Linear space of functionals from $\mathbb R^n$ to $\mathbb R$ is isomorphic to $\mathbb R^n$

Question

Show that linear space of functionals from $\mathbb R^n$ to $\mathbb R$ i.e. $L(\mathbb R^n; \mathbb R)$ is isomorphic to $\mathbb R^n$.

All I know is that I have to find bijection between these two, but what will be the form of it? Since one set contains functionals which transform vectors from $\mathbb R^n$ and the other is just the set of vectors.

Any hints?

Answer 1

Hint:

Pick a basis $\;\{v_1,...,v_n\}\;$ of $\;\Bbb R^n\;$, and define for each $\;j=1,2,...,n\;$ a linear functional :

$$f_jv_k:=\delta_{jk}\;,\;\;\text{with the Kronecker delta}\;\;\delta_{jk}:=\begin{cases}1,&j=k\\0,&j\neq k\end{cases}$$

and extend each $\;f_j\,'$ s definition by linearity.

Answer 2

Not just a bijection. It has to be a bijective linear transformation. Define $\phi:\mathbb{R^n}\to L(\mathbb{R^n};\mathbb{R})$ by $\phi(v)=\psi_v$ where $\psi_v(x)=\langle x,v\rangle$ for all $x\in\mathbb{R^n}$ and $\langle\rangle$ is the standard dot product. I'll leave it to you to check why $\phi$ is a linear transformation and a bijection.