For $n$ positive integer, define $V^n$ by $V^n=\underbrace{V\times...\times V}_{n \ times}$. Prove that $V^n$ and $\mathcal{L}(\mathbf{F}^n,V)$ are isomorphic vector spaces. I would like to know if my proof holds and to have a feedback, please. ($\mathbf{F}$ denotes a field here)

Let $(v_1,...,v_n)$ be a basis of $V$. So, each element in $V$ can be expressed as $\lambda_1 v_1+...+\lambda_n v_n$ for $\lambda_1,...,\lambda_n \in \mathbf{F}$.

Let $\xi:\mathbf{F}^n\to V$, $\xi(\lambda_1,...,\lambda_n)=\lambda_1 v_1+...+\lambda_n v_n$ and define $\psi: V^n\to \mathcal{L}(\mathbf{F}^n,V)$ as $\psi (\lambda_1 v_1+...+\lambda_n v_n,...,\lambda_1 v_1+...+\lambda_n v_n)=\xi(\lambda_1,...,\lambda_n)$.

Clearly $\psi$ is a linear application (it is easy to check). We show now that $\psi$ is injective.

$\psi(\lambda_1 v_1+...+\lambda_n v_n,...,\lambda_1 v_1+...+\lambda_n v_n)=\xi(\lambda_1,...,\lambda_n)=\lambda_1v_1+..+\lambda_nv_n=0 \iff \lambda_1=...=\lambda_n=0$ because $(v_1,...,v_n)$ is linearly independent in $V$. So, $\lambda_1 v_1+...+\lambda_n v_n,...,\lambda_1 v_1+...+\lambda_n v_n=0$ and we conclude that $\psi$ is injective.

Moreover, the dimension of $V^n$ is equal to a dimension of $\mathcal{L}(\mathbf{F}^n,V)$. Thus, by fundamental theorem we conclude that $\psi$ is surjective. Therefore, $\psi$ is an isomorphism

  • $\begingroup$ What is $F$ here? $\endgroup$ Apr 11, 2021 at 18:34
  • $\begingroup$ @mathcounterexamples.net just a field. Sorry I did some mistakes in my proof I'm correcting it right now. $\endgroup$
    – Daniil
    Apr 11, 2021 at 18:34
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    $\begingroup$ $F$ is the field over which the vector space $V$ is defined. $\endgroup$ Apr 11, 2021 at 18:39
  • $\begingroup$ I’m a bit picky. Considering your proof, you’re also making the hypothesis that $V$ is of finite dimension. $\endgroup$ Apr 11, 2021 at 18:41
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    $\begingroup$ You have several issues to fix. First, use different variables for $n$ and the dimension of $V$. As you use the same to denote two things, you’re making confusions. Second, your $\xi$ depends on $(v_1, \dots, v_n)$. You should reflect that in your notations. $\endgroup$ Apr 11, 2021 at 18:53

1 Answer 1


In fact the result is true whatever the dimension of $V$ is.


$$\begin{array}{l|rcl} \Phi : & V^n & \longrightarrow & \mathcal L(F^n,V)\\ & (v_1,\dots,v_n) & \longmapsto & (\lambda_1, \dots, \lambda_n) \mapsto \lambda_1v_1+ \dots + \lambda_n v_n\end{array}$$

$\Phi$ is linear, injective as its kernel is the set consisting of the zero vector and surjective.

  • $\begingroup$ Oh, alright. Thank you. But where do $\lambda_1,...,\lambda_n$ come from if you don't consider a dimension of $V$? And why $(\lambda_1,...,\lambda_n)\to \lambda_1 v_1+...+\lambda_n v_n$ would be well defined in the case if we don't consider the $V's$ dimension? $\endgroup$
    – Daniil
    Apr 11, 2021 at 18:59
  • $\begingroup$ $\lambda_1, \dots, \lambda_n$ are just variables from $F$. The image of an element of $V^n$ being a map under $\Phi$. $\endgroup$ Apr 12, 2021 at 5:55

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