Rebecca Villa

2022-07-15

Let V be a 3 dimensional vector space over a field F and fix $({\mathbf{v}}_{\mathbf{1}},{\mathbf{v}}_{\mathbf{2}},{\mathbf{v}}_{\mathbf{3}})$ as a basis. Consider a linear transformation $T:V\to V$. Then we have

$T({\mathbf{v}}_{\mathbf{1}})={a}_{11}{\mathbf{v}}_{\mathbf{1}}+{a}_{21}{\mathbf{v}}_{\mathbf{2}}+{a}_{31}{\mathbf{v}}_{\mathbf{3}}$

$T({\mathbf{v}}_{\mathbf{2}})={a}_{12}{\mathbf{v}}_{\mathbf{1}}+{a}_{22}{\mathbf{v}}_{\mathbf{2}}+{a}_{32}{\mathbf{v}}_{\mathbf{3}}$

$T({\mathbf{v}}_{\mathbf{3}})={a}_{13}{\mathbf{v}}_{\mathbf{1}}+{a}_{23}{\mathbf{v}}_{\mathbf{2}}+{a}_{33}{\mathbf{v}}_{\mathbf{3}}$

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 $T({\mathbf{v}}_{\mathbf{i}})=\sum _{j}{a}_{ij}{\mathbf{v}}_{\mathbf{j}}$ , then we can identify T by the matrix $({a}_{ij})$. My problem is: isn't the matrix is $({a}_{ji})$ instead of $({a}_{ij})$?

$T({\mathbf{v}}_{\mathbf{1}})={a}_{11}{\mathbf{v}}_{\mathbf{1}}+{a}_{21}{\mathbf{v}}_{\mathbf{2}}+{a}_{31}{\mathbf{v}}_{\mathbf{3}}$

$T({\mathbf{v}}_{\mathbf{2}})={a}_{12}{\mathbf{v}}_{\mathbf{1}}+{a}_{22}{\mathbf{v}}_{\mathbf{2}}+{a}_{32}{\mathbf{v}}_{\mathbf{3}}$

$T({\mathbf{v}}_{\mathbf{3}})={a}_{13}{\mathbf{v}}_{\mathbf{1}}+{a}_{23}{\mathbf{v}}_{\mathbf{2}}+{a}_{33}{\mathbf{v}}_{\mathbf{3}}$

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 $T({\mathbf{v}}_{\mathbf{i}})=\sum _{j}{a}_{ij}{\mathbf{v}}_{\mathbf{j}}$ , then we can identify T by the matrix $({a}_{ij})$. My problem is: isn't the matrix is $({a}_{ji})$ instead of $({a}_{ij})$?

kawiarkahh

Beginner2022-07-16Added 15 answers

Your book has a typo (as did I); it should be

$T({\mathbf{v}}_{i})=\sum _{j}{a}_{ji}{\mathbf{v}}_{j}.$

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

$\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right).$

Applying the algorithm for matrix multiplication,

$\left(\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\end{array}\right)\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c}{a}_{11}\phantom{\rule{negativethinmathspace}{0ex}}\\ {a}_{21}\phantom{\rule{negativethinmathspace}{0ex}}\\ {a}_{31}\phantom{\rule{negativethinmathspace}{0ex}}\end{array}\right)={a}_{11}{\mathbf{v}}_{1}+{a}_{21}{\mathbf{v}}_{2}+{a}_{31}{\mathbf{v}}_{3}=\sum _{j=1}^{3}{a}_{j1}{\mathbf{v}}_{j}.$

$T({\mathbf{v}}_{i})=\sum _{j}{a}_{ji}{\mathbf{v}}_{j}.$

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

$\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right).$

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

Beginner2022-07-17Added 5 answers

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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