Linear Transformation is Injective iff Kernel Contains Only Zero

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Theorem

Let $\mathbf V, \mathbf V'$ be vector spaces, with respective zeroes $\mathbf 0, \mathbf 0'$.

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


Then:

$T$ is injective if and only if $\map \ker T = \set {\mathbf 0}$

where:

$\mathbf 0$ is the zero of the domain of $T$
$\map \ker T$ is the kernel of $T$.


Corollary

Let $\mathbf A$ be in the matrix space $\map {\mathbf M_{m, n} } \R$

Then the mapping:

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

is injective if and only if:

$\map {\mathrm N} {\mathbf A} = \set {\mathbf 0}$

where $\map {\mathrm N} {\mathbf A}$ is the null space of $\mathbf A$.


Proof

Sufficient Condition

That $\mathbf 0 \in \map \ker T$ follows from Kernel of Linear Transformation contains Zero Vector.

That $\map \ker T$ is a singleton follows from the definition of injection.

$\Box$


Necessary Condition

Let $\map \ker T = \set {\mathbf 0}$.

Consider:

$\map T {\mathbf x} = \mathbf b$

where $\mathbf b$ is in the codomain of $T$.

Let this equation have a solution:

$\mathbf x = \mathbf x_1 \in \mathbf V$

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

Clearly:

$\map T {\mathbf x_1} = \map T {\mathbf x_2}$

Observe that:

\(\ds \map T {\mathbf x_1}\) \(=\) \(\ds \mathbf b\)
\(\ds \land \ \ \) \(\ds \map T {\mathbf x_2}\) \(=\) \(\ds \mathbf b\)
\(\ds \leadsto \ \ \) \(\ds \map T {\mathbf x_1} - \map T {\mathbf x_2}\) \(=\) \(\ds \mathbf b - \mathbf b\)
\(\ds \) \(=\) \(\ds \mathbf 0'\)
\(\ds \leadsto \ \ \) \(\ds \map T {\mathbf x_1 - \mathbf x_2}\) \(=\) \(\ds \mathbf 0'\) Definition of Linear Transformation on Vector Space
\(\ds \leadsto \ \ \) \(\ds \paren {\mathbf x_1 - \mathbf x_2}\) \(\in\) \(\ds \map \ker T\) Definition of Kernel of Linear Transformation
\(\ds \leadsto \ \ \) \(\ds \mathbf x_1 - \mathbf x_2\) \(=\) \(\ds \mathbf 0\) Definition of Set Equality: recall $\map \ker T = \set {\mathbf 0}$
\(\ds \leadsto \ \ \) \(\ds \mathbf x_1\) \(=\) \(\ds \mathbf x_2\)

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

$\forall \mathbf x_1,\mathbf x_2 \in \mathbf V: \map T {\mathbf x_1} = \map T {\mathbf x_2} \implies \mathbf x_1 = \mathbf x_2$

and the result follows from the definition of injectivity.

$\blacksquare$


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