Equivalence theorem for injective linear transformations: The null space of T is 0.

Created almost 9 years ago, updated about 8 hours ago

A linear map is injective iff its null space is $\{0\}$. If $T(u)=T(v)$, linearity gives $T(u-v)=0$; a zero kernel forces $u=v$. Conversely, injectivity and $T(0)=0$ imply zero is the only kernel vector.

Like how this is organised?

Build your own knowledge tree on knowen.md: notes, decisions and tasks in one connected markdown graph, shared with your team and your AI agents.