Proof of Rudin's theorem 9.5 
A linear operator A on a finite-dimensional vector space X is one-to-one iff the range of A is all of X.

The proof of => direction looks like this:

Let {$x_1,...,x_n$} be a basis of X. Assume that A is one-to-one and $\sum c_iAx_i=0$. Then, $A\sum c_ix_i=0$, hence $\sum c_ix_i=0$ and so $c_1=...=c_n=0$. Then, {$Ax_1,...,Ax_n$} is independent (=>) {$Ax_1,...,Ax_n$} is a range of X.

Can you please explain how one-to-one property of A is used here? Isn't just the linearity of A sufficient?

$$A\sum c_i x_i = 0$$
We know that $$A0=0$$
Since it is one-to-one, we conclude that 
$$\sum c_i x_i = 0$$
