For a vector space V, the dual space is defined as the space of linear functionals that take a vector from V to the real numbers. Its basis is related to the vector space's basis by the kronecker delta.

How exactly does this construct gets applied? Each vector in V maps to a specific functional. That functional takes a vector and maps it to the real numbers. Geometrically it is interpreted as a level curve. So, each vector in V now corresponds to a level set in the dual space. After this, how does this whole thing get used? I found the below statement in ( math.stackexchange.com/questions/3749/why-do-we-care-about-dual-spaces )

The dual is intuitively the space of "rulers" (or measurement-instruments) of our vector space. Its elements measure vectors.

How is this different from an inner product? ie, inner product is also supposed to measure a vector, right? What additional benefits do we get out of dual spaces? Why can't we define a specific linear functional on need basis (like inner product) instead of defining a whole space of functionals?

  • $\begingroup$ Your question is close to asking "What is the difference between a linear map $V \to V^\ast$ and a bilinear form $V \times V \to \mathbb{R}$?" The answer to that question is there isn't one. You should note that the map $V \to V^\ast$ you describe in the first paragraph depends on the basis. If you choose a different basis you should expect to get a different map. $\endgroup$
    – Max
    Commented Apr 21, 2020 at 17:05
  • $\begingroup$ Are you looking for a proof where one uses the dual space and dual basis? $\endgroup$ Commented Apr 21, 2020 at 22:09

1 Answer 1


On an inner product space, $x \mapsto \langle x,y \rangle$ is a linear functional*. On a Hilbert space, the set of linear functionals of that form is the whole dual space; this is a Riesz representation theorem**.

But many vector spaces have no associated inner product. Some of these are "purely algebraic" vector spaces, which are not really relevant from the point of view of functional analysis, but others are normed vector spaces whose norm doesn't satisfy the parallelogram law. There are many such things, including for instance all the $L^p$ spaces with $p \neq 2$.

Also, as Max mentioned in a comment, it can be useful to note that there is a direct correspondence between a linear map from $V$ to $V^*$ and a bilinear function from $V \times V$ to the base field. This identification uses what computer scientists usually call currying and uncurrying.

* In the complex setting this requires the mathematician's convention for inner products, i.e. linearity in the first argument. In the physicist's convention, you'd use $x \mapsto \langle y,x \rangle$, which they would likely refer to as $\langle y |$.

* * I say "a" Riesz representation theorem because I know at least two theorems that get called "the" Riesz representation theorem for some reason.


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