The endomorphism rings of a vector space and its dual space are isomorphic as algebras over the base field

Let F be a field and let V be a vector space over F of some finite dimension n. Show that the mapping \Omega : \mathsf{End}_F(V) \rightarrow \mathsf{End}_F(\mathsf{Hom}_F(V,F)) given by \Omega(\varphi)(\tau) = \tau \circ \varphi is an F-vector space isomorphism but not a ring isomorphism for n > 1. Exhibit an F-algebra isomorphism \mathsf{End}_F(V) \rightarrow \mathsf{End}_F(\mathsf{Hom}_F(V,F)).

We begin with a lemma.

Lemma: Let R be a unital ring and let M,A,B be left unital R-modules. If \varphi : M \times A \rightarrow B is R-bilinear, then the induced map \Phi : M \rightarrow \mathsf{Hom}_R(A,B) given by \Phi(m)(a) = \phi(m,a) is a well-defined R-module homomorphism. Proof: To see well definedness, we need to verify that \Phi(m) : A \rightarrow B is a module homomorphism. To that end note that \Phi(m)(x+ry) = \varphi(m,x+ry) = \varphi(m,x) + r \varphi(m,y) = \Phi(m)(x) + r\Phi(m)(y). Similarly, to show that \Phi is a module homomorphism, note that \Phi(x+ry)(a) = \varphi(x+ry,a) = \varphi(x,a)+ r\varphi(y,a) = \Phi(x)(a) + r\Phi(y)(a) = (\Phi(x)+r\Phi(y))(a), so that \Phi(x+ry) = \Phi(x) + r\Phi(y). \square

[Note to self: In a similar way, if R is a unital ring and M,N,A,B unital modules, and \varphi : M \times N \times A \rightarrow B is trilinear, then \Phi : M \times N \rightarrow \mathsf{Hom}_R(A,B) is bilinear. (So that the induced map M \rightarrow \mathsf{Hom}_R(N,\mathsf{Hom}_R(A,B)) is a module homomorphism, or unilinear- if you will.) That is to say, in a concrete fashion we can think of multilinear maps as the uncurried versions of higher order functions on modules. (!!!) (I just had a minor epiphany and it made me happy. Okay, so the usual isomorphism V \rightarrow \mathsf{Hom}_F(V,F) is just this lemma applied to the dot product V \times V \rightarrow F… that’s cool.) Moreover, if A = B and if M and \mathsf{End}_R(A) are R-algebras, then the induced map \Phi is an algebra homomorphism if and only if \varphi(m_1m_2,a) = \varphi(m_1,\varphi(m_2,a)) and \varphi(1,a) = a.]

Define \overline{\Omega} : \mathsf{End}_F(V) \times \mathsf{Hom}_F(V,F) \rightarrow \mathsf{Hom}_F(V,F) by \overline{\Omega}(\varphi,\tau) = \tau \circ \varphi. This map is certainly bilinear, and so by the lemma induces the linear transformation \Omega : \mathsf{Hom}_F(V) \rightarrow \mathsf{End}_F(\mathsf{Hom}_F(V)). Since V has finite dimension, and since its dual space \mathsf{Hom}_F(V,F) has the same dimension, to see that \Omega is an isomorphism of vector spaces it suffices to show that the kernel is trivial. To that end, suppose \varphi \in \mathsf{ker}\ \Omega. Then we have \Omega(\varphi)(\tau) = \tau \circ \varphi = 0 for all \tau. In particular, we have \mathsf{im}\ \varphi \subseteq \mathsf{ker}\ \tau for all \tau. If there exists a nonzero element v \in \mathsf{im}\ \varphi, then by the Building-up lemma there is a basis B of V containing v. In particular, there is a linear transformation \tau such that \tau(v) \neq 0. That is, we have \mathsf{im}\ \varphi = 0, so that \varphi = 0. Hence \Omega is injective, and so an isomorphism of vector spaces.

Note that \Omega(\varphi \circ \psi)(\tau) = \tau \circ \varphi \circ \psi, while (\Omega(\varphi) \circ \Omega(\psi))(\tau) = \Omega(\varphi)(\Omega(\psi)(\tau)) = \Omega(\varphi)(\tau \circ \psi) = \tau \circ \psi \circ \varphi. If V has dimension greater than 1, then \mathsf{End}_F(V) is not a commutative ring. Thus these expressions need not be equal in general. In fact, if we choose \tau, \varphi, and \psi such that M(\varphi) = \left[ \begin{array}{c|c} 0 & 1 \\ \hline 0 & 0 \end{array} \right], M(\psi) = \left[ \begin{array}{c|c} 0 & 0 \\ \hline 1 & 0 \end{array} \right], and M(\tau) = [1 | 0], then clearly \tau \circ \varphi \circ \psi \neq 0 and \tau \circ \psi \circ \varphi = 0. In particular, \Omega is not a ring isomorphism if n > 1. On the other hand, if n = 1, then \mathsf{End}_F(V) \cong F is commutative, and \Omega is a ring isomorphism.

On the other hand, these rings are clearly isomorphic since V and \mathsf{Hom}_F(V,F) are vector spaces of the same dimension.

Note that \mathsf{End}_F(V) and \mathsf{End}_F(\mathsf{Hom}_F(V,F)) are both F-algebras via the usual scalar multiplication by F. Fix a basis B of V, and identify the linear transformation \varphi \in \mathsf{End}_F(V) with its matrix M^B_B(\varphi) with respect to this basis. (Likewise for \mathsf{Hom}_F(V,F).) Now define \Theta : \mathsf{End}_F(V) \rightarrow \mathsf{End}_F(\mathsf{Hom}_F(V,F)) by \Theta(M)(N) = NM^\mathsf{T}. It is clear that \Theta is well defined, and moreover is an F-vector space homomorphism. Note also that \Theta(M_1M_2)(N) = N(M_1M_2)^\mathsf{T} = NM_2^\mathsf{T}M_1^\mathsf{T} = \Theta(M_1)(\Theta(M_2)(N)), so that \Theta(M_1M_2) = \Theta(M_1)\Theta(M_2). Thus \Theta is a ring homomorphism; since \Theta(I)(N) = N, we have \Theta(I) = 1, and indeed \Theta is an F-algebra homomorphism. It remains to be seen that \Theta is an isomorphism; it suffices to show injectivity. To that end, suppose \Theta(M)(N) = NM^\mathsf{T} = 0 for all N. Then MN^\mathsf{T} = 0 for all N, and so M = 0. Thus \Theta is an F-algebra isomorphism \mathsf{End}_F(V) \rightarrow \mathsf{End}_F(\mathsf{Hom}_F(V,F)). Note that \Theta depends essentially on our choice of a basis B, and so is not “natural”.

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