A linear transformation $T:V\to V$ is one-to-one if and only if it is onto Let $V$ be a finite dimensional vector space.  Show that a linear transformation $T\colon V \to V$ is one-to-one if and only if it is onto.
The hint I was given was that one only needs to show that $T(e_1),\dots,T(e_n)$ is a basis whenever $e_1,\dots,e_n$ is a basis.
My first attempt is as follows: Let $x \in V$.  Then  $$Tx=[T(x_1)e_1+\dots+T(x_n)e_n]=[x_1T(e_1),\dots,x_nT(e_n)]=x\cdot T(e).$$  I really want $T(e_n)$ to be a basis of $V$, but I'm not sure that it is.  It seems like it is obviously a basis because $T$ maps from $V$ to $V$.  
Is there a property of linear transformations that makes this apparent?
 A: The finite-dimension hypothesis is very important, since this result fails otherwise. Thus we must make use of it in our proof.
I'll try and give a rough sketch of the proof, leaving you to fill in the details. Suppose that $T$ is onto. We want to show it is one-to-one. That is, we want to show that if $T(v) = T(w)$, we have $v = w$. Using linearity, we can rewrite this as $T(v-w) = 0$ implying $v -w = 0$, so that the kernel of $T$ is only zero. How does a non-zero kernel contradict onto-ness? Let $u$ be nonzero, but so that $T(u) = 0$. Then we can extend $u$ to a basis for $V$, and the image of this basis must still form a spanning set, since $T$ is onto. But the image of this basis contains a zero, so we can find a smaller spanning set, impossible in finite dimensions.
Suppose now that $T$ is one-to-one. This means, as before, that if $T(v) = 0$ then $v = 0$. How can we prove that $T$ is onto? Extend $v$ to a basis. The image of this basis by $T$ must still be linearly independent (since $T$ was one-to-one). But if it does not span, we can add a vector to it not in its span and it will still be linearly independent. Now we have a linearly independent set larger than the dimension of the space, impossible in finite dimensions.
A: How do you normally check for a basis?
Well, try that! 
A good definition for this problem is: For every $v \in V$ we can write $v$ as a linear combination of $e$ (your basis).
$T$ is one-to-one: 
$T[e_i] \neq T[e_j]$, use that to check for the basis.
$T$ is surjective:
$v \in V$, then there is an $w \in V$ such that $T[w]=v$. 
Now write $w$ in its basis $e$ and use that $T$ is linear.
A: Hint: Let us write $W:=V$ to make this less confusing. Then, you have an injective linear transformation $T:V\to W$. To prove that it is necessarily surjective, use the fact that injectivity implies $\dim T(V)=\dim V=\dim W$.
To show that surjectivity implies injectivity, note that if $\ker T$ was non-trivial, then $\dim \text{im}(T)$ would necessarily be less than $\dim V$. Since $\dim V=\dim W$, to see a problem with this?
A: Yes $\{T(e_1),T({e_2}),.....,T(e_n)\}$ will be a basis for V because : 
if 
$x_{1}T(e_1)+.....+x_{n}T(e_n)=0$ then $T(x)=0$ such that $x=(x_1,....,x_n)$ 
(hint:$\,\,cT(\alpha)+bT(\beta)=T(c\alpha+b\beta))$
and it is a theorem at linear algebra that if T is one to one Then $\operatorname{ker} n(T)=\{0\}$ and so $x=0$ 
also $x_1=......=x_n=0$
A: Alex's answer is very good, but matrices are nice too. Represent $T$ by $M$, a square matrix. Suppose that $\sum \lambda_i Me_i = 0$. If $M$ is 1-to-1 then it's invertible. For onto $\Rightarrow$ injective, consider the rank.
