0
$\begingroup$

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?

$\endgroup$
6
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @user2628206 Thanks, I have edited now. $\endgroup$
    – Labbsserts
    Jul 11, 2021 at 23:45

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.