Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer
    
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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.