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.

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    $\begingroup$ Perhaps it's a typo, but the dual space for a vector space $V$ over $\mathbb{R}$ should be all linear maps $V \to \mathbb{R}$. $\endgroup$ – Maanroof Jan 29 '15 at 21:58
  • $\begingroup$ @Maanroof thanks, fixed $\endgroup$ – Guerlando OCs Jan 29 '15 at 22:02
  • $\begingroup$ Does it also help your understanding? Assuming we are talking finite-dimensional vector spaces, am I correct you want to know why a dual basis is a basis for $V^*$? $\endgroup$ – Maanroof Jan 29 '15 at 22:05
  • $\begingroup$ @Maanroof yes, I want to know why those linear maps defined that way (that $0$ and $1$ thing) spans the dual space. $\endgroup$ – Guerlando OCs Jan 29 '15 at 22:07
  • $\begingroup$ Well, you need to show two things then: linearly independence and the fact that they span the whole space. For the first, fix a basis $\mathcal{B} = e_1,e_2,...,e_n$ of our space $V$ over $k$, and let $e^1,e^2,....,e^n$ be the associated dual basis. Take an linear combination $\sum_{i=1}^n \lambda_i e^i$ equal to zero and deduce that all $\lambda_i$'s must be zero. As I recall this is fairly straightforward (using the $\delta_i^j$'s). For the second point, use that an arbitrary $\varphi \in V^*$ is linear, and consider what it does on $\mathcal{B}$. $\endgroup$ – Maanroof Jan 29 '15 at 22:15

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 $\,e_1,e_2,e_3$.


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