Let $T\colon V\to V$ be a linear transformation such that $\dim(V)=n<\infty$. Prove that $T$ is bijection >iff T is injective. 
Let $T\colon V\to V$ be a linear transformation such that $\dim(V)=n<\infty$. Prove that
(a)$T$ is bijection
iff
(b)T is injective.

Solution:

*

*show $(a)\implies(b)$
If $T$ is bijection, then it is injective by definition.


*show $(b)\implies (a)$
Let $T$ be injective, then if $v_1\neq v_2 \implies T(v_1) \neq T(v_2)$. Hence, $\exists w\in V$ such that $T(v) \neq w$ for all $v \in V$
Therefore, $\dim(\operatorname{Image}(T))<n \implies \dim(\ker(T))>1$.
So, $\exists z\in V$ such that $T(z)=0$.
Moreover, Let $v$ be any nonzero vector in $V$, $T(v+z) = T(v)+T(z) = T(v)$.
$T$ is not injective, a contradiction.
Is my solution correct?
 A: The formula
$\dim(\operatorname{Ker} T)+\dim(\operatorname{Im} T)=\dim V=n$ from the "rank + nullity theorem" should be enough for this.
If $T$ is injective, then $\dim( \operatorname{Ker} T) =0$ so $\dim(\operatorname{Im} T) =n$, therefore since $\operatorname{Im} T$ is a subspace of $V$ you get $\operatorname{Im} T=V$, because the only subspace of $V$ with the same dimension is $V$ itself. 
The inverse is easier: a bijection is always, by definition, injective. 
A: I think it is essentially correct, but I would write things a bit differently. You are right on using rank-nullity. But then I would say that the only vector $z$ with $T(z)=0$ is the zero vector. Then the kernel is 0-dimensional, i.e., nullity is $0$, so rank is $n$, and then $T(V)$ is an n-dimensional subspace of $V$, and so it must be the whole of $V$ itself (n-dimensional vector spaces do not have proper n-dimensional subspaces.) Then $T(V)=V$.
Since the kernel is $0$ , the map is injective , as you said, f T(z)=T(w); $z \neq w$, then $T(z-w)=0$. Then $T(V)=V$, i.e., $T$ is onto and it is injective.
Just a few things tidied up a bit, but overall correct, I would say.
