# 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.

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there is a typo, the book said $T(\mathbf{v_i})=\sum_j a_{ij}\mathbf{v_j}$, which i think the coefficients should be $a_{ji}$ instead of $a_{ij}$. – Ishigami Jun 30 '13 at 10:20

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.

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isn't $(a_{11},a_{21},a_{31})^T=a_{11}\mathbf{v_1}+a_{21}\mathbf{v_2}+a_{31}\mathbf{v_‌​{3}}$ ? – Ishigami Jun 30 '13 at 10:16
sorry there is a typo, the book said $v_j$, not $v_i$, but I still don't understand why the coefficients are $a_{ij}$ instead of $a_{ji}$. – Ishigami Jun 30 '13 at 10:29
Do you understand the example I gave? – Zev Chonoles Jun 30 '13 at 10:29
I understand $v_1$ corresponds to $(1,0,0)^T$ and that $T(v_1)=A(1,0,0)^T=(a_{11},a_{21},a_{31})^T$, where $A$ is the matrix $(a_{ij})$. But isn't $(a_{11},a_{21},a_{31})^T=a_{11}v_1+a_{21}v_2+a_{31}v_3$? Sorry for asking a dumb question... – Ishigami Jun 30 '13 at 10:48
@Excelsior: Very sorry, that is my mistake. I've corrected my answer. – Zev Chonoles Jun 30 '13 at 11:13

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.

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sorry it is a typo, it should be $v_j$, but i still don't understand why the coefficients are $a_{ij}$ but not $a_{ji}$. – Ishigami Jun 30 '13 at 10:24