What kind of a Linear Map is a Column (and Row) Matrix? Firstly, I understand that all linear maps of the form $\mathbb{R}^n \rightarrow \mathbb{R}^m$ can be represented by a $m\times n$ matrix. 
(Although, I'm unsure whether this entails the converse where all matrices can be construed as linear maps.)
However, in this case, what kind of a linear map is a column and row matrix? 
Let's consider the column matrix $\left[\begin{array}{c}5\\4\\3\end{array}\right]$. 
Assuming that $\mathbb{R}$ is a vector space, with a basis `vector' (1), this matrix represents a function $f$ that maps (1) to the vector $(5,4,3)\in\mathbb{R}^3$, (assuming also that we're using the standard basis in $\mathbb{R}^3$). Thus, $f$ maps all real numbers $c\in\mathbb{R}$ to its scalar multiple $c(5,4,3)\in\mathbb{R}^3$.
This gets confusing however, as column matrices like these are usually construed as just a single vector $(5,4,3)$. Is there a way to make sense of this? 
Thanks for any help!
 A: In what follows, we will use the notation $\mathbb{F}^n_{\operatorname{col}}$ to denote the space of column vectors and similarly $\mathbb{F}^n_{\operatorname{row}}$ to denote the space of row vectors. The spaces are of course isomorphic (with the isomorphism given by the transpose operation) or even considered sometimes to be the same (as no one bothers to say whether there is a mathematical difference between row vectors and column vectors) but it is extremely useful to keep track of the types of vectors involved because certain identifications work only with row/column vectors. It becomes even more useful when working with general vector spaces that don't have natural bases unlike $\mathbb{F}^n$.
Every matrix $A \in M_{m \times n}(\mathbb{F})$ defines a linear map $L_A \colon \mathbb{F}^n_{\operatorname{col}} \rightarrow \mathbb{F}^m_{\operatorname{col}}$ by left multiplication: $L_A(x) = Ax$. The converse holds as well and for every linear map $T \colon \mathbb{F}^n_{\operatorname{col}} \rightarrow \mathbb{F}^m_{\operatorname{col}}$ there exists a (unique) matrix $A \in M_{m \times n}(\mathbb{F})$ such that $T = L_A$. The matrix $A$ is constructed as a matrix whose $i$-th column is the vector $T(e_i)$ where $(e_1,\dots,e_n)$ is the standard basis of $\mathbb{F}^n$. Thus, we get a one-to-one (and even linear) correspondence between linear maps $\mathbb{F}^n_{\operatorname{col}} \rightarrow \mathbb{F}^m_{\operatorname{col}}$ and matrices $A \in M_{m \times n}(\mathbb{F})$.
Under this correspondence, it seems that column vectors $v \in \mathbb{F}^n_{\operatorname{col}}$ can play now two roles: They are elements in the vector space $\mathbb{F}^n_{\operatorname{col}}$ and they (can) represent linear maps $\mathbb{F}_{\operatorname{col}} \rightarrow \mathbb{F}^n_{\operatorname{col}}$ and this is indeed the case. Since $\mathbb{F}_{\operatorname{col}} = \mathbb{F}_{\operatorname{row}} = \mathbb{F}$, we'll stop differentiating between them (from a more abstract point of view, this reflects the fact that $\mathbb{F}$ and $\mathbb{F}^{*}$ are naturally isomorphic). A linear map $T \colon \mathbb{F} \rightarrow \mathbb{F}^n_{\operatorname{col}}$ is determined uniquely by its image on the "standard basis" $(1)$ and conversely, every vector $v$ determines uniquely a linear map $L_v \colon \mathbb{F} \rightarrow \mathbb{F}^n_{\operatorname{col}}$  given by $L_v(a) = av$ (and this is consistent with the notation above). 
It might seem strange to you that a vector $v \in \mathbb{F}^n_{\operatorname{col}}$ can be identified with a linear map but this is precisely one of the major points of linear algebra. A priori, to describe a linear map $T \colon \mathbb{R}^n_{\operatorname{col}} \rightarrow \mathbb{R}^m_{\operatorname{col}}$ one must provide us with infinite amount of data - for each $v \in \mathbb{R}^n$ we must be told what $T$ does to $v$. Thus, it might seem that we need to know the set $\{ (v,Tv) \, | v \in \mathbb{R}^n_{\operatorname{col}} \}$ but since $T$ is linear, it is already determined on a basis (and the converse also holds) and so it is enough to provide us with a finite set of data $(T(e_1), \dots, T(e_n))$ which we can arrange in a matrix. 
This correspondence even plays very well with the relation between composition of linear operators and matrix multiplication. If we are given $L_v \colon \mathbb{F} \rightarrow \mathbb{F}^n_{\operatorname{col}}$ and $L_A \colon \mathbb{F}^n_{\operatorname{col}} \rightarrow \mathbb{F}^m_{\operatorname{col}}$ then the composition $L_A \circ L_v \colon \mathbb{F} \rightarrow \mathbb{F}^m_{\operatorname{col}}$ is determined uniquely by $(L_A \circ L_v)(1) = L_A(v) = Av$ and so can be identified with the vector $Av$. On the other hand, composition of linear operators corresponds to multiplication of the representing matrices resulting also in $Av$!
If you are familiar with linear functionals, you can note now that every linear functional $\varphi \colon \mathbb{F}^n_{\operatorname{col}} \rightarrow \mathbb{F}$ is represented unique by a $1 \times n$ matrix - a row vector and the converse also holds as every row vector $w = (a_1, \dots, a_n)$ defines a linear functional $L_w$ on $\mathbb{F}^n_{\operatorname{col}}$ by
$$ L_w \left( \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix} \right) = (a_1, \dots, a_n) \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix}  = \sum_{i=1}^n a_i x_i. $$
Thus, column vectors are often identified with "regular vectors" while row vectors are identified with linear functionals or "covectors" on the space of regular vectors. This is also consistent with what one does when solving a system of linear equations - each linear equation is represented by a row vector (and indeed, the row vector defines a linear functional whose kernel is precisely the solution space of the equation). The equations are placed as rows in a matrix $A$ which (can) represent a linear map $L_A \colon \mathbb{F}^n_{\operatorname{col}} \rightarrow \mathbb{F}^m_{\operatorname{col}}$ (where there are $m$ equations in $n$ variables). The kernel of $L_A$ is then the solution space of the system of equations.
