 2$  complex matrices when thought of as being functions of the column vectors, namely:

$\det:\mathbb{C}^2\boxplus\mathbb{C}^2\to\mathbb{C}:\left(\begin{bmatrix}z_1 \\ z_2\end{bmatrix},\begin{bmatrix} z_3\\ z_4\end{bmatrix}\right)\mapsto z_1z_4-z_2z_3$

Bilinear Forms

All of these are certainly very important, and commonplace functions the working mathematician encounters on a day-to-day basis. So the question becomes, how do these maps relate, are they specific examples of some larger definable class of maps on vector spaces? Intuition first may tell us that they are linear functionals on the external direct product $\mathscr{U}\boxplus\mathscr{V}$, but a quick check botches this idea, namely:

$\det\left(\begin{bmatrix}1+i\\2+i\end{bmatrix},\begin{bmatrix}3+i \\4+i\end{bmatrix}\right)\ne\det\left(\begin{bmatrix}i \\ i\end{bmatrix},\begin{bmatrix}i\\i\end{bmatrix}\right)+\det\left(\begin{bmatrix} 1 \\ 3\end{bmatrix},\begin{bmatrix} 2 \\4\end{bmatrix}\right)$

But, a little bit of digging reveals that while they aren’t linear in both entries simultaneously they are linear in each entry individually. For example, one can easily check that

$\det\left(\begin{bmatrix}z_1+z_2\\ z_3+z_4\end{bmatrix},\begin{bmatrix}z_5\\z_6\end{bmatrix}\right)=\det\left(\begin{bmatrix}z_1\\z_3\end{bmatrix},\begin{bmatrix}z_5\\z_6\end{bmatrix}\right)+\det\left(\begin{bmatrix}z_2\\z_4\end{bmatrix},\begin{bmatrix}z_5\\z_6\end{bmatrix}\right)$

More generally we call a mapping

$B:\mathscr{U}\boxplus\mathscr{V}\to X$

(where $X$ is any unspecified vector space over $F$bilinear if

$B\left(\alpha_1x_1+\alpha_2x_2,y\right)=\alpha_1B(x_1,y)+\alpha_2B(x_2,y)$

And

$B\left(x,\beta_1y_1+\beta_2y_2\right)=\beta_1B(x,y_1)+\beta_2B(x,y_2)$

If it happens that these are mappings into the ground field $F$, symbolically:

$B:\mathscr{U}\boxplus\mathscr{V}\to F$

we call $B$ a bilinear form. Somethings become immediately apparent from this definition

Remark: From here on out, unless specified otherwise, we will have that $\mathscr{U,V}$ are vector spaces over a field $F$ and $B:\mathscr{U}\boxplus\mathscr{V}\to F$ is a bilinear form.

Theorem: $B(x,0)=B(y,0)=0$ for any $x\in\mathscr{U}$ and $y\in\mathscr{V}$.

Proof: This follows immediately since $B(x,0)=B(x,0+0)=B(x,0)+B(x,0)$, and so $B(x,0)=0$. Similarly, $B(0,y)=B(0+0,y)=B(0,y)+B(0,y)$ and so $B(0,y)=0$. $\blacksquare$

So, we can consider

$\text{Bil}\left(\mathscr{U},\mathscr{V}\right)=\left\{B:\mathscr{U}\boxplus\mathscr{V}\to F\mid B\text{ is bilinear}\right\}$

Consider though that if $B_1,B_2\in\text{Bil}\left(\mathscr{U},\mathscr{V}\right)$ and $\alpha\in F$ then if we define

$B_3:\mathscr{U}\boxplus\mathscr{V}\to F:(x,y)\mapsto B_1(x,y)+B_2(x,y)$

and

$B_4:\mathscr{U}\boxplus\mathscr{V}\to F:(x,y)\mapsto \alpha B_1(x,y)$

we can easily check that $B_3,B_4\in\text{Bil}\left(\mathscr{U},\mathscr{V}\right)$. Thus, we may define $B_1+B_2=B_3$ and $\alpha B_1 =B_4$, and so noticing that

$\bold{0}:\mathscr{U}\boxplus\mathscr{V}\to F:(x,y)\mapsto 0$

is bilinear we can easily conclude that $\text{Bil}\left(\mathscr{U},\mathscr{V}\right)\leqslant F^{\mathscr{U}\times\mathscr{V}}$ (where this latter space is the set of all mappings of the form $\mathscr{U}\times\mathscr{V}\to F$).

Basis and Dimension

With this insight, the first question is, if $\dim_F \mathscr{U}=m$ and $\dim_F\mathscr{V}=n$ what is $\dim_F\text{Bil}\left(\mathscr{U},\mathscr{V}\right)$? But, finding dimension is synonymous (for most finite-dimensional cases)  with finding a basis for the space. Thus, our real goal is to ascertain whether given a basis $\{x_1,\cdots,x_m\}$ for $\mathscr{U}$ and a basis $\{y_1,\cdots,y_n\}$ for $\mathscr{V}$ how we can create a related basis for $\text{Bil}\left(\mathscr{U},\mathscr{V}\right)$ . The first step in this process is the following theorem:

Theorem: For any $\left\{\gamma_{i,j}:i\in[m]\text{ and }j\in [n]\right\}\subseteq F$ there exits a unique $B\in\text{Bil}\left(\mathscr{U},\mathscr{V}\right)$ such that $B(x_i,y_j)=\gamma_{i,j}$.

Proof: We first notice that if $x\in\mathscr{U}$ and $y\in\mathscr{V}$ that $\displaystyle x=\sum_{i=1}^{m}\alpha_i x_i$ and $\displaystyle y=\sum_{j=1}^{n}\beta_j y_j$. Thus:

\displaystyle \begin{aligned}B(x,y) &= B\left(\sum_{i=1}^{m}\alpha_i x_i,\sum_{j=1}^{n}\beta_j y_j\right)\\ &= \sum_{i=1}^{m}\alpha_i B\left(x_i,\sum_{j=1}^{n}\beta_j y_j\right)\\ &=\sum_{i=1}^{m}\alpha_i\left(\sum_{j=1}^{n}\beta_j B(x_i,y_j\right)\\ &=\sum_{i=1}^{m}\sum_{j=1}^{n}\alpha_i \beta_j B(x_i,y_j)\end{aligned}\quad\quad(1)

So, define

$\displaystyle B(x,y)=B\left(\sum_{i=1}^{m}\alpha_i x_i,\sum_{j=1}^{n}\beta_j y_j\right)=\sum_{i=1}^{m}\sum_{j=1}^{n}\alpha_i\beta_j \gamma_{i,j}$

Evidently then $B(x_i,y_j)=\gamma_{i,j}$ and so the existence is clear, and uniqueness follows by considering if $B'$ were another such bilinear form then using $(1)$ we may see that

$\displaystyle B'\left(\sum_{i=1}^{m}\alpha_i x_i,\sum_{j=1}^{n}\beta_j y_j\right)=\sum_{i=1}^{m}\sum_{j=1}^{n}\alpha_i \beta_j B(x_i,y_j)=\sum_{i=1}^{m}\sum_{j=1}^{n}\alpha_i \beta_j \gamma_{i,j}=B\left(\sum_{i=1}^{m}\alpha_i x_i,\sum_{j=1}^{n}\beta_j y_j\right)$

and so $B=B'$. $\blacksquare$

With this we are now able to produce an explicit basis for $\text{Bil}\left(\mathscr{U},\mathscr{V}\right)$ given a basis for $\mathscr{U}$ and one for $\mathscr{V}$.

Theorem: Let $\{x_1,\cdots,x_m\}$ be a basis for $\mathscr{U}$ and $\{y_1,\cdots,y_n\}$ a basis for $\mathscr{V}$. Then,

$\mathscr{B}=\left\{B_{i,j}:i\in[m]\text{ and }j\in[n]\right\}$

is a basis where

$B_{i,j}(x_p,y_q)=\delta_{i,p}\delta_{j,q}$

Proof: The fact that such $B_{i,j}$ exists is a consequence of the prior theorem. Suppose then that

$\displaystyle \sum_{i=1}^{m}\sum_{j=1}^{n}\alpha_{i,j}B_{i,j}=\bold{0}$

Then, evaluating at $(x_k,y_\ell),$ gives

$\displaystyle 0=\sum_{i=1}^{m}\sum_{j=1}^{n}\alpha_{i,j}B_{i,j}(x_k,y_{\ell})=\alpha_{k,\ell}$

Thus, ranging $k$ over $\{1,\cdots,m\}$ and $\ell$ over $\{1,\cdots,n\}$ we find that

$\alpha_{1,1}=\cdots=\alpha_{1,n}=\cdots=\alpha_{m,1}=\cdots=\alpha_{m,n}=0$

and thus $\mathscr{B}$ is linearly independent. To see that $\text{span }\mathscr{B}=\text{Bil}\left(\mathscr{U},\mathscr{V}\right)$ we let $B\in\text{Bil}\left(\mathscr{U},\mathscr{V}\right)$ be arbitrary. We claim that

$\displaystyle B=\sum_{i=1}^{m}\sum_{j=1}^{n}B(x_i,y_j)B_{i,j}$

To see this we note that if $\displaystyle x=\sum_{i=1}^{m}\alpha_i x_i\in\mathscr{U}$ and $\displaystyle y=\sum_{j=1}^{n}\beta_j y_j\in\mathscr{V}$ that

\displaystyle \begin{aligned}\sum_{i=1}^{m}\sum_{j=1}^{n}B(x_i,y_j)B_{i,j}(x,y)&=\sum_{i=1}^{m}\sum_{j=1}^{n}B(x_i,y_j)\left(\sum_{k=1}^{m}\sum_{\ell=1}^{n}\alpha_k\beta_{\ell} B_{i,j}(x_k,y_{\ell})\right)\\ &= \sum_{i=1}^{m}\sum_{j=1}^{n}\alpha_i\beta_j B(x_i,y_j)\\ &= B(x,y)\end{aligned}

Thus, $\mathscr{B}$ is linearly independent and $\text{span }\mathscr{B}=\text{Bil}\left(\mathscr{U},\mathscr{V}\right)$, thus $\mathscr{B}$ is a basis for $\text{Bil}\left(\mathscr{U},\mathscr{V}\right)$ as required. $\blacksquare$

From this we may gather the following rather aesthetically appealing corollary:

Corollary: $\displaystyle \dim_F\text{Bil}\left(\mathscr{U},\mathscr{V}\right)=\dim_F\mathscr{U}\cdot\dim_F\mathscr{V}$

References:

1. Halmos, Paul R. “Bilinear Forms.” Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print.

October 23, 2010 -

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