At the bottom of page 3 of Spivak's Calculus on Manifolds Book the author mentions a linear transformation of a basis: $T:\bf{R^n}\rightarrow \bf{R}^m$ is a matrix $A$ with $m$ rows and $n$ columns. But the transformation formulae of the individual basis, $e_i$, is confusing:
$T(e_i) = \sum_{j=1}^m a_{ji}e_j$
and the coefficients of $T(e_j)$ are the columns of A.
To take a simple example if A is a 3*2 matrix $\begin{bmatrix}3&1\\2&3\\1&5\end{bmatrix}$ and we want to transform a $\bf{R}^2 \rightarrow \bf{R}^3 $ then $T(e_1)$ should be $\begin{equation} \begin{bmatrix}3&1\\2&3\\1&5\end{bmatrix} * \begin{bmatrix}1\\0\end{bmatrix} \end{equation}$ which is product of the rows of A and the first basis column vector giving the column vector $\begin{bmatrix}3\\2\\1\end{bmatrix}$ (similarly for $e_2$), but then the formulae for this should be $T(e_i) = \sum_{j=1}^{n} a_{ij}*e_j$ not what the author has written out. What am I missing?
I think you're mixing up the typical order of indexing. When we write a linear transformation $T$ in terms of some ordered basis $\{e_i\}$, it generally looks (and is indexed) like so:
$$ T = \begin{pmatrix} a_{11} & a_{12} & \cdots \\ a_{21} & a_{22} & \cdots \\ \vdots & \vdots & \ddots \end{pmatrix} $$
So if we apply this to some basis vector, say the first standard basis vector $e_1$:
$$ T(e_1) = \begin{pmatrix} a_{11} & a_{12} & \cdots \\ a_{21} & a_{22} & \cdots \\ \vdots & \vdots & \ddots \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ \vdots \end{pmatrix} $$
Following the rules for matrix multiplication, as well as breaking the result vector into a linear combination of basis vectors, we get
$$ T(e_1) = \begin{pmatrix} a_{11} \\ 0 \\ \vdots \end{pmatrix} + \begin{pmatrix} 0 \\ a_{21} \\ \vdots \end{pmatrix} + \cdots \\ = a_{11} e_1 + a_{21} e_2 + \cdots \\ = \sum_{j=1}^{n} a_{j1}e_j $$
We used $i=1$ in our example to keep things simple, but we could've used any index, so the general result holds.