# Show that $V$ and $\mathcal{L} (\mathbb{F},V)$ are isomorphic vector spaces.

In LADR by Axler, there are two theorems stated as following:

Theorem 1: Two finite-dimensional vector spaces over $$\mathbb{F}$$ are isomorphic $$\iff$$ They have the same dimension.

Theorem 2: $$V$$ and $$W$$ finite dimensional $$\implies$$ $$\mathcal{L}(V,W)$$ is finite and $$\dim\ (\mathcal{L}(V,W))=(\dim\ > V)(\dim\ W)$$

I suppose one way to show isomorphism is by showing a bijection, but in light of the two theorems above should it not be possible to just state that they have the same dimension instead?

• It is as you say, if $V$ is finite then you are done. You can't use your given theorem two though if $V$ is a non-finite vector space and must argue differently Jul 11, 2021 at 23:34
• In your problem, $V$ doesn't have to be finite-dimensional, so theorem 1 and 2 are not aplicable. As a hint, note that if $T \in \mathcal L(\mathbb F,V)$, then for every $a \in \mathbb F$ we have $T(a) = T(a1) = aT(1)$, so $T$ depends uniquely of its value in $1$. Jul 11, 2021 at 23:36
• I presume In your title you meant $\mathcal L(\mathbb F,V)$ and not $\mathcal L \in (\mathbb F,V)$ Jul 11, 2021 at 23:37
• To add to @azif00 's hint, It is worth noting explicitly that the result you wish to prove is valid regardless of the dimension of the space, i.e. even when $V$ is infinite. Jul 11, 2021 at 23:38
• @user2628206 Thanks, I have edited now. Jul 11, 2021 at 23:45

As discussed in the comments, the askers proof is valid in the case that $$V$$ is a finite vector space. unfortunately, if $$V$$ is infinite one must argue differently. Fortunately, there is canonical bijection between the two spaces $$V$$ and $$\mathcal L(\mathbb F,V)$$ that is simple to construct. We perform such a construction in the result below which yields the required result as an immediate corollary:

## Result:

For all fields $$\mathbb F$$ and vector spaces $$V$$ thereover, the map $$\phi:\mathcal L(\mathbb F,V) \to V \ T \mapsto T(1_\mathbb F)$$ is an isomorphism of vector spaces.

## Proof

Suppose that $$\mathbb F$$ and $$V$$ are as given. We first prove that $$\phi$$ is a linear map. To that end suppose that $$T,T^\prime \in \mathcal L(\mathbb F,V)$$ and $$\lambda, \lambda^\prime \in \mathbb F$$. Remark by algebraic manipulation that: \begin{align} \phi(\lambda \cdot T + \lambda^\prime \cdot T^\prime) &= \left(\lambda \cdot T + \lambda^\prime \cdot T^\prime\right)(1_\mathbb F)\\ &= \lambda \cdot T(1_\mathbb F) + \lambda^\prime \cdot T(1_{\mathbb F})\\ &= \lambda \cdot \phi(T) + \lambda^\prime \cdot \phi(T^\prime). \end{align} This immediately verifies that $$\phi$$ is a linear map. To complete the proof it remains to prove that $$\phi$$ is a bijection of underlying sets. We do this by showing that $$\phi$$ is both injective and surjective.

To se that $$\phi$$ is injective, suppose that $$T,T^\prime \in \mathcal L(\mathbb F,V)$$ are such that $$\phi(T) = \phi(T^\prime)$$. It suffices to prove that $$T = T^\prime$$. In this case suppose that $$a \in \mathbb F$$. Observe equation $$(1)$$ follows by the linerity of $$T$$ and $$T^\prime$$ and the hypothesis that $$\phi(T) = \phi(T^\prime)$$:

\begin{align} T(a) &= T(a \cdot 1_{\mathbb F})\\ &= a \cdot T(1_\mathbb F)\\ &= a \cdot \phi(T)\\ &= a \cdot \phi(T^\prime)\\ &= a \cdot T^\prime(1_\mathbb F)\\ &= T^\prime(a \cdot 1_\mathbb F)\\ &= T^\prime(a). \tag{1} \end{align} Remark that we have shown the statement $$\forall a \in \mathbb F: T(a) = T^\prime(a)$$. Thus, by function extentionality, $$T = T^\prime$$. This confirms that $$\phi$$ is injective.

We complete the proof by showing that $$\phi$$ is surjective. Suppose that $$a \in \mathbb V$$. Let $$T: \mathbb F \to V \ x \mapsto x \cdot a$$. It is a very quick exercise to see that $$T$$ is linear. Note now that $$\phi(T) = T(1_\mathbb F) = a$$. Remark we have shown the statement that $$\forall a \in V \ \exists T \in \mathcal L(\mathbb V,\mathbb F): \phi(T) = a$$. This is exactly to say that $$\phi$$ is surjective and so the result is proved.