Dimension and Injectivity Theorem

• Suppose that $V,W$ are finite dimensional vector spaces
• If $dim(W)<dim(V)$
• Then, $T:V\to W$ is not Injective

Intuition

There is no injective linear transformation of $T:R^{100}\to R^2$ that remains injective

Proof

• The Rank Nullity Theorem gives: $dim(T)=dim(Ker(T))+dim(Image(T))$
• $⟹dim(T)=dim(Ker(T))+dim(W)$ as $Image(T)=W$
• Note that $dim(W)<dim(V)$ by assumption
• Thus, $⟹1\le dim(ker(T))$
• This means we are guaranteed some non-zero vector in the kernel. $ker(T)\ne \{\overrightarrow{0}v\}$
• Thus, $T$ is not injective