matrix representations of linear transformation I have a indexing problem about the matrix representation of linear transformation.
Let $V$ be a $3$ dimensional vector space over a field $F$ and fix $(\mathbf{v_1},\mathbf{v_2},\mathbf{v_3})$ as a basis. Consider a linear transformation $T: V \rightarrow V$.
Then we have
$$T(\mathbf{v_1})=a_{11}\mathbf{v_1}+a_{21}\mathbf{v_2}+a_{31}\mathbf{v_3}$$
$$T(\mathbf{v_2})=a_{12}\mathbf{v_1}+a_{22}\mathbf{v_2}+a_{32}\mathbf{v_3}$$
$$T(\mathbf{v_3})=a_{13}\mathbf{v_1}+a_{23}\mathbf{v_2}+a_{33}\mathbf{v_3}$$
So that we can identify $T$ by the matrix 
$$\begin{pmatrix}
a_{11}&a_{12}&a_{13}\\
a_{21}&a_{22}&a_{23}\\
a_{31}&a_{32}&a_{33}\\
\end{pmatrix}$$
But then when I read several linear algebra book, it said
if $T(\mathbf{v_i})=\sum_j a_{ij}\mathbf{v_j}$, then we can identify $T$ by the matrix $(a_{ij})$. My problem is: isn't the matrix is $(a_{ji})$ instead of $(a_{ij})$? Could someone please explain the subtle difference, thanks in advance.
 A: Your book has a typo (as did I); it should be 
$$T(\mathbf{v}_i)=\sum_ja_{ji}\mathbf{v}_j.$$
Here is an example: in the basis $\{\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\}$, we know that $\mathbf{v}_1$ corresponds to the column vector
$$\begin{pmatrix}
1\\ 0\\ 0\end{pmatrix}.$$
Applying the algorithm for matrix multiplication,
$$\begin{pmatrix}
a_{11}&a_{12}&a_{13}\\
a_{21}&a_{22}&a_{23}\\
a_{31}&a_{32}&a_{33}\\
\end{pmatrix}\begin{pmatrix}
1\\ 0\\ 0\end{pmatrix}=\begin{pmatrix}
a_{11}\!\\ a_{21}\!\\ a_{31}\!\end{pmatrix}=a_{11}\mathbf{v}_1+a_{21}\mathbf{v}_2+a_{31}\mathbf{v}_3=\sum_{j=1}^3a_{j1}\mathbf{v}_j.$$

I had instinctively been thinking of the formula
$$(AB)_{ij}=\sum_k A_{ik}B_{kj}$$
for multiplying two matrices together, where the index of the summation appears on the right in the part of the expression for $A$ - that is, $A_{i\hspace{0.02cm}\large\mathbf{k}}$. However, this is expressing the entries of $AB$; to express a column of $AB$ itself as a summation of basis vectors, the index variable would be the one on the left.
A: They're written it wrong.  They should've written $T(v_j) = \sum_i a_{ij} v_i$--edited so that this is correct now. I saw that $i$ wasn't free in the original way it was written but not the transpose part. I think a form that doesn't transpose is inherently easier to work with.
