Rebecca Villa

2022-07-15

Let V be a 3 dimensional vector space over a field F and fix $\left({\mathbf{v}}_{\mathbf{1}},{\mathbf{v}}_{\mathbf{2}},{\mathbf{v}}_{\mathbf{3}}\right)$ as a basis. Consider a linear transformation $T:V\to V$. Then we have
$T\left({\mathbf{v}}_{\mathbf{1}}\right)={a}_{11}{\mathbf{v}}_{\mathbf{1}}+{a}_{21}{\mathbf{v}}_{\mathbf{2}}+{a}_{31}{\mathbf{v}}_{\mathbf{3}}$
$T\left({\mathbf{v}}_{\mathbf{2}}\right)={a}_{12}{\mathbf{v}}_{\mathbf{1}}+{a}_{22}{\mathbf{v}}_{\mathbf{2}}+{a}_{32}{\mathbf{v}}_{\mathbf{3}}$
$T\left({\mathbf{v}}_{\mathbf{3}}\right)={a}_{13}{\mathbf{v}}_{\mathbf{1}}+{a}_{23}{\mathbf{v}}_{\mathbf{2}}+{a}_{33}{\mathbf{v}}_{\mathbf{3}}$
So that we can identify T by the matrix
$\left(\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\end{array}\right)$
But then when I read several linear algebra book, it said if $T\left({\mathbf{v}}_{\mathbf{i}}\right)=\sum _{j}{a}_{ij}{\mathbf{v}}_{\mathbf{j}}$ , then we can identify T by the matrix $\left({a}_{ij}\right)$. My problem is: isn't the matrix is $\left({a}_{ji}\right)$ instead of $\left({a}_{ij}\right)$?

### Answer & Explanation

kawiarkahh

Your book has a typo (as did I); it should be
$T\left({\mathbf{v}}_{i}\right)=\sum _{j}{a}_{ji}{\mathbf{v}}_{j}.$
Here is an example: in the basis $\left\{{\mathbf{v}}_{1},{\mathbf{v}}_{2},{\mathbf{v}}_{3}\right\}$, we know that ${\mathbf{v}}_{1}$ corresponds to the column vector
$\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right).$
Applying the algorithm for matrix multiplication,
$\left(\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\end{array}\right)\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c}{a}_{11}\phantom{\rule{negativethinmathspace}{0ex}}\\ {a}_{21}\phantom{\rule{negativethinmathspace}{0ex}}\\ {a}_{31}\phantom{\rule{negativethinmathspace}{0ex}}\end{array}\right)={a}_{11}{\mathbf{v}}_{1}+{a}_{21}{\mathbf{v}}_{2}+{a}_{31}{\mathbf{v}}_{3}=\sum _{j=1}^{3}{a}_{j1}{\mathbf{v}}_{j}.$

Patatiniuh

They're written it wrong. They should've written $T\left({v}_{j}\right)=\sum _{i}{a}_{ij}{v}_{i}$ - 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.