Let $T: V\to W$ be a linear transformation and let {$v_1,...,v_n$}, {$w_1,...,w_m$} be ordered basis of $V$ and $W$ then

$$ \left\{ \begin{array}{c} a_{11}w_1+a_{12}w_2+\cdots+a_{1m}w_m=T(v_1) \\ a_{21}w_1+a_{22}w_2+\cdots+a_{2m}w_m=T(v_2) \\ \vdots \\ a_{n1}w_1+a_{n2}w_2+\cdots+a_{nm}w_m=T(v_n) \\ \end{array} \right. $$

For me the "natural" way to define the matrix of the transformation $T$ would be: $$ \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1m} \\ a_{21} & a_{22} & \cdots & a_{2m} \\ \vdots & \vdots & \vdots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nm} \\ \end{pmatrix} $$

but instead we work with the transpose of the matrix, but why? Can someone please clarify this to me?

  • 3
    $\begingroup$ It's to do with whether you multiply by a column matrix (on the right) or a row matrix (on the left). $\endgroup$ – James Oct 6 '14 at 23:31

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