Verifying the examples of Dual Space

As Pedro said in his answer, the dual space of a vector space $V$ is a set whose elements are linear functions from $V$ to what Pedro called the "underlying field". If this is unfamiliar to you, that probably means this mysterious "underlying field" is $\mathbb{R}$, the real numbers. So the dual space (sometimes denoted $V^*$) is the set of all linear functionals, i.e. all linear maps $$T: V \to \mathbb{R}$$

In other words, once you pick a basis for $V$, the dual space is just all functions represented by a row matrix $$(a_1 \ a_2 \ \dots \ a_n)$$ For example, if $V$ is the set of all $2 \times 2$ matrices, then the trace map $Tr: V \to \mathbb{R}$ is a linear functional (ie a linear map from $V$ to $\mathbb{R}$), since $$Tr(xA + yB) = x \ Tr(A) + y \ Tr(B)$$ where $x$ and $y$ are real numbers ("scalars") and $A$ and $B$ are matrices. Mind you, this is not the entire dual space! This is just one element of the dual space. So $Tr \in V^*$. If you pick the basis $$\{ \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix} \begin{pmatrix}0 & 0 \\ 0 & 1 \end{pmatrix}\}$$ for $V$, then the map $Tr$ is given by the row matrix $$(1 \ 0 \ 0 \ 1)$$

Footnote: The reason for saying "underlying field" is that linear algebra can be done over many different "objects": the rational numbers, complex numbers, and finite fields - integers modulo a prime number $p$, and oftentimes even the integers or other "rings". But if that's not a point of interest now, then that's fine.

You're thinking about a linear mapping $V\to k$ where $k$ is the underlying field of your vector space. This is usually called a linear functional. These are the elements of the dual space of a vector space $V$, the linear mappings $V\to k$ (so the dual space of a vector space $V$ is not a mapping, it is the set of all linear mapings $V\to k$). As an example, you have above the linear functional that takes an element in $V=k[X]$, a polynomial with coefficients in your field, and returns the value of that polynomial at $0$ -- this is usually called an evaluation map (for obvious reasons), and if we're evaluating at $t\in k$; we usually denote it by ${\rm ev}_t$. So ${\rm ev}_t(p)=p(t)$.

You can indeed check that $p\mapsto p(0)$ where $p$ is a polynomial is linear, and $p(0)\in k$, so you have an element of $V^\ast:={\rm Hom}_k(V,k)$ (that's the notation for the dual space of $V$). The second example is the zero map from $V\to k$; it sends every element of $V$ to $0\in k$ (the additive identity of your field.) This is again linear, so it is an element of $V^\ast$.