Rank-Nullity of Linear Functional. I read in a text that rank of linear functional is $1$ and Nullity=$n-1$ if we consider a nonzero $n$ dimensional vector space $V$ to some scalar field. But I can not visualize it.
I want to see a few examples of maybe $\mathbb{R^{2}}$ and $\mathbb{R^{3}}$ Field.
In my opinion,
$T:V\to \mathbb{R^{2}}$, then $T(v)=(x,y)$, why should the rank be $1$ here?
 A: If $V$ is a vector space over $F$ then , a linear map $T:V\to F$ is called a linear functional.  You need to view the $F$ in the right hand side as a vector space of $1$ dimension over $F$. That is $\{1\}$ generates the vector space $F$ over $F$.
If the map is non-zero. That is if $T(v)\neq 0$ for some $v\in V$. Then the rank of this map is $1$. as $T(v)$ spans the entire space $F$ over $F$.
From the rank-nullity theorem it then follows that $\dim(\ker(T))=n-1$.
In your example the image($\mathbb{R^{2}})$ is not a field .
An example of a linear functional would be i-th coordinate map.
$\pi_{i}:\mathbb{R}^{n}\to\mathbb{R}$ such that $\pi_{i}(c_{1},c_{2},...,c_{n})=c_{i}$.
Or even for a finite dimensional vector space with basis($\{v_{1},v_{2},..v_{n}\}$) you can define like this:-
$f_{i}(\sum_{i=1}^{n}c_{i}v_{i})=c_{i}$.
If you want from an infinite dimensional vector space, consider the evaluation map.
$ev_{a}:P(\mathbb{R})\to\mathbb{R}$ such that $ev_{a}(f(x))=f(a)$. Where $P(\mathbb{R})$ denotes the space of real polynomials and $a\in\mathbb{R}$.
