Hi there.

I've a doubt in getting the image of a linear transformmation.

given this LT: $\displaystyle R^2 \rightarrow R^3\ T(x,y)=(x-2y,y-x,3x)$

the book I'm using always attempts to do as following:

$\displaystyle (x-2y,y-x,3x)=(a,b,c)$ and solve the system $\displaystyle \begin{pmatrix} +1-2=a \\-1+1=b \\+3+0=c\end{pmatrix}$

$\displaystyle \Rightarrow \\ c-3a+6a+6b = 0 $

$\displaystyle c+3a+6b = 0\\ \Rightarrow base=\{(1,0,-3);(0,1,-3)\}$

I wonder if I get an equivalent result if I do it the following manner:

$\displaystyle (x-2y,y-x,3x) = x(1,-1,3)+y(-2,1,0) \Rightarrow base=\{(1,-1,3);(-2,1,0)\}$

of course, in this case I'll need to check if are them linearly independent if I want to get a base.

thanks in advice

VIKKO