I am required to prove the following theorem is the argument presented below Correct?

Given that $n\in\mathbf{Z^+}$. The vector spaces $V^n$ and $\mathcal{L}(\mathbf{F}^n,V)$ are isomorphic, where we define $V^n$ by $V^n = \underbrace{V\times V\times\cdot\cdot\cdot\times V}_{n\text{ terms}}$.

Proof. Consider the following definition of the map $\Delta:V^n\to\mathcal{L}(\mathbf{F}^n,V)$. $$\Delta(v_1,v_2,.\ .\ .\ ,v_n) = T\text{ where }T(e_j) = v_j\text{ for all }j\in I_n$$ and $I_n = \{i\in\mathbf{Z^+}|i\leq n\}$ and $e_1,e_2,.\ .\ .\ ,e_n$ denotes the standard basis of $\mathbf{F}^n$.

Assume for arbitrary $v = (v_1,v_2,.\ .\ .\ ,v_n)$ and $w = (w_1,w_2,.\ .\ .\ ,w_n)$ in $V^n$ that $\Delta(v) = T_1 = T_2 = \Delta(w)$ then in particular $\forall j\in I_n\left(T_1(e_j) = T_2(e_j)\right)$ equivalently $\forall j\in I_n\left(v_j = w_j\right)$ consequently $v=w$ thus $\Delta$ is injective.

Let $T$ be any transformation in $\mathcal{L}(\mathbf{F}^n,V)$ using $\textbf{3.5}$ we may uniquely identify $T$ by considering the vectors $T(e_1) = v_1,T(e_2) = v_2,.\ .\ .\ ,T(e_n) = v_n$ then from the above definition it is apparent that $\Delta(v_1,v_2,.\ .\ .\ ,v_n) = T$ thus $T$ is surjective.

To demonstrate that the above map is Linear we must show that $\Delta(av+bw) = a\Delta(v)+b\Delta(w)$ where $v,w\in\mathbf{F}^n$. We leave it to the reader to workout the argument.

Taking all the above arguments together it follows that $\Delta$ is an isomorphism from $V^n$ to $\mathcal{L}(\mathbf{F}^n,V)$.


NOTE: theorem $3.5$ is the result that a linear transformation $T:V\to W$ is uniquely determined by the images of the basis vectors of $V$ under $T$.


Your argument is correct.

This is how I would do it, however. Let $\Gamma:\mathcal L(\mathbb F^n,V)\to V^n$, be given by $\Gamma(T)=(T(e_1),\ldots,T(e_n))$. Then $\Gamma$ is linear, and it is straightforward that $$ \Gamma\circ\Delta=\text{id}_{V^n},\ \ \ \Delta\circ\Gamma=\text{id}_{L(\mathbb F^n,V)}. $$ So $\Delta$ is an isomorphism.

Note also that if you only care about proving that there is an isomorphism, it is enough to show that they have the same dimension. So another proof would be to show that $$ \dim V^n=n\,\dim V=\dim L(\mathbb F^n,V). $$

  • $\begingroup$ Thanks @MartinArgerami but would the latter argument be valid if $V$ was infinite dimensional ? $\endgroup$ – Atif Farooq Mar 1 '18 at 5:57
  • $\begingroup$ No. Cardinality of the basis is an invariant for infinite-dimensional spaces too. $\endgroup$ – Martin Argerami Mar 1 '18 at 12:43

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