3
$\begingroup$

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.

$\endgroup$
6
  • 1
    $\begingroup$ 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
4
$\begingroup$

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 $\,e_1,e_2,e_3$.

$\endgroup$

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.