Multilinear Forms

Definition 44   (Multilinear and Alternating Forms)

1. $\Phi : V^{r} \longrightarrow K$ is a multilinear form if $\Phi$ is linear on each variable, i.e., $\Phi(\ldots, c \cdot v_{i} + u_{i}, \ldots) = c \cdot \Phi(\ldots, v_{i}, \ldots) + \Phi(\ldots, u_{i}, \ldots)$, for each $i,\: 1 \leq i \leq r$.
   
   

   In the above equality, the parts represented by $\cdots$ denote the variables which are kept fixed.

2. $\Phi$ is an alternating form if $\Phi$ is a multilinear form and $\Phi(v_{1}, \ldots, v_{r}) = 0$ whenever $v_{i} = v_{j}$ for some $i \not= j$.

Remarks:

1. For a fixed $r$, a multilinear form is also called a $r$-linear form. A $1$-linear form is nothing but a linear functional. A $2$-linear form is also called a bilinear form.
2. Notations:

   $\begin{array}{cl} (a) & M^{r}(V) = \{\phi : V^{r} \longrightarrow K \;\,\rule[... ...0.005in}{.2in}\,\; \phi \mbox{ is $r$-linear and alternating}\}. \end{array}$

3. It is easy to check that $M^{r}(V),\: \Lambda^{r}(V)$ are subspaces of ${\cal F}(V^{r}, K)$.
4. It is easy to check that given $\phi \in M^{r}(V)$, $\phi \in \Lambda^{r}(V)$ if and only if
   $\phi(v_{1}, \ldots, v_{r}) = - \phi(v_{1}, \ldots, v_{i-1}, v_{j}, v_{i+1}, \ldots, v_{j-1}, v_{i}, v_{j+1}, \ldots, v_{r})$
   
   

   Pf:

   $\Longrightarrow$

   : By multilinearity,
   
   $\phi(v_{1}, \ldots, v_{i-1}, v_{i} + v_{j}, v_{i+1}, \ldots, v_{j-1}, v_{i} + v_{j}, v_{j+1}, \ldots, v_{r})$

   $\begin{array}{ccl} & = & \phi(v_{1}, \ldots, v_{i-1}, v_{i}, v_{i+1}, \ldot... ...i+1}, \ldots, v_{j-1}, v_{j}, v_{j+1}, \ldots, v_{r}) \\ & = & 0. \end{array}$

   The first and the fourth items of the middle term are 0, since $\phi$ is alternating. The conclusion is then obvious.

   $\Longleftarrow$

   : Conversely, if $\phi \in M^{r}(V)$ and

   $\phi(v_{1}, \ldots, v_{r}) = - \phi(v_{1}, \ldots, v_{i-1}, v_{j}, v_{i+1}, \ldots, v_{j-1}, v_{i}, v_{j+1}, \ldots, v_{r})$ , then $\phi(v_{1}, \ldots, v_{i-1}, v_{i}, v_{i+1}, \ldots, v_{j-1}, v_{i} , v_{j+1}, \ldots, v_{r}) = 0$. Here, we use the assumption that the characteristic of $K$ is not $2$.

Example: Let $V$ be an $n$-dimensional vector space, ${\cal B} = \{v_{1}, \ldots, v_{n}\}$ be a fixed basis for $V$. For each $\beta \in M^{2}(V)$, $\:\exists\:!\:$ square matrix $M_{n \times n}$, such that if $u = \sum x_{i} \cdot v_{i}$, $v = \sum y_{i} \cdot v_{i}$, then $\beta(u, v) = [x_{1}, \ldots, x_{n}] \cdot M_{n \times n} \cdot \left[\!\!\begin{array}{c} y_{1} \\ \vdots \\ y_{n} \end{array}\!\!\right].$

Proof:

Let $\beta(v_{i}, v_{j}) = b_{ij}$.

$$\begin{array}{ccl} \beta(u, v) & = & \beta(\sum x_{i} \cdot ... ... & = & \sum x_{i} \cdot \sum b_{ij} \cdot y_{j}. \end{array}$$

Q.E.D.

Remarks:

1. From the last example, it is easy to show that $\dim M^{2}(V) = n^{2}$.
2. Notice that a bilinear map $\beta : V \times V \longrightarrow K$ is usually not symmetric, i.e., $\beta(u, v) \not= \beta(v, u)$ in general:

   
   

   $$[1, 2]\cdot \left[\!\!\begin{array}{rr} 1 & 3 \\ -3 & 2 \... ...left[\!\!\begin{array}{c} 1 \\ 2 \end{array}\!\!\right] = 25.$$
   
   

3. Recall the inner product $<\!\!x_{1}, \ldots, x_{n}\!\!> \cdot <\!\!y_{1}, \ldots, y_{n}\!\!> = \sum_{i=1}^{n} x_{i} \cdot y_{i}$. It is easy to verify that the inner product is a bilinear form. What is its corresponding square matrix?