Dual Endomorphism Ring (11.3.1)

Dummit and Foote Abstract Algebra, section 11.3, exercise 1:

Let $V$ be a vector space of finite dimension $n$. Prove that the map $$\psi : \text{End}(V) \rightarrow \text{End}(V^*)$$ $$\psi(\varphi) = \varphi^*$$ is an isomorphism of vector spaces. Show $\psi$ is not a ring isomorphism when $n \geq 2$. Exhibit an $F$-algebra isomorphism from $\text{End}(V)$ to $\text{End}(V^*)$.

Proof: Recall that $\text{dim } V = \text{dim } V^*$ implying $\text{dim End}(V) = \text{dim End}(V^*)$, so that it suffices to show $\psi$ is a nonsingular linear transformation. To show linearity, we must show $$\psi(\varphi_1 + \alpha \varphi_2) = \psi(\varphi_1)+\alpha \psi(\varphi_2)$$ $$(\varphi_1 + \alpha \varphi_2)^*(f) = \varphi_1^*(f) + (\alpha \varphi_2^*)(f)~~~~~\forall f \in V^*$$ $$(\varphi_1 + \alpha \varphi_2)^*(f)(v) = \varphi_1^*(f)(v) + (\alpha \varphi_2^*)(f)(v)~~~~~\forall v \in V$$ $$f \circ (\varphi_1 + \alpha \varphi_2)(v) = f \circ \varphi_1^*(v) + f \circ \alpha \varphi_2^*(v)$$ which follows from the linearity of $f$. To show nonsingularity, suppose $\psi(\varphi)=0$, which is to say $\varphi^*(f) = 0$ for all $f \in V^*$, which is to say $\varphi^*(f) (v) = f \circ \varphi (v) = 0$ for all $v \in V$. Assuming $\varphi$ is nonzero implies $\varphi(v) = \sum a_ie_i$ where some $a_k$ is nonzero. Letting $f$ be the element that sends $e_k$ to $1$ and all other $e_j$ to zero, we obtain a contradiction for $f \circ \varphi (v) = f(\sum a_ie_i) = a_k$.

Suppose $n \geq 2$ and $\psi$ is a ring isomorphism $$\psi(\varphi_1 \circ \varphi_2) = \psi(\varphi_1) \circ \psi(\varphi_2)$$ $$\psi(\varphi_1 \circ \varphi_2)(f) = \psi(\varphi_1) \circ \psi(\varphi_2)(f)~~~~~\forall f \in V^*$$ $$f \circ (\varphi_1 \circ \varphi_2)= f \circ (\varphi_2 \circ \varphi_1)$$ Choosing $\varphi_1,\varphi_2$ as $n \times n$ matrices such that $\varphi_2 \circ \varphi_1 = 0$ but $\varphi_1 \circ \varphi_2$ is nonzero, we can choose $f$ that maps a single basis element involved in the nonzero image of an element $v$ under the latter to $1$ so that the equality is violated and $\psi$ is not a ring isomorphism. Note that this approach does prove $\psi$ is a ring isomorphism when $n=1$ as $1 \times 1$ matrices commute.

Let $V$ have basis $e_1,...,e_n$, so that $V^*$ has basis $e_1^*,...,e_1^*$, $\text{End}(V)$ has basis $e_{ab}$ (considered as the $n \times n$ matrix with $1$ in position $a,b$ and zeros elsewhere) and $\text{End}(V^*)$ has basis $e_{ab}^*$ for $1≤a,b≤n$. Define $ψ : \text{End}(V) → \text{End}(V^*)$ by its action on the basis of $\text{End}(V)$ via $ψ(e_{ab})=e_{ab}^*$. This extends to a linear transformation that sends basis to basis and as such is nonsingular. All that remains is to demonstrate multiplicativity. We may observe this $$ψ(φ_1 \circ φ_2)=ψ(φ_1) \circ ψ(φ_2)$$ as the composition seen as matrix multiplication remains the same through the transformation.$~\square$