Let $V$ be a $K$-vector space, let $f:V\to V$ be a linear operator, show equivalence between $f$ is injective, surjective and bijective

Question

I have some issues understandig a proof of the following theorem.

Let $V$ be a finite-dimensional $K$-vector space and $f: V \to V$ a linear operator. Then the following are equivalent:

$f$ is injective

$f$ is surjective

$f$ is bijective.

Proof

3 => 1, 2: Trivial

1 => 3:

Let $\dim V = n\in \mathbb{N}$. Consider $\mathscr{B}=\{ b_1,\ldots, b_n\}$ a basis for $V$.

Now consider $T = \{ f(b_1), \ldots, f(b_n)\}$. Since $f$ is injective this set is linear independent.

If $T$ spans the whole space, then $T$ would be a basis and $f$ would be bijective.

But why does $T$ span the whole space? Lecture notes refer to the following theorem:

Let $T\subseteq S\subseteq V$ where $T$ is linear independant, $\text{span}(S) =V$ then there exists a basis $\mathscr{B}$ for $V$ such that $T\subseteq \mathscr{B}\subseteq S$.

I don't see how this theorem can be used here.

How many elements has $T$, and how many elements has a basis? — Daniel Fischer, Aug 05 '15 at 17:26
Well $n$. So I guess the reasoning is the following? Since $\dim V = n$, each basis of $V$ should have $n$ elements. And since $T$ is linear independant and consists of $n$ elements, no elements should be added to make it span the space. Which means it is a basis? — dietervdf, Aug 05 '15 at 17:32
Right. By the way, do you know the rank formula? [You might know it as the rank-nullity theorem.] — Daniel Fischer, Aug 05 '15 at 17:34
Rank-nullity theorem is the standard tool for this kind of statements. — A.Γ., Aug 05 '15 at 17:35
It's all about "rank" in the next chapter. I'll check it out :) — dietervdf, Aug 05 '15 at 17:36

Let $V$ be a $K$-vector space, let $f:V\to V$ be a linear operator, show equivalence between $f$ is injective, surjective and bijective

Proof

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