Linear Transformation Maps Zero Vector to Zero Vector

Theorem

Let $\mathbf V$ be a vector space, with zero $\mathbf 0$.

Likewise let $\mathbf V\,'$ be another vector space, with zero $\mathbf 0'$.

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

Then:

$T: \mathbf 0 \mapsto \mathbf 0'$

Corollary

$\mathbf 0 \in \map \ker T$

where $\map \ker T$ is the kernel of $T$.

Proof 1

From the vector space axioms we have that $\exists \mathbf 0 \in \mathbf V$.

It remains to be proved that $\map T {\mathbf 0} = \mathbf 0'$:

\(\ds \map T {\mathbf 0}\)

\(=\)

\(\ds \map T {\mathbf 0 + \mathbf 0}\)

\(\ds \)

\(=\)

\(\ds \map T {\mathbf 0} + \map T {\mathbf 0}\)

Definition of Linear Transformation on Vector Space

\(\ds \leadsto \ \ \)

\(\ds \mathbf 0'\)

\(=\)

\(\ds \map T {\mathbf 0}\)

subtracting $\map T {\mathbf 0}$ from both sides

$\blacksquare$

Proof 2

From the vector space axioms we have that $\exists \mathbf 0 \in \mathbf V$.

What remains is to prove that $\map T {\mathbf 0} = \mathbf 0'$:

\(\ds \map T {\mathbf 0}\)

\(=\)

\(\ds \map T {0 \, \mathbf 0}\)

Zero Vector Scaled is Zero Vector

\(\ds \)

\(=\)

\(\ds 0 \, \map T {\mathbf 0}\)

Definition of Linear Transformation on Vector Space

\(\ds \)

\(=\)

\(\ds \mathbf 0'\)

Vector Scaled by Zero is Zero Vector

$\blacksquare$