Originally Posted by

**HallsofIvy** ??? I had just said "That is, the image is all of $\displaystyle R^2$ and has dimension 2." I have no idea how you conclude from the fact that a particular set of vectors is not a basis, that the dimension must be 0.

The set of vectors {(0, 5), (0, 0), (1, 0), (0, 3)} is not independent, first because a set of independent vectors **cannot** contain the 0 vector. Obviously, 0(0, 5)+ 1(0, 0)+ 0(1, 0)+ 0(0, 3)= 0 without all coefficients being 0. If you throw the 0 vector out, you have {(0, 5), (1, 0), (0, 3)} which is not independent because (0,5) and (0,3) are multiples of each other. Obviously, again, 3(0, 5)+ 0(1, 0)+ (-5)(0, 3)= (0, 0) without all coefficients being 0. If you throw out either (0, 5) or (0, 3) you get {(1, 0), (0, 3)} (or {(0, 5), (1, 0)}) which [b]is[\b] an independent set and is a basis for the image.