Why do i have to transpose the coefficients of a system of linear transformations to get its matrix form? I have been reading a book about linear algebra that is very popular in my country and  in the chapter about the matrix of a linear transformation he states the following:
$``$ Be $\ U\ $ and $\ V\ $ two vector spaces, over $\ {\rm I\!R} \ $, of dimensions $n$ and $m$, respectively. Consider the linear transformation $\ F:U \rightarrow V \ $. Given the basis $\ B=\{u_1, ...,u_n\}\ $ of $\ U \ $ and the basis $\ C = \{v_1,...,v_m\} \ $  of $\ V \ $ ,  therefore each one of the vectors $\ F(u_1),...,F(u_n)\ $ are in $\ V \ $ and consequently are linear combinations of the basis $\ C \ $:
$$F(u_1) = \alpha_{11}v_1 + \alpha_{21}v_2 + ... + \alpha_{m1}v_m$$
$$F(u_2) = \alpha_{12}v_1 + \alpha_{22}v_2 + ... + \alpha_{m2}v_m$$
$$\vdots$$
$$F(u_n) = \alpha_{1n}v_1 + \alpha_{2n}v_2 + ... + \alpha_{mn}v_m$$
$Definition \ \ 4$ - A matrix $\ m\times n \ $ over $\ {\rm I\!R} \ $.
$$ \qquad \quad \begin{pmatrix}
    \alpha_{11} & \alpha_{12} & \dots  & \alpha_{1n} \\
    \alpha_{21} & \alpha_{22}  & \dots  & \alpha_{2n} \\
    \vdots & \vdots  & \ddots & \vdots \\
    \alpha_{m1} & \alpha_{m2}  & \dots  & \alpha_{mn}
\end{pmatrix}=(\alpha_{ij})$$
that is obtained from the previous considerations is called matrix of $F$ in relation to the basis $B$ and $C$ . $"$
So why is it the transpose of the matrix of the coefficients of the linear relations instead of the usual matrix ?
 A: Let $M$ be the matrix of $F$. To represent $u_1$ in the basis $B$, we use the vector $e_1 = \begin{bmatrix}1\\0\\ \vdots\\0\end{bmatrix}$.
Similarly we represent $u_i$ as the $n$-dimensional standard basis vector $e_i$ (all zeros except "$1$" in the $i$th entry).
Also, we represent $v_i$ as the $m$-dimensional standard basis vector $e_i$.
So, the equation
$$F(u_1) = \alpha_{11} v_1 + \cdots + \alpha_{m1} v_m$$
becomes
$$M \begin{bmatrix}1\\0\\ \vdots\\0\end{bmatrix} = \begin{bmatrix}\alpha_{11}\\ \alpha_{21}\\ \vdots\\ \alpha_{m1}\end{bmatrix}.$$
From here, it is clear that the first column of $M$ is
$\begin{bmatrix}\alpha_{11}\\ \alpha_{21}\\ \vdots\\ \alpha_{m1}\end{bmatrix}$. The other columns can be determined similarly.
A: a typical vector in $U$ has the form:
$$
x = x_1u_1 + x_2u_2 + \dots + x_nu_n
$$
so
$$
F(x_1u_1) = x_1(\alpha_{11}v_1 + \alpha_{21}v_2 + ... + \alpha_{m1}v_m) \\
F(x_2u_2) = x_2(\alpha_{12}v_1 + \alpha_{22}v_2 + ... + \alpha_{m2}v_m) \\
\vdots \\
F(x_nu_n) = x_n(\alpha_{1n}v_1 + \alpha_{2n}v_2 + ... + \alpha_{mn}v_m)
$$
add these together to obtain $F(x)$, and regroup the terms by the basis vectors of $V$, giving:
$$
F(x) = (\alpha_{11}x_1+\alpha_{12}x_2+ \dots + \alpha_{1n} x_n)v_1 + \dots
$$
A: You can represent that transformation either way, with or without transposing. It all depends on whether you want to multiply your matrix by a row vector on the left, or by a column vector on the right. Each author chooses one convention or the other. It's good to recognize that it's just a convention, and not a mathematical necessity.
