A linear transformation $T:V\to V$ is one-to-one if and only if it is onto

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.

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.