Kernels and Injectivity Theorem

• Let $T:V\to W$ be a linear map
• Then, the following are equivalent:
  1. $T$ is injective
  2. $ker(T)=0v$
  3. $dim(ker(T))=0$

Proof

Proving $1⟹2$

• Suppose $T:V\to W$ is linear and injective.
• We know $\{\overrightarrow{0}v\}\subset ker(T)$ because $T(\overrightarrow{0}v)=\overrightarrow{0}w$ for all linear maps
• Suppose $\overrightarrow{v}\in ker(T)$. This gives: $\overrightarrow{0}w=T(\overrightarrow{v})=T(\overrightarrow{0}v)$
• Since $T$ is injective, we get $v=\overrightarrow{0}v$
• Thus, $ker(T)\subset \{\overrightarrow{0}v\}$
• Thus, $ker(T)=\{\overrightarrow{0}v\}$

Proving $1⟹3$

• Suppose $ker(T)=\{\overrightarrow{0}v\}=span(\varnothing )$
• Thus, $\varnothing$ is a basis of $ker(T)$
• It follows that $dim(ker(T))=0$

Proving $3⟹1$

• Suppose $dim(ker(T))=0$
• Therefore, $\varnothing$ is a basis of $ker(T)$
• This gives $ker(T)=span(\varnothing )=\{\overrightarrow{0}v\}$
• Now, we argue that $T$ is injective
  • Suppose $T(\overrightarrow{x})=T(\overrightarrow{y})$
  • $⟺T(x)-T(y)=0$
  • $⟺T(x-y)=\overrightarrow{0}$ by linearity
  • Thus, $x-y\in ker(T)=\{\overrightarrow{0}v\}$
  • Thus, $x-y=\overrightarrow{0}v⟹x=y$
  • It follows that $T$ is injective