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.
Thanks in advance!
 A: 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
$$End_k(V) \cong End_k(V^*)^{op}$$
A: 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.

