# Prove that $V^n$ and $\mathcal{L}(\mathbf{F}^n,V)$ are isomorphic vector spaces

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

• What is $F$ here? Apr 11, 2021 at 18:34
• @mathcounterexamples.net just a field. Sorry I did some mistakes in my proof I'm correcting it right now. Apr 11, 2021 at 18:34
• $F$ is the field over which the vector space $V$ is defined. Apr 11, 2021 at 18:39
• I’m a bit picky. Considering your proof, you’re also making the hypothesis that $V$ is of finite dimension. Apr 11, 2021 at 18:41
• 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. Apr 11, 2021 at 18:53

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.
• 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? Apr 11, 2021 at 18:59
• $\lambda_1, \dots, \lambda_n$ are just variables from $F$. The image of an element of $V^n$ being a map under $\Phi$. Apr 12, 2021 at 5:55