# Basis of image of linear transformation

1. Apr 20, 2008

### karnten07

1. The problem statement, all variables and given/known data
Find a basis of the image im(LA) of the linear transformation

LA: R^5 $$\rightarrow$$R^3, x$$\mapsto$$Ax

where

A =
1 -2 2 3 -1
-3 6 -1 1 -7
2 -4 5 8 -4

and hence determine the dimension of im(LA)

3. The attempt at a solution
Using the equation LA= Ax,

can i set up an augmented matrix of A and say, y1, y2 and y3 on the end? Then perform row reduction to get it into row reduced echelon form. Is that the right way to go about it? I will perform the calculation and update with it but any direction would be greatly appreciated.

2. Apr 20, 2008

### karnten07

A has row reduced form of:

1 -2 0 -1 3 y1-2y3
0 0 1 2 -2 y3-2y1
0 0 0 0 0 y2-3y1-5y3

I have the solutions to the basis of the null space for A, how is this case different?

3. Apr 20, 2008

### karnten07

I rearranged the equations to get:

y3=2x1-4x2+5x3+4x5
y1=3x1-6x2+10x3-x4+7x5