Why is Hom(V,W)\text{Hom}(V,W) the same thing as V∗⊗WV^* \otimes W?

I have a couple of questions about tensor products:

Why is Hom(V,W) the same thing as VW?

Why is an element of VmVn the same thing as a multilinear map VmVn?

What is the general formulation of this principle?

Answer

The result is generally wrong for infinite-dimensional spaces: see this question.

For finite dimensional space V, let’s build an isomorphism f:VWhom(V,W) by defining

f(ϕw)(v)=ϕ(v)w

This clearly defines a linear map VWhom(V,W) (it’s bilinear in V×W). Reciprocally, take a basis (ei) of V, then define g:hom(V,W)VW by:

g(u)=ieiu(ei)

Where (ei) is the dual basis to (ei) (I will use a few of its properties in red below). This is well-defined because V is finite-dimensional (the sum is finite). Let’s check that f and g are inverse to each other:

  • For u:VW, f(g(u))(v)=ei(v)u(ei)=u(ei(v)ei)=u(v) and so f(g(u))=u.

  • For ϕwVW, g(f(ϕw))=eif(ϕw)(ei)=eiϕ(ei)w=ϕ(ei)eiw=ϕw

And so f and g are isomorphisms, inverse to each other.


It is known that for finite dimensional V, then (V)m=(Vm). Then an element of VmVn is an element of (Vm)Vn=hom(Vm,Vn). So by definition / universal property of the tensor product, it’s a multilinear map VmVn.

Attribution
Source : Link , Question Author : Eric Auld , Answer Author : Community

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