Point of post: In this post we will discuss the concept of multilinear forms as is described in section 29 of Halmos.
Way back when we discussed the concept of bilinear forms. Namely, for vector spaces over a vector space we decided that a bilinear form
was a mapping which is linear in each of it’s “slots”. Put more precisely we described it as being such that
for all fixed . We gave a similar quality for the “other slot”. We then remarked then that many commonplace functions satisfy these qualities, the inner product on 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
In other words, we thought of the matrix as being a function of it’s row vectors. The question is then, can we interpret the determinant of a complex matrix as a bilinear form? Well, not that I’m aware of. To me it’s non-obvious how one takes a matrix, for example, and divvies it up into two vectors in . 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 vectors in (and by imaginative extension, on matrices) by considering
And this, is the archetypal example of a multilinear form.
Let be vector spaces over some field . We call
a multilinear form (also called an -form if it’s on spaces) if for each has the property that
where . It follows immediately from this definition that
Unsurprisingly we may define the linear combination of multilinear forms in the obvious way, namely: if are multilinear forms on then is the multilinear form which takes the values
It’s evident that with this definition of addition and scalar multiplication that the set of all multilinear forms on denoted 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 has dimension then . We prove this in two parts.
Lemma: Let be vector spaces over with bases respectively. Then, if there is a unique for which .
Proof: Note that for each we have that
It’s evident then that if we define
then, has the desired property and if is another -form such that we see that
and thus by the arbitrariness of the vectors it follows that .
Theorem: Let be vector spaces over with bases respectively. Then, where
is a basis for .
Proof: It’s clear from our lemma that such a class of -linear forms exists. Now, to prove it’s linearly independent we merely note that if
then we note that this implies that
for each , from where linear independence of follows. To prove that spans we let be arbitrary. We claim that
To prove this we merely note that for each that
from where the conclusion follows by the uniqueness portion of the lemma.
1. Halmos, Paul R. Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print