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I'm currently taking an honors course in linear algebra and am trying to get a feel for what the dual space actually is and what the motivation behind creating such a space is. From what I understand, the dual space naturally appears when we're trying to solve linear equations:

$$f_1(x_1,x_2,.....x_n) = a_{11} x_{1} + a_2 x_2 + .... a_{1n} x_n = b_1$$ $$\vdots$$ $$\vdots$$ $$f_n(x_1,x_2,.....x_n)=a_{11} x_{1} + a_2 x_2 + .... a_{1n} x_n = b_n$$

where $f_i$ would essentially take the role of a row vector of a matrix. Thus, $f_i$ is the linear map from a vector in $\mathbb{R}^n$ to a linear equation $\mathbb{R}$.

I also understand that this is particularly helpful for defining linear functions from a vector space to $\mathbb{R}$ when there is no canonical basis for your vector space.

What I don't understand is what the dual space is telling us for vector spaces like $\mathbb{R}[x]$ (with the standard basis). I understand that one dual basis for this vector space are the linear differential operators $D^k = \frac{1}{k!}\frac{d^k}{dx^k}\biggr\rvert_{x=0}$ which obviously again maps this the $\mathbb{R}[x]$ to a linear equation in $\mathbb{R}$, but what is the meaning of this linear equation? Does defining the dual space for this vector space give us any insight into $\mathbb{R}[x]$? Does defining the dual space for this vector space make any previously difficult problems easier?

Any extra insight into the use and general purpose of dual spaces would also be greatly appreciated!

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  • $\begingroup$ Reading this question and the answers will help you quite a lot. $\endgroup$ – Dietrich Burde Mar 29 '17 at 19:54
  • $\begingroup$ Thanks a bunch @DietrichBurde, this actually did help me a lot. $\endgroup$ – hijasonno Mar 29 '17 at 20:14
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One way of thinking about the dual space is that elements of the dual space give $\textit{units}$ to the elements of the vector space.

So for example if we are thinking of $\mathbb{R}^3$ as a vector space then considering $(1,0,0)$ what does that $1$ really mean? 1 what?

If we have two elements of the dual $\phi_1, \phi_2$ with $\phi_1=c \cdot\phi_2$ then we can think of $\phi_1$ and $\phi_2$ as just measuring the same thing but in different units.

This point of view is quite useful in the study of smooth manifolds where the cotangent vectors serve the role as a sort of "units of integration".

As far as $\mathbb{R}[x]$ specifically, this point of view is not that useful. However, there are useful facts you can get from the dual space.

For example:

Consider $x_0\in\mathbb{R}$. Let $\phi_0$ be in the dual of $\mathbb{R}[x]$ defined by $\phi_0(p(x)) = p(x_0)$ (ie. $\phi_0$ is just point evaluation at $x_0$. We know from the abstract theory that if we restrict to polynomials of degree $\leq n$ then the dimension of the vector space, and thus the dual space, is $n+1$.

Also it is not hard to show that if $x_i$ are distinct points then the corresponding $\phi_i$ will be linearly independent elements of the dual space. Thus we get that an $n$-degree polynomials is uniquely determined by $n+1$ points.

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  • $\begingroup$ This was super helpful, thanks! Your last point is a really cool way to prove the uniqueness of interpolation polynomials. $\endgroup$ – hijasonno Mar 29 '17 at 20:41
  • $\begingroup$ also, I understand that the point of view of linear differential operators isn't useful, but does it tell us anything about $\mathbb{R}[x]$? $\endgroup$ – hijasonno Mar 29 '17 at 21:19

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