# Dual space of vector space

I am trying to prove the following theorem and I am getting stuck. Please take a look at this theorem what I have try so far.

Theorem: Let $$V$$ be a finite dimensional vector space over field F. Prove there is a ring isomorphism of $$End(V)$$ to $$End(V^{*})$$, where $$End(V)$$ denotes the linear transformation of $$V$$ to itself and $$V^{*}$$ is the dual space of $$V$$, i.e $$V^{*}= Hom(V, F)$$.

Proof:

Define the map $$\Phi$$: $$End(V)$$ $$\to$$ $$End(V^{*})$$ : $$\Phi (\varphi) (f)= f \circ \varphi$$ where $$\varphi$$ is an element of $$End(V)$$ and $$f$$ is an element of $$V^{*}$$.

Let $$\varphi _1$$ and $$\varphi_2$$ in $$End(V)$$

-Addition preserving :$$\Phi (\varphi_1 + \varphi_1) (f)=f \circ (\varphi_1 + \varphi_2)= f \circ \varphi_1 + f \circ \varphi_2= \Phi (\varphi_1) (f) + \Phi (\varphi_2) (f)$$.

-Multiplication preserving: $$\Phi (\varphi_1 \varphi_2) (f)=f \circ (\varphi_1 \varphi_2)= f \circ \varphi_1 \circ \varphi_2 = f \circ (\varphi_1 \circ \varphi_2)= \Phi(\varphi_1 \varphi_2(f))$$.

-Unit preserving: $$\Phi(i)(f)= f \circ i =f$$ $$\Rightarrow$$ $$\Phi(i)$$ is the identity in $$End(V^{*})$$.

-Injectivity: $$\Phi(\varphi)=0 \Rightarrow f \circ \varphi =0$$ for all $$f \in End(V^{*}) \Rightarrow \varphi =0$$. Hence $$ker \Phi =0 \Rightarrow \Phi$$ is injective.

This is what I have tried so far, I can not prove $$\Phi$$ is subjective. Can anybody give me the solutions for this ? Altarnative proof is also very helpful for me.

• When you proof your statement End$(V) \simeq F$ you say that $\Gamma(g(1)) = g$. But what is $g(1)$? $g$ is a map on $V$ and we don't necessarily have an element $1$ in the vector space $V$. You don't even need that statement for your proof. For the surjectivity of $\Phi$ you could use a dimension argument, namely that dim$(V)=$ dim$(V^*)$. Jun 7, 2021 at 10:25
• @Lukas You are right. I made a mistake, let me fix it. Jun 7, 2021 at 11:00
• @Lukas By the way, is $End(V) \cong F$ in general ? Jun 7, 2021 at 11:03
• As vector spaces this is definitely not true, because dim$(F)=1$ and dim$(\text{End}(V)) = (\text{dim}(V))^2$ which will not be the same if dim$(V)>1$. I also think that they are not isomorphic as rings. But I am not 100% sure about that. Jun 7, 2021 at 19:26

You almost done with that last attempt, but let me make some observations about your solution:

• Finish the addition-preserving argument: since you already prove that $$\Phi(\varphi_1+\varphi_2)(f) = \Phi(\varphi_1)(f) + \Phi(\varphi_2)(f)$$ holds for every $$f \in V^*$$, and $$\Phi(\varphi_1)(f) + \Phi(\varphi_2)(f)$$ is the map $$\Phi(\varphi_1)+\Phi(\varphi_2)$$ evaluated at $$f$$, it follows that $$\Phi(\varphi_1+\varphi_2)$$ and $$\Phi(\varphi_1)+\Phi(\varphi_2)$$ coincide at every point, meaning that they are equal.
• What is $$\Phi(\varphi_1\varphi_2(f))$$? To start, $$\Phi$$ is not multiplication preserving: since for any $$f \in V^*$$ we have \begin{align} \Phi(\varphi_1 \circ \varphi_2)(f) &= f \circ (\varphi_1 \circ \varphi_2) \\ &= (f \circ \varphi_1) \circ \varphi_2 \\ &= \Phi(\varphi_2)(f \circ \varphi_1) \\ &= \Phi(\varphi_2)(\Phi(\varphi_1)(f)) = \big( \Phi(\varphi_2) \circ \Phi(\varphi_1) \big)(f) \end{align} it follows that $$\Phi(\varphi_1 \circ \varphi_2) = \Phi(\varphi_2) \circ \Phi(\varphi_1)$$. So, what you are defining here is a ring homomorphism $$\Phi: \operatorname{End}(V)^{\rm op} \to \operatorname{End}(V^*)$$, that is, a ring anti-homomorphism $$\Phi: \operatorname{End}(V) \to \operatorname{End}(V^*)$$.
• It would be nice if you write that $$i$$ is the identity map in $$V$$, and by the way, $$\Phi(i)$$ is not the identity map in $$\operatorname{End}(V^*)$$, it is the identity map in $$V^*$$ because $$\Phi(i)(f) = f$$ for every $$f \in V^*$$.
• If you already know $$\bigcap_{f \in V^*} \ker f = 0$$ (why the injectivity of $$\Phi$$ follows from this?), then the injectivity argument is fine, except that it should be "for all $$f \in V^*$$" instead of "for all $$f \in \operatorname{End}(V^*)$$".

Also:

• In finite dimensions, any injective linear map between two vector spaces of the same dimension is an isomorphism, so, it is enough to show that $$\Phi$$ is also a linear map, since $$\dim V = \dim V^* \implies \dim \operatorname{End}(V) = (\dim V)^2 = (\dim V^*)^2 = \dim \operatorname{End}(V^*).$$
• No, in general, $$\operatorname{End}(V)$$ and $$F$$ are not isomorphic. What is true is that $$\operatorname{Hom}(F,V) \cong V$$ via $$\varphi \mapsto \varphi(1)$$. I think this is the reason about your confusion.
• Thank you for a great answer ! Jun 11, 2021 at 3:16

Question: "Theorem: Let V be a finite dimensional vector space over field F. Prove there is a ring isomorphism of End(V) to End(V∗), where End(V) denotes the linear transformation of V to itself and V∗ is the dual space of V, i.e V∗=Hom(V,F)."

Answer: If $$dim(V):=n$$ and $$V:=k\{e_1,..,e_n\}$$ with $$V^*:=k\{x_1,..,x_n\}$$

with $$x_i:=e_i^*$$ the dual basis you get an isomorphism

$$\phi: V^*\otimes_k V \cong End_k(V)$$

defined by $$\phi(x_i\otimes e_j)(u):=x_i(u)e_j$$. This map sends the element $$z:=\sum_i x_i \otimes e_i$$ to the identity endomorphism $$Id_V\in End_k(V)$$. Hence you may use the ring structure on $$End_k(V)$$ to construct an associative product on $$V^*\otimes_k V$$:

$$f_1 \otimes v_1 \bullet f_2 \otimes v_2:=f_2 \otimes f_1(v_2)v_1.$$

With this definition it follows $$V^*\otimes_k V$$ is an associative unital $$k$$-algebra with the element $$z$$ as a multiplicative unit, inducing an isomorphism

$$V^*\otimes_k V\cong End_k(V)$$

of associative $$k$$-algebras.

It follows there are canonical isomorphisms (of vector spaces)

$$\rho: End_k(V) \cong V^*\otimes_k V \cong V\otimes_k V^* \cong V^{**}\otimes_k V^* \cong End_k(V^*).$$

You must check if this isomorphism is a ring isomorphism. It could be (since the rings are non-commutative) you must change the order of the product:

If $$f_i\otimes v_i \in V^*\otimes V$$ you get

$$f_1\otimes v_1 \bullet f_2 \otimes v_2:= f_2\otimes f_1(v_2)v_1$$

and

$$\rho(f_1\otimes v_1 \bullet f_2 \otimes v_2)= f_1(v_2) v_1 \otimes f_2 \in V\otimes V^*.$$

You moreover get

$$\rho(f_1\otimes v_1)\bullet \rho(f_2\otimes v_2)=v_2 \otimes f_2(v_1)f_1.$$

You also get

$$\rho(f_2\otimes v_2) \bullet \rho(f_1 \otimes v_1):= v_1\otimes f_1(v_2)f_2= \rho(f_1\otimes v_1 \bullet f_2 \otimes v_2).$$

Hence it seems the canonical map "switches" the order of the multiplication, giving a canonical isomorphism of rings

$$End_k(V) \cong End_k(V^*)^{op}$$