Is there a reason vectors in space are represented as column vectors (in that nothing works with row vectors)? As opposed to row vectors? It would seem that whenever performing operations on vectors in space (applying a matrix/linear transformation to it, for example) does not work unless the vector is in it's column form (since lots of things, such as matrix multiplication, are dependent on dimensionality). Why is it that things work with column vectors but not row vectors?
 A: It seems that you consider the vector space $K^m$ of all $m$-tuples1 $\mathbf x = (x_1,\dots, x_m)$ with coordinates $x_i$ in a field $K$, for example $K  = \mathbb R$.
One can very well argue that it is just a notational issue (i.e. a mere convention) how to write the elements  $\mathbf x \in K^m$. The only requirement is that we can uniquely identify the coordinates of each vector. Expressed more formally, the chosen notation must come along with well-defined coordinate projections $p^m_i : K^m \to K$.
Some standard representation variants are

*

*Tupel: $\mathbf x = (x_1,x_2,\dots, x_m)$

*Row vector: $\mathbf x = (x_1 \quad x_2 \quad \dots \quad x_m)$

*Column vector: $\left(\begin{array}{c}x_1\\x_2\\\vdots\\x_m\end{array}\right)$
The difference between tupel and row vector representation seems perhaps to be somewhat artifical since we only changed the separating character between the elements $x_i$ (comma / blank space), but we should be aware that in both row and column vector representation $\mathbf x$ is written as a matrix2.
As long as we only consider vectors in $K^m$, the choice of a representation is  absolutely irrelevant. We can add vectors and take scalar multiples in the obvoius way. However, the concrete representation variant becomes important if we consider linear maps $f : K^m \to K^n$. The point is that such a linear map is usually expressed in form of a matrix obtained by assembling the images $f(\mathbf e^m_i) \in K^n$ of the standard basis vectors3 $\mathbf e^m_i \in K^m$. This can be done based on row and on column vector representation.

*

*In row vector representation: The $m$ row vectors $f(\mathbf e^m_1),\dots,f(\mathbf e^m_m)$ are written one below the other which gives a matrix $\mathbf M_{row}(f) \in K(m,n)$.

*In column vector representation: The $m$ column vectors $f(\mathbf e^m_1),\dots,f(\mathbf e^m_m)$ are written next to each other from left to right which gives a matrix $\mathbf M_{col}(f) \in K(n,m)$.

Clearly these matrices are transposed, i.e. $(\mathbf M_{row}(f))^T = \mathbf M_{col}(f)$. It is well-known that the image vectors $f(\mathbf x)$ can now be computed via matrix multiplication 4. We have

*

*In row vector representation:
\begin{equation}
f(\mathbf x) = \mathbf x \cdot \mathbf M_{row}(f) . \tag{1}
\end{equation}

*In column vector representation5:
\begin{equation}
f(\mathbf x) = \mathbf M_{col}(f) \cdot \mathbf x .  \tag{2}
\end{equation}
These two equations explain why the column vector form is the most popular variant when working with matrix representations of vectors and linear maps: In $(2)$ the vector $\mathbf x$ occurs right of the operator on both sides of the equation, in $(1)$ the order is reversed.
To distinguish visibly between $\mathbf x \in K^m$ and its notational variants in row and column vector representation I suggest to write $\mathbf x_{row} \in K(1,m)$ and  $\mathbf x_{col} \in K(m,1)$. Doing so we get the two unambiguous formulas
\begin{equation}
f(\mathbf x)_{row} = \mathbf x_{row} \cdot \mathbf M_{row}(f) \tag{3}.
\end{equation}
\begin{equation}
f(\mathbf x)_{col} = \mathbf M_{col}(f) \cdot \mathbf x_{col} . \tag{4}
\end{equation}
Note that $\mathbf x_{row}^T = \mathbf x_{col}$.
Whatever our preference is, we have to make a fundamental choice either for the row or the column vector form.  This representation convention must then be strictly applied to all vectors of $K^m$ and all linear maps $f : K^m  \to K^n$; we never decide case-by-case.
To emphasize it, both possible representation conventions work perfectly.
In the sequel we  focus on the column vector representation and simply write $\mathbf M(f) = \mathbf M_{col}(f)$. The column vectors of $\mathbf M(f)$ are the images $f(\mathbf e^m_i)$ of the standard basis vectors of $K^m$, i.e. are regarded as ordinary vectors in $K^n$. What about the row vectors $(\mathbf M(f))_{(i)}\in K(1,m)$ of $\mathbf M(f)$? Clearly $(\mathbf M(f))_{(i)}$ is the matrix representation of the $i$-th coordinate function $f_i = p^m_i \circ f : K^m \to K$. Therefore the row vectors must not be regarded as ordinary vectors of $K^m$ written in row vector form, but as matrix representations of linear funtionals on $K^m$. Note that these are elements of the dual space $(K^m)^*$ of $K^m$. Thus our representation convention gives a simple rule of thumb:

*

*Column vectors stand for the elements of some $K^n$.

*$n \times m$ matrices stand for linear maps $K^m \to K^n$.

*Row vectors stand for linear functionals on some $K^m$.

This seems to be very clear, but indeed we get two different, not to say inconsistent, interpretations of column vectors: On the one hand they stand for elements of $K^n$, on the other hand for linear maps $K \to K^n$. The correspondence between elements of $K^n$ and linear maps $K \to K^n$ is obvious: $\mathbf x \in K^n$ is identified with the linear map $f_{\mathbf x} : K \to K^n, f_{\mathbf x}(a) = a\mathbf x$. In fact, $\mathbf M({f_{\mathbf x}}) = f_{\mathbf x}(e^1_1) = f_{\mathbf x}(1) = \mathbf x$ in column vector representation.
So we have a "hybrid interpretation" of column vectors (and analogically of row vectors if we decide to adopt column vector representation). We must therefore admit that our above three vector representation variants are not only notational issues of complete arbitrariness; it requires a context-related explanation what a row or column vector stands for.

Footnotes

*

*Formally we can use the definition $K^m =$ set of all functions $\mathbf x : \{1,\dots,m\} \to K$. Then $x_i =  \mathbf x(i)$.


*The vector space of $p \times q$ matrices with entries in $K$ is denoeted by $K(p,q)$. Formally we can use the definition $K(p,q) =$ set of all functions $\mathbf A : \{1,\dots,p\} \times \{1,\dots,q\} \to K$. With $a_{ij} = \mathbf A(i,j)$ one usually writes $\mathbf A = (a_{ij})$. The $i$-th row of $\mathbf A$ will be denoted by $\mathbf A_{(i)} \in K(1,q)$ (lower index) and the $j$-th column by $\mathbf A^{(j)} \in K(p,1)$ (upper index).


*The standard ordered basis of $K^m$ is the $m$-tupel of vectors $\mathcal E^m = (\mathbf e^m_1,\dots, \mathbf e^m_m)$ with $\mathbf e^m_i = (\delta_{i1},\dots,\delta_{im}) \in K^m$, where $\delta_{ij} = 1$ for $i = j$ and $\delta_{ij} = 0$ for $i \ne j$. For $\mathbf x \in K^m$ with coordinates $x_i$ we have $\mathbf x = \sum_{i=1}^m x_i \mathbf e^m_i$.


*Let us recall how matrix multiplication $\mu : K(p,q) \times K(q,r) \to K(p,r)$ works. First observe that for a row vector $\mathbf v \in K(1,q)$ with coordinates $v_j$ and a column vector $\mathbf w \in K(q,1)$ with coordinates $w_j$ we take $\mathbf v \cdot \mathbf w = \sum_{j=1}^q v_jw_j \in K$. Now let $\mathbf A = (a_{ij}) \in K(p,q)$ and $\mathbf B = (b_{jk}) \in K(p,q)$. Then $\mathbf A$ consists of $p$ row vectors  $\mathbf A_{(i)} \in K(1,q)$ and $\mathbf B$ consists of $r$ column vectors $\mathbf B_{(k)} \in K(q,1)$. The matrix $\mathbf C = \mathbf A \cdot \mathbf B$ has then the entry $\mathbf A_{(i)} \cdot \mathbf B^{(k)}$ at position $(i,k)$. Thus we get the well-known representation $\mathbf C = (c_{ik})$ with $c_{ik} = \sum_{j=1}^q a_{ij}b_{jk}$. Clearly, the $i$-th row of $\mathbf A \cdot \mathbf B$ is $\mathbf A_{(i)} \cdot \mathbf B$ and the $k$-th column is $\mathbf A \cdot \mathbf B^{(k)}$. Moreover, for a row vector $\mathbf a \in K(1,q)$ with coordinates $a_j$ we get $\mathbf a \cdot \mathbf B = \sum_{j=1}^q a_j \mathbf B_{(j)}$ and for a column vector $\mathbf b \in K(q,1)$ with coordinates $b_j$ we get $\mathbf A \cdot \mathbf b = \sum_{j=1}^q b_j \mathbf A^{(j)}$


*We have $\mathbf M_{col}(f) \cdot \mathbf x = \sum_{j=1}^m x_j (\mathbf M_{col}(f))^{(j)} = \sum_{j=1}^m x_j f(\mathbf e^m_j) = f(\sum_{j=1}^m x_j \mathbf e^m_j) = f(\mathbf x).$ Formula $(1)$ is obtained similarly.
