Linear Transformation is Injective iff Kernel Contains Only Zero/Corollary

Corollary to Linear Transformation is Injective iff Kernel Contains Only Zero
Let $\mathbf A$ be in the matrix space $\mathbf M_{m, n} \left({\R}\right)$

Then the mapping:


 * $\R^n \to \R^m: \mathbf x \mapsto \mathbf {A x}$

is injective $\operatorname {N} \left({\mathbf A}\right) = \left\{ {\mathbf 0}\right\}$

where $\operatorname {N} \left({\mathbf A}\right)$ is the null space of $\mathbf A$.

Proof
From Matrix Product as Linear Transformation, $\mathbf x \mapsto \mathbf {Ax}$ defines a linear transformation.

The result follows from Linear Transformation is Injective iff Kernel Contains Only Zero and the definition of null space.

Also see

 * Null Space Contains Only Zero Vector iff Columns are Independent