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?

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    $\begingroup$ 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 $\endgroup$ Jul 11, 2021 at 23:34
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    $\begingroup$ 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 $1$. $\endgroup$
    – azif00
    Jul 11, 2021 at 23:36
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    $\begingroup$ I presume In your title you meant $\mathcal L(\mathbb F,V)$ and not $\mathcal L \in (\mathbb F,V)$ $\endgroup$ Jul 11, 2021 at 23:37
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    $\begingroup$ 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. $\endgroup$ Jul 11, 2021 at 23:38
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    $\begingroup$ @user2628206 Thanks, I have edited now. $\endgroup$
    – Labbsserts
    Jul 11, 2021 at 23:45

1 Answer 1


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:


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.


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.


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