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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?

Thank you in advance for your help.

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    $\begingroup$ Let $V$ be a vector space over the field $\Bbb K.$ Then a linear functional $f$ on $V$ is a linear map $f : V \longrightarrow \Bbb K.$ $\endgroup$ – math maniac. May 19 '20 at 4:14
  • $\begingroup$ 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 $V^*.$ $\endgroup$ – math maniac. May 19 '20 at 4:21
  • $\begingroup$ 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 $B=0$. $\endgroup$ – Ivo Terek May 19 '20 at 4:48
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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.

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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.

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