Why do we care about dual spaces?

When I first took linear algebra, we never learned about dual spaces. Today in lecture we discussed them and I understand what they are, but I don't really understand why we want to study them within linear algebra.

I was wondering if anyone knew a nice intuitive motivation for the study of dual spaces and whether or not they "show up" as often as other concepts in linear algebra? Is their usefulness something that just becomes more apparent as you learn more math and see them arise in different settings?

Edit

I understand that dual spaces show up in functional analysis and multilinear algebra, but I still don't really understand the intuition/motivation behind their definition in the standard topics covered in a linear algebra course. (Hopefully, this clarifies my question)

Solution 1:

Let $V$ be a vector space (over any field, but we can take it to be $\mathbb R$ if you like, and for concreteness I will take the field to be $\mathbb R$ from now on; everything is just as interesting in that case). Certainly one of the interesting concepts in linear algebra is that of a hyperplane in $V$.

For example, if $V = \mathbb R^n$, then a hyperplane is just the solution set to an equation of the form $$a_1 x_1 + \cdots + a_n x_n = b,$$ for some $a_i$ not all zero and some $b$. Recall that solving such equations (or simultaneous sets of such equations) is one of the basic motivations for developing linear algebra.

Now remember that when a vector space is not given to you as $\mathbb R^n$, it doesn't normally have a canonical basis, so we don't have a canonical way to write its elements down via coordinates, and so we can't describe hyperplanes by explicit equations like above. (Or better, we can, but only after choosing coordinates, and this is not canonical.)

How can we canonically describe hyperplanes in $V$?

For this we need a conceptual interpretation of the above equation. And here linear functionals come to the rescue. More precisely, the map

$$\begin{align*} \ell: \mathbb{R}^n &\rightarrow \mathbb{R} \\ (x_1,\ldots,x_n) &\mapsto a_1 x_1 + \cdots + a_n x_n \end{align*}$$

is a linear functional on $\mathbb R^n$, and so the above equation for the hyperplane can be written as $$\ell(v) = b,$$ where $v = (x_1,\ldots,x_n).$

More generally, if $V$ is any vector space, and $\ell: V \to \mathbb R$ is any non-zero linear functional (i.e. non-zero element of the dual space), then for any $b \in \mathbb R,$ the set

$$\{v \, | \, \ell(v) = b\}$$

is a hyperplane in $V$, and all hyperplanes in $V$ arise this way.

So this gives a reasonable justification for introducing the elements of the dual space to $V$; they generalize the notion of linear equation in several variables from the case of $\mathbb R^n$ to the case of an arbitrary vector space.

Now you might ask: why do we make them a vector space themselves? Why do we want to add them to one another, or multiply them by scalars?

There are lots of reasons for this; here is one: Remember how important it is, when you solve systems of linear equations, to add equations together, or to multiply them by scalars (here I am referring to all the steps you typically make when performing Gaussian elimination on a collection of simultaneous linear equations)? Well, under the dictionary above between linear equations and linear functionals, these processes correspond precisely to adding together linear functionals, or multiplying them by scalars. If you ponder this for a bit, you can hopefully convince yourself that making the set of linear functionals a vector space is a pretty natural thing to do.

Summary: just as concrete vectors $(x_1,\ldots,x_n) \in \mathbb R^n$ are naturally generalized to elements of vector spaces, concrete linear expressions $a_1 x_1 + \ldots + a_n x_n$ in $x_1,\ldots, x_n$ are naturally generalized to linear functionals.

Solution 2:

Since there is no answer giving the following point of view, I'll allow myself to resuscitate the post.

The dual is intuitively the space of "rulers" (or measurement-instruments) of our vector space. Its elements measure vectors. This is what makes the dual space and its relatives so important in Differential Geometry, for instance. This immediately motivates the study of the dual space. For motivations in other areas, the other answers are quite well-versed.

This also happens to explain intuitively some facts. For instance, the fact that there is no canonical isomorphism between a vector space and its dual can then be seen as a consequence of the fact that rulers need scaling, and there is no canonical way to provide one scaling for space. However, if we were to measure the measure-instruments, how could we proceed? Is there a canonical way to do so? Well, if we want to measure our measures, why not measure them by how they act on what they are supposed to measure? We need no bases for that. This justifies intuitively why there is a natural embedding of the space on its bidual. (Note, however, that this fails to justify why it is an isomorphism in the finite-dimensional case).

Solution 3:

There are some very beautiful and easily accessible applications of duality, adjointness, etc. in Rota's modern reformulation of the Umbral Calculus. You'll quickly gain an appreciation for the power of such duality once you see how easily this approach unifies hundreds of diverse special-function identities, and makes their derivation essentially trivial. For a nice introduction see Steven Roman's book "The Umbral Calculus".

Solution 4:

Absolutely yes to the last question. There are already many applications listed in the Wikipedia article. I will give a few more.

First of all, together with tensor products, dual spaces can be used to talk about linear transformations; the vector space $\text{Hom}(V, W)$ of linear transformations from a vector space $V$ to a vector space $W$ is canonically isomorphic to the tensor product $V^{\ast} \otimes W$. I assume you care about linear transformations, so you should care about tensor products and dual spaces as well. (A really important feature of this decomposition is that it is true in considerable generality; for example, it holds for representations of groups and is a natural way to prove the orthogonality relations for characters.)

Dual spaces also appear in geometry as the natural setting for certain objects. For example, a differentiable function $f : M \to \mathbb{R}$ where $M$ is a smooth manifold is an object that produces, for any point $p \in M$ and tangent vector $v \in T_p M$, a number, the directional derivative, in a linear way. In other words, a differentiable function defines an element of the dual to the tangent space (the cotangent space) at each point of the manifold.

Solution 5:

I still have very little mathematical experience (mainly linear algebra and a bit or real analysis) so the only answer I can give you is that dual spaces are already useful in linear algebra itself.

LEMMA dimension of the annihilator Let $V$ be a vector space on the field $\mathbb{F}$ with $\operatorname{dim}V = n$ and $W \subseteq V$ a subspace, if $W^{0} = \{ \phi \in V^{*} | \phi_{|W} = 0 \}$ is the annihilator of $W$, then $\operatorname{dim}W + \operatorname{dim}W^{0} = n$

Let $(w_1, \ldots, w_k)$ be a basis of $W$ and $(w_1,\ldots,w_k,v_{k+1},\ldots,v_n)$ a completion to basis of $V$, if $(w_1^{*},\ldots,w_k^{*},v_{k+1}^{*},\ldots,v_n*)$ is the corresponding dual base of $V^{*}$ then it's easy to prove by double inclusion that $W^{0}=\langle v_{k+1}^{*},...,v_n^{*}\rangle$ and thus $\operatorname{dim}W^{0}= n-k$

THEOREM dimension of the orthogonal space Let $V$ be a vector space on the field $\mathbb{F}$ with $\operatorname{dim}V = n$, endowed with a nondegenerate scalar product $\langle , \rangle$, then if $W \subseteq V$ is a subspace $\operatorname{dim}W + \operatorname{dim}W^{\bot} = n$

First, notice that the statement is not obvious. In fact if a scalar product is not positive-definite $V = W \oplus W^{\bot}$ could be false (you could have $v \in W$ such that $\langle v, v\rangle$ = 0 and therefore $W\cap W^{\bot}\neq \{ 0\}$). Now by mean of the scalar product you can define a very nice isomorphism $\psi : V \to V^{*}$ such that $\psi (v) = \langle \cdot , v\rangle$ for all $v \in V$, you can think about it as the product by $v$. This is an isomorphism, in fact $\operatorname{dim}V = \operatorname{dim}V^{*}$ and $\operatorname{ker}\psi = 0$ because $\langle , \rangle$ is nondegenerate. Now observe that $\psi (W^{\bot}) = W^{0}$ and thus $\operatorname{dim}W^{\bot}=\operatorname{dim}W^{0}=n-\operatorname{dim}W$ by the lemma.