Linear Transformation is Injective iff Kernel Contains Only Zero

Theorem
Let $\mathbf V$ be $\R^n$, or any other vector space, with a zero $\mathbf 0$.

Likewise let $\mathbf V\,'$ be $\R^m$, or any other vector space, with a zero $\mathbf 0\,'$.

Let $T: \mathbf V \to \mathbf V\,'$ be a linear transformation.

Then:


 * $T$ is injective $\iff \ker\left({T}\right) = \left\{ {\mathbf 0}\right\}$

where:


 * $\mathbf 0$ is the zero of the domain of $T$


 * $\ker \left({ T }\right)$ is the kernel of $T$.

Corollary
Let $\mathbf A$ be in the matrix space $\mathbf M_{m,n}\left({\R}\right)$

Then the mapping:


 * $\R^n \to \R^m, \mathbf x \mapsto \mathbf {Ax}$

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

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

Sufficient Condition
That $\mathbf 0 \in \ker\left({T}\right)$ follows from the corollary to Linear Transformation Maps Zero Vector to Zero Vector.

That $\ker\left({T}\right)$ is a singleton follows from the definition of being one-to-one.

Necessary Condition
Suppose $\ker\left({T}\right) = \left\{ {\mathbf 0}\right\}$.

Consider the equation $T\left({\mathbf x}\right) = \mathbf b$, where $\mathbf b$ is in the codomain of $T$.

Suppose this equation has a solution $\mathbf x = \mathbf x_1 \in \mathbf V$.

Suppose $\mathbf x = \mathbf x_2 \in \mathbf V$ is also a solution.

Clearly $T\left({\mathbf x_1}\right) = T\left({\mathbf x_2}\right)$.

Now, observe that:

As $\mathbf x_1, \mathbf x_2$ were arbitrary:


 * $\forall \mathbf x_1,\mathbf x_2 \in \mathbf V: T\left({\mathbf x_1}\right) = T\left({\mathbf x_2}\right) \implies \mathbf x_1 = \mathbf x_2$

and the result follows from the definition of injectivity.

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

The result follows from the definition of null space.

Also see

 * Null Space Contains Only Zero Vector iff Columns are Independent