Prove that $\Bbb R^n$ and $\mathcal L(\Bbb R,\Bbb R^n)$ are isomorphic.

Question

Is this question asking me to prove there exists an invertible linear map between $\Bbb R^n$ and the set of all linear maps between $\Bbb R$ and $\Bbb R^n$? And if so, how would I do that?

Answer 1

More generally, for any field $K$ and any $K$-vector space $E$, you have an isomorphism: $$E\simeq \mathscr L(K,E).$$

Hint: What is the dimension of the base field over itself? What do you need to know for a linear map from the base field into a vector space to be entirely determined?

Answer 2

Yes, I think that is exactly what is being asked here.

Let

$\alpha \in \mathcal L(\Bbb R, \Bbb R^n) \tag 1$

be linear; then for $t \in \Bbb R$ we have

$\alpha(t) = \alpha(t \cdot 1) = t\alpha(1) \in \Bbb R^n; \tag 2$

thus each $\alpha \in \mathcal L (\Bbb R, \Bbb R^n)$ determines a vector

$\alpha(1) \in \Bbb R^n; \tag 3$

likewise, for any

$\alpha \in \Bbb R^n, \tag 4$

we may define an element

$\alpha(t) \in \mathcal L(\Bbb R, \Bbb R^n) \tag 5$

by

$\alpha(t) = t\alpha; \tag 6$

we are thus motivated to define a mapping

$\phi:\Bbb R^n \to \mathcal L(\Bbb R, \Bbb R^n) \tag 7$

by

$\phi(\alpha) = t\alpha = \alpha(t) \in \mathcal L(\Bbb R, \Bbb R^n); \tag 8$

it is easy to see that $\phi$ is linear, for

$\phi(a \alpha + \beta) = t(a \alpha + \beta) = a(t\alpha) + t\beta = a\alpha(t) + \beta(t) = a\phi(\alpha) + \phi(\beta), \; a \in \Bbb R, \alpha, \beta \in \Bbb R^n; \tag 9$

furthermore, $\phi$ is clearly surjective, since

$\phi(\alpha(1)) = \alpha(t) \in \mathcal(\Bbb R, \Bbb R^n), \tag{10}$

as we have seen above; finally, if

$\alpha \in \ker \phi, \tag{11}$

then

$\forall t \in \Bbb R, \; \phi(\alpha) = t\alpha = 0 \Longleftrightarrow \alpha = 0, \tag{12}$

whence

$\ker \phi = \{0\}, \tag{13}$

implying $\phi$ is injective as well; thus $\phi$ is a linear isomorphism 'twixt $\Bbb R^n$ and $\mathcal L(\Bbb R, \Bbb R^n)$.

Nota Bene: As pointed out by our colleague Bernard, the above argument goes through if $\Bbb R$ is replaced by any field $\Bbb K$, and $\Bbb R^n$ is replaced by any vector space $V$ over $\Bbb K$. End of Note.