Prove that if T is a finite operator then T is surjective iff T is injective Prove that if $V$ is finite dimensional and $T\in L(V)$, $T$ is injective iff $T$ is surjective. 
From the direction of Injectivity to surjectivity, I was able to prove the relation. With the oppsite direction, I tried the following.
Assume that $T$ is surjective.
Let $(e_{1},e_{2},...e_{n})$ be a basis of $V$
Let $v_{1},v_{2} \in V$ ,
Then $v_{1} = a_{1}e_{1}+a_{2}e_{2}....a_{n}e_{n}$ for $a\in K$ where $K$ is a field.
Similarly, $v_{2} = b_{1}e_{1}+b_{2}e_{2}....b_{n}e_{n}$ for $b\in K$
Now, suppose that $v_{1} = v_{2}$, then
$a_{1}e_{1}+a_{2}e_{2}....a_{n}e_{n} = b_{1}e_{1}+b_{2}e_{2}....b_{n}e_{n}$
$\therefore a_{j} = b_{j}$  for  $j=1,2,...n$
Now, 
$T(v_{1})=T(a_{1}e_{1}+a_{2}e_{2}....a_{n}e_{n})= a_{1}T(e_{1})+a_{2}T(e_{2})....a_{n}T(e_{n})$
Similarly,
$T(v_{2})=T(b_{1}e_{1}+b_{2}e_{2}....b_{n}e_{n})= b_{1}T(e_{1})+b_{2}T(e_{2})....b_{n}T(e_{n})$
But since $a_{j} = b_{j}$
$a_{1}T(e_{1})+a_{2}T(e_{2})....a_{n}T(e_{n}) = b_{1}T(e_{1})+b_{2}T(e_{2})....b_{n}T(e_{n})$
So, 
$T(a_{1}e_{1}+a_{2}e_{2}....a_{n}e_{n}) = T(b_{1}e_{1}+b_{2}e_{2}....b_{n}e_{n})$
Which follows that, 
$T(v_{2}) = T(v_{2})$ Which makes $T$ injective. 
Now I know something must be seriously wrong since I didn't mention anything about the surjectivity of $T$.
It looks as though I have deduced that any linear mapping of finite dimensional vector spaces is injective.  
Can anyone please tell me how this line of reasoning is wrong?
 A: The rank-nullity theorem says that
$$
\dim V=\dim\ker T+\dim\operatorname{im}T
$$
Moreover, $\dim\operatorname{im}T\le V$. You also know that


*

*$T$ is injective if and only if $\dim\ker T=0$

*$T$ is surjective if and only if $\dim\operatorname{im}T=\dim V$


Now you can finish:


*

*If $T$ is injective, then $\dim\ker T=0$, so…

*If $T$ is surjective, then $\dim\operatorname{im}T=\dim V$, so…

About your proof, you indeed have not applied the assumption of surjectivity. Also you should not assume $v_1=v_2$, but rather $T(v_1)=T(v_2)$ and prove from this that necessarily $v_1=v_2$.

Note. It's obvious that if $v_1=v_2$ you get $T(v_1)=T(v_2)$. Injectivity means the converse holds: if $T(v_1)=T(v_2)$ then $v_1=v_2$.

Start from the codomain: let $\{e_1,\dots,e_n\}$ be a basis of $V$; since $T$ is surjective, there are $f_1,\dots,f_n$ such that $T(f_i)=e_i$, for $i=1,2,\dots,n$. Then $\{f_1,\dots,f_n\}$ is linearly independent (prove it), so a basis for $V$.
Suppose $v_1=a_1f_1+\dots+a_nf_n$ and $v_2=b_1f_1+\dots+b_nf_n$ with $T(v_1)=T(v_2)$. Then
$$
a_1e_1+\dots+a_ne_n=b_1e_1+\dots+b_ne_n
$$
and we conclude $a_i=b_i$, for $i=1,2,\dots,n$, hence $v_1=v_2$.
A: Although K.Power's proof is much more concise, I believe i have also completed the proof with the same method.  
Now instead of $v_{1} = v_{2}$ , Assume $ T(v_{1}) = T(v_{2})$
then, 
$ T(a_{1}e_{1}+a_{2}e_{2}....a_{n}e_{n}) = T(b_{1}e_{1}+b_{2}e_{2}....b_{n}e_{n})$
$a_{1}T(e_{1})+a_{2}T(e_{2})....a_{n}T(e_{n}) = b_{1}T(e_{1})+b_{2}T(e_{2})....b_{n}T(e_{n}) $
by rearranging the terms, 
$0=(b_{1}-a_{1})T(e_{1})+(b_{2}-a_{2})T(e_{2})....(b_{n}-a_{n})T(e_{n})$
Now , assuming the surjectivity of $T$,
 $(T(e_{1}),T(e_{2}),...T(e_{n}))$ spans $ Im\ T=V $ where $dim V = n$
Since there are n elements in $(T(e_{1}),T(e_{2}),...T(e_{n}))$ that spans n dimensional space, $(T(e_{1}),T(e_{2}),...T(e_{n}))$ is linearly independent.
Now for $0=(b_{1}-a_{1})T(e_{1})+(b_{2}-a_{2})T(e_{2})....(b_{n}-a_{n})T(e_{n})$ to hold, $b_{j}-a_{j}$ must equal $0$ so $b_{j} = a_{j}$
So we can conclude that
$a_{1}e_{1}+a_{2}e_{2}....a_{n}e_{n} = b_{1}e_{1}+b_{2}e_{2}....b_{n}e_{n}$
$ \therefore v_{1} = v_{2}$
