# 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

$$\sum c_i x_i = 0$$

• oh, makes sense. Thank you! Commented Jul 13, 2017 at 18:53

Injectivity is used in this implication: $$A\sum c_ix_i=0\implies \sum c_ix_i=0.$$ Since by linearity $A(0)=0$, by injectivity it follows that $Az=0$ implies $z=0$.

We know that A(0)=0 And if A is one to one then A(X)=0 implies X=0 Here your $x= \sum c_ix_i$ $A \sum c_ix_i=0 \implies \sum c_i x_i=0$