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 $V^{*}= Hom(V, F)$.


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!

  • 1
    $\begingroup$ 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$(V^*)$. $\endgroup$
    – Lukas
    Jun 7, 2021 at 10:25
  • $\begingroup$ @Lukas You are right. I made a mistake, let me fix it. $\endgroup$
    – Huy Nguyen
    Jun 7, 2021 at 11:00
  • $\begingroup$ @Lukas By the way, is $End(V) \cong F$ in general ? $\endgroup$
    – Huy Nguyen
    Jun 7, 2021 at 11:03
  • $\begingroup$ 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. $\endgroup$
    – Lukas
    Jun 7, 2021 at 19:26

2 Answers 2


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 $f \in \operatorname{End}(V^*)$".


  • 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.
  • $\begingroup$ Thank you for a great answer ! $\endgroup$
    – Huy Nguyen
    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

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


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

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.