So the question asks: Let $V$ be a finite-dimensional vector space, and let $T : V\to V$ be a linear transformation. Prove that if $T$ is surjective, then $T$ is an isomorphism. Show that the statement may be false if $V$ is not finite-dimensional.
So so far I got:
Since $T$ is surjective, $\forall v_1\in V$, $\exists v_2\in V$ $: T(v_2) = v_1$.
$T$ carries linearly independent subsets of $V_1$ onto linearly independent subsets of $V_2$.
Assume that $T(x) = 0$.
If the set $\{x\}$ is linearly independent, then by assumption $\{0\}$ is linearly independent, which is a contradiction.
Hence, the set $\{x\}$ is linearly dependent.
Then $x = 0$. That is, $N(T) = \{0\}$. Therefore, $T$ is one-to-one.
Since $T$ is onto and one-to-one,
So $T$ is isomorphism.
But I think the proof will still hold if they are not finite -dimensional. Why is $T$ not isomorphism in that case?
Use the dimensions theorem: if $\;\dim V=n\;$ and $\;T:V\to V\;$ is a linear map, then
$$n=\dim\ker T+\dim\text{ Im }T$$
Thus, we get that
$$T\;\;\text{is surjective}\;\iff\dim\text{ Im }T=n\stackrel{\text{ by above}}\iff\dim\ker T=0$$
and thus $\;T\;$ is surjective iff it is injective iff it is bijective.
Now, let $\;V=\{\;\{x_n\}_{n=1}^\infty\subset\Bbb R\}\;$ be the real linear space of all real sequences, with sum and scalar multiplication indexwise, and define
$$T:V\to V\;\;,\;\;\;\;T\{x_n\}:=\{x_2,x_3,....\}$$
Clearly $\;T\;$ is surjective but it is not injective.