# How is the Dual Space a Vector Space?

Trying to understand the idea of a Dual Space and I'm having a little trouble understanding what is meant by the idea of a linear functional. My understanding is that a functional maps from either an n-tuple or a polynomial to some field, but I don't understand how these concepts are supposed to be linear.

Like, if we consider an easier example, say R^2, what would a linear functional look like? Would it have to map to the field of real numbers, R, or any field? How can a functional be linear if you need two variable inputs, like x and y for R^2?

And finally, if we define the Dual Space to be the set of all possible linear functionals from the Vector Space to a field, how exactly is this itself a Vector Space? Are the vectors the functionals themselves? And if so, what is the field over which the Dual Space is a Vector Space?

• Let $V$ be a vector space over the field $\Bbb K.$ Then a linear functional $f$ on $V$ is a linear map $f : V \longrightarrow \Bbb K.$ May 19, 2020 at 4:14
• The set of all linear functionals on $V$ form a vector space over the same field $\Bbb K$ w.r.t. the addition ($+$) and the scalar multiplication ($\cdot$) defined as follows $:$ \begin{align*} (1)\ (f+g) (v) & = f(v) + g(v),\ \text {for all}\ v \in V. \\ (2)\ (cf) (v) & = c f(v), \text {for all}\ v \in V\ \text {and}\ \text {for all}\ c \in \Bbb K. \end{align*} This vector space is known as the dual space of $V$ and it is denoted by $V^*.$ May 19, 2020 at 4:21
• About the $\Bbb R^2$ remark: being linear in the pair-variable $(x,y)$ is not the same as being bilinear in $x$ and $y$. In fact, one can show that if $B\colon V\times W \to Z$ is bilinear and linear simultaneously, then $B=0$. May 19, 2020 at 4:48

Given a vector space $$V$$ over a field $$\mathbb F$$, the dual space $$V^\vee$$ consists of linear functionals on $$V$$. Linear functionals on $$V$$ are $$\mathbb F$$-linear maps $$V\to\mathbb F$$.

For example, $$\mathbb R^2$$ is a vector space over the field $$\mathbb R$$, so its dual space consists of $$\mathbb R$$-linear maps $$\mathbb R^2\to\mathbb R$$. An example of such a linear functional is $$f(x,y):=x+y$$.

The dual space of $$V$$ is equipped with the following $$\mathbb F$$-vector space structure. Given $$f,g\in V^\vee$$ and $$\alpha\in\mathbb F$$, we define $$f+g$$ as the map $$V\to\mathbb F$$ via $$(f+g)(v):=f(v)+g(v)$$, and we define $$\alpha f$$ as the map $$V\to\mathbb F$$ via $$(\alpha f)(v):=\alpha\cdot f(v)$$; this is for all $$v\in V$$. It is easy to see that $$f+g,\alpha f\in V^\vee$$, and it is easy to check that all the vector space axioms are satisfied.

If you think of linear functionals as mappings $$l:V\to \mathbb{F}$$ where $$V$$ is an $$n$$-dimensional vector space and $$\mathbb{F}$$ is a field, then you can think of $$l$$ as having a matrix representation as a $$1\times n$$ matrix if you specify a preferred basis for $$V$$. For instance, in $$\mathbb{R}^2$$, and linear function will be of the form $$l(v) = av_1 + bv_2$$ where

$$v \;\; =\;\; \left [ \begin{array}{c} v_1 \\ v_2 \\ \end{array} \right ]$$

in the standard basis. This then implies that

$$l \;\; \equiv \;\; \left [ \begin{array}{cc} a & b \\ \end{array} \right ]$$

with respect to this basis. In fact, we can realize $$V^*$$, the dual $$V$$, as having a vector space structure if we specify a basis $$e_1, \ldots, e_n$$ of $$V$$. Then we obtain a basis of dual vectors (i.e. functionals) where we have $$l_i(e_j) = \delta_{ij}$$. In the matrix representation with respect to the $$e$$-basis, we have that each $$l_i$$ is a $$n$$-dimensional row vector with $$i$$th entry 1 and all other entries zero. This clearly forms a vector space and it's easy to show that the $$l_i$$ possess linear independence and the spanning property.