# Quick doubt about linear application and its matrix

1. Jun 7, 2013

### Felafel

1. The problem statement, all variables and given/known data
Let $f:\mathbb{R}^3\to \mathbb{R}^3$ such that $v_1=(1,0,1) , v_2=(0,1,-1), v_3=(0,0,2)$ and $f(v_1)=(3,1,0), f(v_2)=(-1,0,2), f(v_3)=(0,2,0)$
find $M^{E,E}_f$ where $E=(e_1,e_2,e_3)$ is the canonical basis.

3. The attempt at a solution
i see
$v_1=e_1+e_3$
$v_2=e_2-e_3$
$v_3=2e_3$
thus
$f(e_1)+f(e_3)=(3,1,0)$
$f(e_2)-f(e_3)=(-1, 0 ,2)$
$2f(e_3)=(0,2,0)$
solving the system i get
$f(e_3)=(0,1,0)$
$f(e_1)=(3,0,0)$
$f(e_2)=(-1,1,2)$

and so I thought the matrix was:

(3 -1 0)
(0 1 1)
(0 2 0)

but according to my book the solution should be:
(3 0 0)
(-1 1 2)
(0 1 0)

why? shouldn't the image of the vector form the columns instead of the rows?

2. Jun 7, 2013

### Dick

Your answer looks right assuming the vectors are column vectors. Perhaps the book is treating them as row vectors and multiplying the matrix on the right?

3. Jun 7, 2013

### Felafel

i don't see why it would do that though.
it's just wrong, probably
thank you :)