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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 ?

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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: $$ F(x) = (\alpha_{11}x_1+\alpha_{12}x_2+ \dots + \alpha_{1n} x_n)v_1 + \dots $$

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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.

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  • $\begingroup$ But wouldn't that be from ${\rm I\!R}^n $ to $V$ instead of $U$ to $V$ ? $\endgroup$ – Felipe Dilho Jun 23 '17 at 3:30
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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.

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