If the null space contains only the zero vector, the map is one-to-one

Question

How does finding out if the null space has only the zero vector prove one-to-one?

One-to-one means that there are distinct images for each distinct vector input.

$$\mathbb R^n \to \mathbb R^m$$

For each distinct vector in $\mathbb R^n$ we have a distinct vector in $\mathbb R^m$.

If we use Reduced row echelon form to solve a matrix A and each $x=(x_1,x_2,...x_n)$ equals $0$, then that shows that there is a unique solution, and that we have a linearly independence.

With what I know I am not making the connection here.

Answer 1

If $f(v)=f(w)$ then $f(v-w)=0$, so $v-w=0$.

Answer 2

Look at this it may help. It is proven by using contradiction as

Suppose that $v_1 \neq 0_v$ were in $N(A)$, then $Av_1= 0_w$. Also, $A(0_v)=0_w$, hence, since $v1 \neq 0_v$, A is not one to one.

By assumption $N(A)=0_v$. If $v_1 \neq v_2$ then $v_1-v_2 \neq 0_v$, consequently $0_w \neq A(v_1 -v_2)=Av_1 -Av_2$. In other words, $v1 \neq v2$ implies that $A v_1 \neq Av_2$, i.e., A is one to one.

(The book is A second course on linear systems by C.A. Desoer)