## 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”.