- #1

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Sketch of proof: If nullity(T) = 0, then ker(T) = {0}. So T(x) = T(y) --> T(x) - T(y) = 0 --> T(x-y) = 0 --> x-y = 0 --> x = y, which shows that T is injective. For the other direction, if T is injective, then then 0 must be the only element in the kernel, since it is always true for linear transformations that T(0) = 0, but since T is injective there is no other element in V that maps to the zero vector in W.

So there's the proof, but I still don't intuitively understand why the kernel only containing the zero vector means that T is injective, and vice versa. In contrast, the relation between the image of T and condition of being surjective is easy to see, since in order to map to all of the elements of W the image of T must have the same dimension as W. This can intuitively be seen with a diagram of the mapping from V to W, for example. I can't really imagine a diagram that plainly shows the injective condition.