Halmos Section 24 and 25: Tensor Product and Product Bases
Point of post: This will be the solutions to the problems in Halmos’s book corresponding to this post.
Problem: If and are vectors in and respectively, find the coordinates of in with respect to the product basis where and .
Remark: For those working out of Halmos as well, I’d be surprised if you understood this on a first read. What I’ve determined it to mean is that since is a basis for (where the are the canonical basis elements for ) then how can we express in terms of this basis.
Proof: We notice that and and so
Problem: Let , , and be defined as in my post on tensor products. Prove there exists an isomorphism
Proof: Note that and are bases for and respectively. So, define
Clearly this is a linear bijection and so we may extend to
by linearity. By a previous post we may conclude that is an isomorphism. Also, we note that if and we see that
Problem: To what extent is the formation of tensor products commutative and associative? What about the distributive laws?
Proof: It’s true that , besides a clear dimension argument one immediately sees that if and are bases for respectively then the canonical map determined by
is evidently an isomorphism. Similarly, via the natural isomorphism which is determined by
Lastly, the above distributivity property holds vai the isomorphism determined by
Regardless, all of these are trivially true by a dimension argument.
Problem: If is a finite-dimensional vector space, and if is it true that ? Not always, take then notice that
a) Suppose that is a finite-dimensional real vector space, and of all complex numbers regarded as a real vector space. Form the tensor product . Prove that there is a way of defining products of complex numbers with elements of so that whenever .
b) Prove that with respect to vector addition, and with respect to scalar multiplication as defined in a) the space is a complex vector space.
c) Find the dimension of the complex vector space in terms of the dimension of the real vector space .
d) Prove that the vector space is isomorphic to a subspace in .
a) For and define
Let be a basis for and let and so extend this to a map
This map is evidently associative and . To prove distributivity it suffices to prove that and . The first one is proven as follows:
The second one follows by definition of the extension.
b) This is all pretty standard stuff. In other words, all of the axioms are just rote definitions.
c) We know that when we consider to be a real vector space that is a basis. We claim that is a basis for . To see this suppose that
but, since is linearly independent (over ) it follows that . Also, noting (again as above) that
and it follows that . Thus, is a basis for over and we get the pleasant result that
c) We can clearly see that is an embedding whether we consider to be a real or complex vector space, but in the latter case it’s an isomorphism.
1. Halmos, Paul R. “Bilinear Forms.” Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print
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