I've been introduced to the concept of dual space in linear algebra. I can understand perfectly that the dual space of the space $V$ is a space $V^*$ made of all possible linear maps from $V$ to $\mathbb{R}$. So, for example, let $V$ be $\mathbb{R^3}$, then we have, as elements of the dual space, for example:
$$F(x,y,z) = x + y\\F(x,y,z) = -x + z\\F(x,y,z) = 3z\\F(x,y,z) = 3x + 4y + 5z\\ \cdots$$
What I don't understand, it's why these things called functionals, span the dual space. I've seen a proof but I didn't understand. I know (at least have an intuition) that the dual space can be represented as the space of all possible linear combinations of $x,y,z$ like:
$$F(x,y,z) = ax + by + cz$$
but how to prove that this is suficient to generate the entire space? And why that rule that maps to $0$ and $1$ form a basis to this space?
Sorry by all these questions, but this concept seemed a lot strange for me, and I can't understand why dual spaces and finding its basis are so important.
Best Answer
Any linear map $fu$ from $V=\mathbf R^3$ to $\mathbf R$ is determined by its values on the vectors of a base $\mathcal B =(e_1, e_2, e_3)$. For if $v=\lambda e_1+\mu e_2+\nu e_3$, then $f(v)=\lambda f(e_1)+\mu f(e_2)+\nu f(e_3)$.
Now if $f(e_1)=\alpha_1$, $f(e_2)=\alpha_2$, $f(e_3)=\alpha_3$ and if $e_1^*, e_2^*,e_3^*$ is the dual basis of $\mathcal B$, it's easy to check that $$f=\alpha_1 e_1^*+\alpha_2 e_2^*+\alpha_3 e_3^*$$ since both sides take the same value for $\,e_1,e_2,e_3$.