Linear Transformation is Injective iff Kernel Contains Only Zero
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:
- $\map \ker T$ is the kernel of $T$.
Corollary
Let $\mathbf A$ be in the matrix space $\map {\mathbf M_{m, n} } \R$
Then the mapping:
- $\R^n \to \R^m: \mathbf x \mapsto \mathbf {A x}$
- $\map {\mathrm N} {\mathbf A} = \set {\mathbf 0}$
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.
- $\map T {\mathbf x_1} = \map T {\mathbf x_2}$
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.
$\blacksquare$
Sources
- 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): Appendix $\text{A}$ Preliminaries: $\S 1.$ Linear Algebra