"If T is a linear transformation on a finite dimensional vector space, then T is one-one implies T must be onto. Also T is onto implies T must be one-one."
I do not understand the proof of this.
Does it come from this --
For a finite dimensional vector space,
- T is invertible iff T is one-one.
- T is invertible iff T is onto.
So from statement 1 and 2, we can say that T is one-one if T is onto and vice-versa.
Does proof of this kind is valid, if A implies B and B in turn implies A , and similarly A implies C and C in turn implies A, then can we say that B implies C?
let $T \in L(U,V)$ to be one-one, where $dim(U)$ and $dim(V)$ are finite.
$T$ is one-one $\iff ker(T)=\{\underline{0}\} \iff dim(ker(T))=0$
So by Rank–nullity theorem, $$rank(T)+dim(ker(T))=dim(U)$$ $$\implies rank(T)=dim(U)$$
So now all vectors from $range(T)$ can be mapped by some vectors from $U$, that means it is onto. But we assumed that $T$ is one-one, so some vectors means one vector only.
Now, suppose $T$ is onto. Then all $\underline{y} \in V$ can be written as
$T(\underline{x})= \underline{y}$ , where $\underline{x} \in U$
So it is clear that $rank(T)=dim(V)$.
Hence $dim(ker(T)) = dim(U)-rank(T) = dim(U)-dim(V) = 0$ $$\iff ker(T)=\{\underline{0}\}$$ $\iff T $ is one-one