# Linear Transformation: finding the image.

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• Oct 4th 2009, 08:33 AM
viko
Linear Transformation: finding the image.
Hi there.

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

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

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

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

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

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

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

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
• Oct 5th 2009, 06:25 AM
HallsofIvy
Quote:

Originally Posted by viko
Hi there.

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

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

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

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

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

Actually, there is an error here: with a= 0, b= 1, c= -6, not -3. A basis for the image is {(1, 0, -3); (0, 1, -6)}

Quote:

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

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

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
Yes, that is perfectly valid. Any you can check that a= 1, b= -1, c= 3 and a= -2, b= 1, c= 0 also satisfy c+3a+6b = 0.

(If they were not independent that linear transformation would be singular and map $R^2$ into a one dimensional subspace of $R^3$.)
• Oct 6th 2009, 08:55 PM
viko
thank you very much

yoour answer was very useful for me