Spivak's Notation on Linear Transformation 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?
 A: 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.
A: In your example, note that $ T (e_1) $ is in $ \mathbb{R} ^ 3 $ so can not be written as such a linear combination. 
$T(e_1)= 3\begin{bmatrix}1\\0\\0\end{bmatrix}+2\begin{bmatrix}0\\1\\0\end{bmatrix}+1\begin{bmatrix}0\\0\\1\end{bmatrix}$.
The $e_j$ on the right side of the formula refers to the base of $\mathbb{R}^m$.
