I have this from my text:
So I understand that the $\ker(T)$ is the set of vectors in V in $T: V \rightarrow W$ that satisfy $T(v) = 0.$ In english, to me this means the vectors that when transformed (have the function applied) brings you a zero vector.
I don't really understand the converse part. So we supposed the kernel of T consists of only the 0 vector. What are u and v? What is the point of this part of the proof? They are just random vectors right that don't have to be part of the kernel right?
The surprising part of this proof to me is that you can judge the one-to-oneness of a transformation using only the zero vector. What if the zero vector is one to one but other vectors in V can be transformed to the same vector in W?
The book does not give an example and this seems too abstract. Can someone provide an example?
Best Answer
The theorem says that $T$ is one-to-one if and only if $\ker(T)=\{\mathbf{0}\}.$ So, if $\ker(T)=\{\mathbf{0}\},$ we have to show that $T$ is one-to-one. What does that mean? It means no two different vectors map to the same vector. That is to say, if $\mathbf{u}\neq\mathbf{v},$ then $T(\mathbf{u})\neq T(\mathbf{v}).$ This is equivalent to the statement that if $T(\mathbf{u})= T(\mathbf{v}),$ then $\mathbf{u}=\mathbf{v},$ because this is the contrapositive of the first statement.
As to your statement, "The surprising part of this proof to me is that you can judge the one-to-oneness of a transformation using only the zero vector," yes indeed; that's why the theorem is worth proving, and powerful. As to your last question, "What if the zero vector is one to one but other vectors in V can be transformed to the same vector in W," this can't happen. That's what the theorem is all about.
You're probably thinking about this as if $T$ were just an arbitrary function, but $T$ is linear. That's a very strong condition.