Linear Transformations - Finding the basis for the image

Nezero
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Homework Statement


Find a basis for the image of the linear transformation T: R^4 -->R^3 given by the formula T(a,b,c,d) = (4a+b -2c - 3d, 2a + b + c - 4d, 6a - 9c + 9d)


Homework Equations




The Attempt at a Solution



Well this question followed asking about the basis for the kernel which was easy enough. Unfortunately my notes on this aren't very clear and I don't know where to start.
 
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You have 3 equations involving 4 "unknown" parameters. I recommend applying T to (1,0,0,0), (0,1,0,0), (0,0,1,0), and (0,0,0,1) in turn. That will give you 4 vectors in R^3 which clearly cannot be independent. A basis will be a subset of that set of 4 vectors. What space do they span?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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