Prove: $T:V \to V$ is bijective if and only if $T$ is injective. I know that if $T$ is bijective then $T$ is injective.
But how to prove when $T$ is injective, then $T$ is bijective?
I know that that $R(T)$ is a subspace of $V$, and for each vector in $R(T)$, let's say $u$ , there is a $v$ in $V$ that $T(v)=u$. But, what's next? I don't know how to prove $R(T) =V$.
Am I on the right track? Thanks
 A: This is true if you assume $V$ is a finite $n \in \mathbb{Z}_+$ dimensional vector space i.e. $V$ has a basis set of $n$ vectors $B = \{b_1, \ldots, b_n\}$. You also need that $T : V \to V$ is linear. Otherwise, the statement is just false: take $V = \mathbb{R}$ and $T(x) = e^x$. This is an injection that is not a surjection.
Now if $T : V \to V$ is linear, consider the set $T_B = \{T(b_n), \ldots, T(b_n)\}$. $T$ is injective and $B$ and $T_B$ are finite sets; so $T_B$ has exactly $n$ vectors also. Next, using linearity, you can show that $T_B$ is linearly independent because $B$ is. So, $T_B$ forms a basis for $V$ and any $v \in V$ can be written as
$$
v = \alpha_1 T(b_1) + \cdots + \alpha_n T(b_n) = T(\alpha_1 b_1 + \cdots + \alpha_n b_n)
$$ demonstrating the surjectivity of $T$.
P.S. This fact is not necessarily true for infinite dimensional vector spaces. The set of all infinite real sequences $\mathbb{R}^\mathbb{N}$ is a real vector space and the shift map $\sigma : \mathbb{R}^\mathbb{N} \to \mathbb{R}^\mathbb{N}$ given by
$$
\sigma : \langle x_0, x_1, x_2, x_3, \ldots \rangle \mapsto \langle 0, x_0, x_1, x_2, \ldots \rangle
$$ is injective and linear. But it is not surjective: it misses sequences with non-zero first elements.
A: If a linear transformation $T$ is injective, then $\ker(T)=\{0\}$. Applying the rank-nullity theorem, the rank of $T$ is exactly $\dim(V)$. The only full-dimensional subspace of a vector space $V$ is $V$ itself, thus $T$ is surjective as well.  
A: $R(T)\subseteq V$ is a subspace of $V$ that means   
$\dim R(T) \leq \dim V$
Assume $T$ is not bijective we want to show  $\dim R(T) < \dim V$
$T$ is not bijective so there are $v_1 \neq v_2$ in $V$ and $T(v_1)= T(v_2)$ 
$\Rightarrow T(v_1-v_2)=0\Rightarrow \ker T\neq \{0\}\Rightarrow \dim (\ker T)\geq 0$ 
Now because $\dim(\ker T)+\dim(Im(T)) = \dim(V)$
It follows that $\dim (Im(T))< \dim(V)$ 
Which means $R(T)=Im(T)\neq V$ 
Contradiction to the fact $T$ is injective 
