Abstract Nonsense

Crushing one theorem at a time

Multilinear Forms


Point of post: In this post we will discuss the concept of multilinear forms as is described in section 29 of Halmos.

Motivation

Way back when we discussed the concept of bilinear forms. Namely, for vector spaces \mathscr{U},\mathscr{V} over a vector space F we decided that a bilinear form

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

was a mapping which is linear in each of it’s “slots”. Put more precisely we described it as being such that

B(\alpha x+\beta x',y_0)=\alpha B(x,y_0)+\beta B(x',y_0)

for all fixed y_0\in\mathscr{V}. We gave a similar quality for the “other slot”. We then remarked then that many commonplace functions satisfy these qualities, the inner product on \mathbb{R}^n and the determinant being two such examples. The next obvious question is, why stop at two “slots”? We recalled that the way we interpreted the determinant as a bilinear form was

\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_2

In other words, we thought of the matrix \begin{bmatrix}z_1 & z_2\\ z_3 & z_4\end{bmatrix} as being a function of it’s row vectors. The question is then, can we interpret the determinant of a n\times n complex matrix as a bilinear form? Well, not that I’m aware of. To me it’s non-obvious how one takes a 3\times 3 matrix, for example, and divvies it up into two vectors in \mathbb{C}^3. But, if we extend our idea of bilinear forms, to say trilinear forms then we should be ok. In other words, we can extend the determinant to act on n vectors in \mathbb{C}^n (and by imaginative extension, on n\times n matrices) by considering

\det:\mathbb{C}^n\boxplus\cdots\boxplus\mathbb{C}^n\to\mathbb{C}

And this, is the archetypal example of a multilinear form.

Multilinear Forms

Let \mathscr{V}_1,\cdots,\mathscr{V}_n be vector spaces over some field F. We call

K:\mathscr{V}_1\boxplus\cdots\boxplus\mathscr{V}_n\to F

multilinear form (also called an n-form if it’s on n spaces) if for each k\in\{1,\cdots,n\} K has the property that

K (v_1,\cdots,\underbrace{\alpha v+\beta v'}_{k^{\text{th}}\text{ coordinate}},\cdots,v_n)=\alpha K\left(v_1,\cdots,v,\cdots,v_n\right)+\beta K\left(v_1,\cdots,v',\cdots,v_n\right)

where v_j\in\mathscr{V}_j,\text{ }j\in\{1,\cdots,n\}-\{k\}. It follows immediately from this definition that

\displaystyle K\left(\sum_{j_1=1}^{m_1}\alpha_{1,j_1}x_{1,j_1},\cdots,\sum_{j_n=1}^{m_n}\alpha_{n,j_n}x_{n,j_n}\right)=\sum_{j_1=1}^{m_1}\cdots\sum_{j_n=1}^{m_n}\prod_{k=1}^{n}\alpha_{k,j_k}K\left(x_{1,j},\cdots,x_{n,j}\right)

Unsurprisingly we may define the linear combination of multilinear forms in the obvious way, namely: if K_1,K_2  are multilinear forms on \mathscr{V}_1\boxplus\cdots\boxplus \mathscr{V}_n then \alpha K_1+\beta K_2 is the multilinear form which takes the values

(\alpha K_1+\beta K_2)(\bold{x})=\alpha K_1(\bold{x})+\beta K_2(\bold{x}),\quad \bold{x}\in\mathscr{V}_1\boxplus\cdots\boxplus\mathscr{V}_n

It’s evident that with this definition of addition and scalar multiplication that the set of all multilinear forms on \mathscr{V}_1\boxplus\cdots\boxplus\mathscr{V}_n denoted \text{Mult}\left(\mathscr{V}_1,\cdots,\mathscr{V}_n\right) is a vector space.

Since Halmos does not prove much about this (this isn’t his goal) I won’t go that much out of the realm of the book. Except, to prove the theorem which he states. Namely that if \mathscr{V}_j,\text{ }j=1,\cdots,n has dimension m_j then \dim\text{Mult}\left(\mathscr{V}_1,\cdots,\mathscr{V}_n\right)=m_1\cdots m_n.  We prove this in two parts.

Lemma: Let \mathscr{V}_1,\cdots,\mathscr{V}_n be vector spaces over F with bases \{x_{1,1},\cdots,x_{1,m_1}\},\cdots,\{x_{n,1},\cdots,x_{n,m_n}\} respectively. Then, if \left\{\beta_{i_1,\cdots,i_n}:i_1\in[m_1],\cdots,i_n\in[m_n]\right\}\subseteq F there is a unique K\in\text{Mult}\left(\mathscr{V}_1,\cdots,\mathscr{V}_n\right) for which K\left(x_{1,\ell_1},\cdots,x_{n,\ell_n}\right)=\beta_{\ell_1,\cdots,\ell_n}.

Proof: Note that for each \displaystyle \sum_{j_1=1}^{m_1}\alpha_{1,j_1}x_{1,j_1}\in\mathscr{V}_1,\cdots,\sum_{j_n=1}^{m_n}\alpha_{n,j_n}x_{n,j_n} we have that

\displaystyle \sum_{j_1}^{m_1}\cdots\sum_{j_n=1}^{m_n}\prod_{k=1}^{n}\alpha_{k,j_k}K\left(x_{1,j_1},\cdots,x_{n,j_n}\right)

It’s evident then that if we define

\displaystyle K_0\left(\sum_{j_1=1}^{m_1}\alpha_{1,j_1}x_{1,j_1},\cdots,\sum \alpha_{n,j_n}\right)=\sum_{j_1=1}^{m_1}\cdots\sum_{j_n=1}^{m_n}\prod_{k=1}^{n}\alpha_{k,j_k}\beta_{j_1,\cdots,j_m}

then, K_0 has the desired property and if K is another n-form such that K\left(x_{1,\ell_1},\cdots,x_{n,\ell_n}\right) we see that

\displaystyle \begin{aligned}K\left(\sum_{j_1=1}^{m_1}\alpha_{1,j_1}x_{1,j_1},\cdots,\sum_{j_n=1}^{m_n}\alpha_{n,j_n}x_{n,j_n}\right) &= \sum_{j_1=1}^{n}\cdots\sum_{j_n=1}^{m_n}\prod_{k=1}^{n}\alpha_{k,j_k}K\left(x_{1,j_1},\cdots,x_{n,j_n}\right)\\ &= \sum_{j_1=1}^{m_1}\cdots\sum_{j_n=1}^{m_n}\prod_{j=1}^{n}\alpha_{k,j_k}\beta_{j_1,\cdots,j_n}\\ &= K_0\left(\sum_{j_1=1}^{m_1}\alpha_{1,j_1}x_{1,j_1},\cdots,\sum_{j_n=1}^{m_n}\alpha_{n,j_n}x_{n,j_n}\right)\end{aligned}

and thus by the arbitrariness of the vectors it follows that K=K_0. \blacksquare

Theorem: Let \mathscr{V}_1,\cdots,\mathscr{V}_n be vector spaces over F with bases\mathscr{B}_1=\{x_{1,1},\cdots,x_{1,m_1}\} \cdots\mathscr{B}_n=\{x_{n,1},\cdots,x_{n,m_n}\} respectively. Then, \mathscr{B}=\left\{K_{p_1,\cdots,p_n}:i_1\in[m_1],\cdots,i_n\in[m_n]\right\} where

\displaystyle K_{p_1,\cdots,p_n}\left(x_{1,q_1},\cdots,x_{n,q_n}\right)=\prod_{k=1}^{n}\delta_{p_n,q_n}

is a basis for \text{Mult}\left(\mathscr{V}_1,\cdots,\mathscr{V}_n\right).

Proof: It’s clear from our lemma that such a class of n-linear forms exists. Now, to prove it’s linearly independent we merely note that if

\displaystyle \sum_{j_1=1}^{m_1}\cdots\sum_{j_1=1}^{m_n}\alpha_{j_1,\cdots,j_n}K_{j_1,\cdots,j_n}=\bold{0}

then we note that this implies that

\displaystyle \alpha_{\ell_1,\cdots,\ell_n}=\sum_{j_1=1}^{m_1}\cdots\sum_{j_n=1}^{m_n}\alpha_{j_1,\cdots,j_n}K_{j_1,\cdots,j_n}\left(x_{1,\ell_1},\cdots,x_{n,\ell_n}\right)=0

for each x_{1,\ell_1}\in\mathscr{B}_1,\cdots,x_{n,\ell_n}\in\mathscr{B}_n, from where linear independence of \mathscr{B} follows. To prove that \mathscr{B} spans \text{Mult}\left(\mathscr{V}_1,\cdots,\mathscr{V}_n\right) we let K\in\text{Mult}\left(\mathscr{V}_1,\cdots,\mathscr{V}_n\right) be arbitrary. We claim that

\displaystyle K=\sum_{j_1=1}^{m_1}\cdots\sum_{j_n=1}^{m_n}K\left(x_{1,j_1},\cdots,x_{n,j_n}\right)K_{j_1,\cdots,j_n}

To prove this we merely note that for each x_{1,\ell_1}\in\mathscr{B}_1,\cdots,x_{n,\ell_n}\in\mathscr{B}_n that

\displaystyle \sum_{j_1=1}^{m_1}\cdots\sum_{j_n=1}^{m_n}K\left(x_{1,j_1},\cdots,x_{n,j_n}\right)K_{j_1,\cdots,j_n}\left(x_{1,\ell_1},\cdots,x_{n,\ell_n}\right)=K\left(x_{1,\ell_1},\cdots,x_{n,\ell_n}\right)

from where the conclusion follows by the uniqueness portion of the lemma. \blacksquare

Corollary: \displaystyle \dim_F\text{Mult}\left(\mathscr{V}_1,\cdots,\mathscr{V}_n\right)=\prod_{j=1}^{n}\dim_F \mathscr{V}_j

References:

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

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November 11, 2010 - Posted by | Algebra, Halmos, Linear Algebra, Uncategorized | , ,

4 Comments »

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