Basis of image of linear transformation

[karnten07]

Homework Statement

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)

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.

 

[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?

 

[karnten07]

I rearranged the equations to get:

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

Does this even have any use, I am just getting lost with this thing. Please help, urgent.

 

[HallsofIvy]

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?

That is not what I get! You had to divide by 5 to get that "1" in the second row didn't you? In the last column, I get, for the second row, (y3- 2y1)/5 and, for the third row (5y3- 13y1-y2)/5. In order that there exist x such that Ax= y you must have 5y3- 13y1- y2= 0. That is the same as y2= 5y3- 13y1 so if you take y1= 1, y3= 0, y2 must be -13. <1, -13, 0> is in the image. If you take y1= 0, y3= 1, y2 must be 5. <0, 5, 1> is in the image.